Page 1 :

MATHEMATICS, Part -, , PATTERN NINE, , 9, , 2, , 61.00

Page 2 :

The Constitution of India, Chapter IV A, , Basic Duties, ARTICLE 51A, Basic Duties - It is the duty of every citizen of India (a) to abide by the Constitution and respect its ideals and institutions, the national flag and the National Anthem;, , (b), , to cherish and uphold the noble ideals which inspired our national liberation struggle;, , (c), , to uphold and protect the sovereignty, unity and integrity of India;, , (i.e.), , to uphold the country and render national service when called;, , (and), , to promote harmony and the spirit of common brotherhood among all the people of India, transcending religious, linguistic and regional boundaries going beyond , or sectional diversity, to renounce to practices harmful to the dignity of women;, , (f), , to value and preserve the rich heritage of our composite culture;, , (g), , to protect the natural environment and verbalize eaters, including forests, lakes, rivers and wildlife , and have compassion for living creatures;, , (h), , develop the t scientific ethos, humanism, and the spirit of investigation and reform;, , (i), , the public to protect property and renounce violence;, , (j), , the pursuit of excellence in all spheres of individual activity and collectively, so that the nation continually rises to higher levels of endeavor and achievement;, , (k), , who is a parent or guardian to provide an education for his or her child or ward, as the case may be, between the ages of six and fourteen years.

Page 11:

1, , Basic Geometry Concepts, , Let's study., , • Point, line and plane, • Intermediate, • Coordinates of a point and distance • Conditional statements, • Demonstration, Did you recognize the figure on the side? It is an image of the pyramids in Egypt built 3,000 years before the Christian era. How could people in such ancient times build such large structures? It is not possible to build such large structures without developed knowledge of geometry and engineering. The very word geometry suggests the origin of the subject. It is composed of the Greek words geo (earth) and metria (measurement). Therefore, it can be assumed that the topic must have developed out of the need to search the land, i. H. Country. Geometry was developed in many nations at different times and for different constructions. The first Greek mathematician, Thales, went to Egypt. He is said to have determined the height of a pyramid by measuring its shadow and using properties of similar triangles. The ancient Indians also had a deep understanding of geometry. In the Vedic period, people used geometric properties to build altars. The Shulba-sutra book describes how to build different shapes by measuring with a string. Throughout the ages, mathematicians Aaryabhat, Varahamihir, Bramhagupta, Bhaskaracharya and many others have made valuable contributions to geometry. Likewise, we also do not define a point, a line or a plane. These are some basic concepts in geometry. Lines and planes are sets of points. Remember that the word "line" is used in the sense of "straight line"., , 1

Page 12:

Point coordinates and distance, consider the following number line., , B C, A, -5 -4 -3 -2 -1, , O, 0, , D, 1, , 2, , E, 3, , 4, , 5, , Fig. 1.1, Here, point D on the number line denotes the number 1. Therefore, we say that 1 is the coordinate of point D. Point B denotes the number - 3 beyond the line. Therefore, the coordinate of point B is - 3. Likewise, the coordinates of points A and E are - 5 and 3, respectively. Point E is 2 units away from point D. This means that the distance between points D and E is 2. So we can find the distance between two points on a number line by counting the number of units. The distance between points A and B on the number line above is also 2. Now let's see how to find the distance using the coordinates of the points. To find the distance between two points, consider their coordinates and subtract the smaller coordinate from the larger one. The coordinates of points D and E are 1 and 3, respectively. We know that 3 > 1. Therefore, the distance between points E and D = 3 - 1 = 2. The distance between points E and D is given by d ( E ,D). This is the same as l(ED), that is, the length of the segment ED., d (C, D) = 1 - (- 2), d (E, D) = 3 - 1 = 2, = 1+ 2 =3 , \ l(ED) = 2, \ d (C, D) = l(CD) = 3, d (E, D) = l(ED) = 2, similar to d (D, C) = 3 , similar to d(D,E) = 2, Now let's find d(A,B). The coordinate of A is -5 and that of B is -3; - 3 > - 5, \d (A, B) = - 3 - (- 5) = - 3+5 = 2. From the examples above, it is clear that the distance between two different points is always a positive number., Note that if the two points are not distinct, the distance between them is zero., , , Remember !, , • The distance between two points is obtained by subtracting the smaller coordinate from •, , the larger coordinate. , The distance between any two points is a non-negative real number., , 2

Page 13:

Let's learn., Betweenness, If P, Q, R are three different collinear points, then there are three possibilities., , P, , Q, , R Q, Fig. 1.2, , R, , P, , R, , (i ) the point Q is between, P and R, , P, , Q, , (ii) the point R is between, (iii) the point P is between, P and Q, R and Q, , if d (P, Q) + d(Q,R) = d(P,R) then it says that the point Q is between P and R. The intermediate is represented as P - Q - A. Worked examples, Ex. The points A, B and C are such that, , d(A,B) = 5, d(B,C) = 11, , and d(A,C) = 6., , Which of the points is between the other two ?, Solution : Which of the points A, B and C is between the other two can be decided as follows., , B, , , d(B,C) = 11 . 🇧🇷 🇧🇷 🇧🇷 (I), , d(A,B) + d(A,C)= 5+6 = 11 . 🇧🇷 🇧🇷 🇧🇷 (II), , \ d(B,C) = d(A,B) + d(A,C) . 🇧🇷 🇧🇷 🇧🇷 [from (I) and (II)], , 5, , A, , 6, Fig. 1.3, , C, , Point A is between point B and point C. Ex. (2) U, V, and A are three cities on a straight road. The distance between U and A is 215 km, between V and A 140 km and between U and V 75 km. Which of these is between the other two?, Solution: d(U,A) = 215; d (V,A) = 140;, , d (U,V) = 75, , , d (U,V) + d (V,A) = 75 + 140 = 215;, , d (U,A) = 215, , \ d (U,A) = d (U,V) + d (V,A), , \ City V lies between cities U and A., , 3

Page 14:

Ex (3) The coordinate of point A on a number line is 5. Find the coordinates of the points on the same number line that are 13 units away from A. Solution: As shown in the figure, take points T and D on the left and to the right of A, respectively, at a distance of 13 units., , 5-13, , , , T, -8, , 5+13, D, 18, , A, 5, Fig. 1.4, , The coordinate of the point T that is to the left of A is 5 - 13 = - 8, The coordinate of the point D that is to the right of A is 5 + 13 = 18, , \ the coordinates of the points are 13 units away from A - 8 and 18. Check your answer: d (A,D) = d(A,T) = 13, , activity, (1 ) Points A , B, C are placed aside. Using a stretched string, check whether the three points, , A, , are collinear or not. If they are collinear, write which one is between the other two. (2) Four points P, Q, R and S are given on each side. Check which three of them are collinear and which three are not , , the other two., , C, , Q, , S, R, , collinear. In the case of three collinear points, indicate which one is between , , B, , P, , . (3) Students are invited to line up for mass exercise. How will you check whether the standing students are lined up or not? (4) How did you verify that light rays travel in a straight line? Recall a science experiment you performed in an earlier pattern., , 4

Page 15:

Exercise set 1.1, 1., , Find the distances using the number line below., , Q P K J H, -5 -4 -3 -2 -1, , O A, 0 1, Fig. 1.5, , 2., , 3. , , B, 2, , C, 3, , D, 4, , E, 5, , 6, , (i) d(B,E), , (ii) d( J, A), , (iii) d(P, C) , , (iv) d(J, H), , (v) d(K, O), , (vi) d(O, E) , , (vii) d(P , J) , , (viii) d(Q, B), , If the coordinate of A is x and that of B is y, find d(A, B) ., (i) x = 1, y = 7, , (ii) x = 6, y = - 2, , (iii) x = - 3, y = 7, , (iv) x = - 4, y = - 5, , (v) x = - 3 , y = - 6, , (vi) x = 4, y = - 8, , Use the information below to find out which of the points is between the other two., If the points are not collinear, indicate this., , 4., , (i) d(P, R) = 7, , , d(P, Q) = 10, , , d(Q, R) = 3, , (ii) d(R, S) = 8 , , , d(S, T) = 6, , , d(R, T) = 4, , (iii) d(A, B) = 16,, , d(C, A) = 9, , , d (B, C ) = 7 , , (iv) d(L, M) = 11, , , d(M, N) = 12,, , d(N, L) = 8, , (v) d(X , Y) = 15, , , d(Y, Z) = 7, , , d(X, Z) = 8, , (vi) d(D, E ) = 5, , , d(E, F) = 8, , , d(D, F) = 6, , On a number line, the points A, B and C are such that d(A,C) = 10 , d(C,B) = 8, Find d( A, B) considering all possibilities. , , 5., , The points X, Y, Z are collinear so that d(X,Y) = 17, d (Y, Z) = 8, find d(X,Z) ., , 6., , Sketch the correct figure and write the answers to the following questions ., , , , (i) If A - B - C , , and l(AC) = 11,, , l(BC) = 6.5, then l(AB ) = ?, , (ii) If R - S - T, , and l(ST) = 3.7 ,, , l(RS) = 2.5 ,, , then l(RT) =?, , (iii) If X - Y - Z and l(XZ) = 3 7 , l (XY) = 7 , then l (YZ) =?, 7., , Which figure is formed by three non-collinear points ?, , 5

Page 16:

Let's learn., In the book Mathematics - Part I for std IX we learned union and intersection of sets in the topic about sets. Now let's describe a segment, a ray and a line as sets of points., (1) segment :, The union of point A, point B and points between A and B is called segment AB. Segment AB is written as Segment AB for short. Seg AB stands for Seg BA. Points A and B are called the endpoints of Seg AB. The distance between the ends of a segment is called the segment length. That is, l(AB) = d(A,B), l(AB) = 5 is also written as AB = 5., (2) Ray AB :, Suppose A and B are two distinct points. The union of all the points on the segment AB and the points P such that A - B - P is called the radius AB. Here, point A is called the starting point of the ray AB., , A, , Fig. 1.6, , B, , BP, Fig. 1.7, , A, , (3) Line AB :, The union of points on the radius AB and the ray opposite the radius AB is called the line AB., The set of points on the segment AB is a subset of the points on the line AB. , (4) Congruent segments :, If the lengths of two segments are equal, then the two segments are congruent., If l(AB) = l(CD) then sec AB @ sec CD, , A, , B, , C , , Fig. 1.8, , (5) properties of congruent segments :, , D, , (i) reflexivity: seg AB @ seg AB, (ii) symmetry: if seg AB @ seg CD then seg CD @ seg AB, (iii) transitivity: if seg AB @ seg CD and seg CD @ seg EF, then seg AB @ seg EF, (6) midpoint of a segment:, if A-M-B and seg AM @ seg MB, then M is called the midpoint of the AB section. Each segment has one and only one midpoint., , 6, , A, , M, Fig. 1.9, , B

Page 17:

(7) segment comparison :, , If the length of segment AB is less than the length of segment CD, it is written as seg AB < sec CD or sec CD > sec AB., , B, , A, C, , The comparing segments depends on their length., , D, , Fig. 1.10, , (8) Perpendicularity of segments or rays :, , When lines containing two segments, two rays or a ray and a segment are perpendicular to each other , except the two segments, two rays or the, segment and ray must be perpendicular to each other., , C, B, , A, D, , In figure 1.11, sec AB ^ line CD ,, , seg AB ^ ray CD., , Fig. 1.11, , (9) Distance of a point from a line :, If sec CD ^ line AB and point D lies on, , C, , line AB, then the length of sec CD is called the distance from point C of the line AB., , A, , The point D is called the foot of the perpendicular. a', of the line AB., , DB, Fig. 1.12, , Exercise Theorem 1.2, 1. The following table shows the points on a number line and their coordinates. Decide whether the pair of segments below the table is congruent or not., Point, Coordinate, , A, , -3, , B, , C, , 5, , 2, , D, , -7, , E, , 9 , , (i) Section DE and Section AB (ii) Section BC and Section AD (iii) Section BE and Section AD, 2. Point M is the midpoint of Section AB. If AB = 8, find the length of AM. 3. Point P is the midpoint of section CD. If CP = 2.5, find l(CD)., 4. If AB = 5 cm, BP = 2 cm, and AP = 3.4 cm, compare the segments., , 7

Page 18:

5. Write the answers to the following questions referring to Figure 1.13., (i) write the name of the anti-ray of the RP ray, (ii) write the intersection of the PQ ray and the RP ray., , T, , ( iii ) Write the union of sec PQ and sec QR., , S R, , P, Fig. 1.13, , Q, , (iv) Give the rays of which sec QR is a subset., (v) Write the pair of opposite rays with common endpoint R., (vi) Write any two rays with common endpoint S. , (vii) Write the intersection of radius SP and radius ST., 6. Answer the questions using Figure 1.14. , , R, -6, , U, , Q, -4, , L, , P, -2, , A, 0, Fig. 1.14, , B, 2, , C, 4, , V, , D, 6, , (i) Find the points that are equidistant from point B. , (ii) Write a pair of points that are equidistant from point Q., ( iii) Find d(U,V), d(P,C), d(V,B), d(U,L ) The part of the The statement that follows "if" is called the antecedent, and the part that follows "then" is called a consequent The statement can be written in conditional form as: 'If the given quadrilateral is a rhombus, then its diagonals are perpendicular bisectors to each other. ' If the antecedent and consequent are reversed in a given conditional statement, the resulting statement is called the inverse of the given statement. If a conditional statement is true, its inverse is not necessarily true. Study the following examples., Conditional statement: If a quadrilateral is a rhombus, then its diagonals are perpendicular to each other., , 8

Page 19:

Inverse : If the diagonals of a quadrilateral are perpendicular bisectors, then it is a rhombus., In the example above, the statement and its inverse are true., Now consider the following example, Conditional statement : If a number is prime, then it is even or odd., On the other hand, if a number is even or odd, then it is prime., In this example, the statement is true, but its reciprocal is false., , Let's learn., Prove , We have many properties of angles , triangles and quadrilaterals are studied through activities., In this pattern, we will look at the subject of geometry from a different angle, which goes back to the Greek mathematician Euclid, who in the 3rd century before the Christian era. He collected and rationalized the knowledge of geometry prevalent in his day. He took some self-evident geometric statements that were accepted by everyone and called them postulates. He showed that based on postulates some more properties can be proved logically., Properties proved logically are called theorems., Some of Euclid's postulates are given below., (1) There are infinitely many straight lines passing through a point. (2) There is one and only one line passing through two points., (3) A circle of given radius can be drawn by taking each point as the center., (4) All right angles are coincident., (5) If two interior angles made on one side of a through two lines add up to less than two right angles, so lines made in that direction, Euclid, intersect. We verify some of these postulates through activities. A property is said to be true if it is logically provable. It is then called a theorem. The logical argument used to prove a theorem is called a proof. If we want to prove that a conditional statement is true, its antecedent is called the "given part" and its consequent the "given part". the part to be proved'., There are two types of proofs, direct and indirect., Let's give a direct proof of the property of angles formed by two intersecting lines., , 9

Page 20:

Theorem : The opposite angles of two intersecting lines are equal Given : The line AB and the line CD intersect at the point O such that A - O - B, C - O - D., , A, , Prove : ( i ) ÐAOC = ÐBOD , (ii) ÐBOC =, , Proof :, , , , ÐAOD, , D, , C, O, , Fig. 1.15, , B, , ÐAOC + ÐBOC = 180° . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 (I) (Angles in the linear pair), ÐBOC + ÐBOD = 180° . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 (II) (angles in the linear pair), ÐAOC + ÐBOC = ÐBOC + ÐBOD . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 [from (I) and (II)], \ÐAOC = ÐBOD. 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 Elimination of ÐBOC., Similarly, one can prove that ÐBOC = ÐAOD., , Indirect Proof :, This type of proof starts with the assumption that the consequent is false. With it and the previously accepted properties, we begin to reason step by step and arrive at a conclusion. The conclusion contradicts the antecedent or an already accepted property. So the assumption that the consequence is wrong goes wrong. Therefore, the consequent is assumed to be true. Examine the following example. Claim: A prime number greater than 2 is odd. Conditional statement: If p is a prime number greater than 2, then it is odd. Given: , , p is a prime number greater than 2. That is, 1 and p are the only divisors of p., , To prove: p is an odd number., Prove :, , Assume that p is not an odd number. , Then p is an even number. , , \ a divisor of p is 2..... (I), but since p is a prime greater than 2., , \ 1 and p are the only divisors of p ..... (II), the statements (I) and (II) are contradictory., \ the assumption that p is not odd is false., This proves that a prime number greater than 2 is odd., , 10, , ....(given)

Page 21:

Exercise Theorem 1.3, 1. Write the following statements in if-then form., (i) The opposite angles of a parallelogram are congruent., (ii) The diagonals of a rectangle are congruent., (iii) In an isosceles triangle, the segment joining the vertex and the midpoint of the base is perpendicular to the base., 2. Write the opposites of the following statements., (i) The alternate angles formed by two parallel lines and their transversals are congruent . , (ii) If a pair of interior angles formed by a transversal of two lines are complementary, then the lines are parallel., (iii) The diagonals of a rectangle are congruent., , Problem Set 1, 1., , Choose Find the correct alternative from the answers to the questions below., , (i) How many midpoints does a segment have?, , (A) only one, (B) two, (C) three, , (D) many, , ( ii) How many points are there at the intersection of two different lines?, (A) infinity, (B) two, (C) one, (D) not one, (iii) How many lines are determined by three different lines p points? , , (A) two, (B) three, (C) one or three, (D) six, (iv) Find d(A, B) when the coordinates of A and B are - 2 and 5 (A ) - 2, (B) 5, (C) 7, (D) 3, (v) If P - Q - R and d(P,Q) = 2, d(P,R) = 10, then find d(Q, R)., (D) 20, , (A) 12, (B) 8, (C) 96, 2nd, , On a number line, the coordinates of P, Q, R are 3, - 5, respectively 6. Justify whether the following gen statements are true or false., (i) d(P,Q) + d(Q,R) = d(P,R)(ii) d(P,R) + d(R ,Q ) = d(P ,Q), (iii) d(R,P) + d(P,Q) = d(R,Q), (iv) d(P,Q) - d(P,R ) = d(Q,R ), , 3., , The coordinates of some pairs of points are given below. So find the distance between each pair., (i) 3, 6, (v) x + 3, x - 3, , (ii) - 9, - 1, (vi) - 25, - 47, , (iii ) - 4, 5, (vii) 80, - 85, , 11, , (iv) 0, - 2

Page 22:

4., , the coordinate of point P on a number line is - 7. Find the coordinates of the points on the number line that are 8 units away from point P., , 5., , Answer the following questions., (i ) If A - B - C and d(A,C) = 17, d(B,C) = 6.5 then d(A,B) = ?, (ii) If P - Q - R and d(P, Q ) = 3.4, d(Q,R)= 5.7 then d(P,R) = ?, , 6., , The coordinate of point A on a number line is 1. What are the coordinates of the Points on the number line that are 7 units of A ?, , 7., , Write the following statements in conditional form., (i) Every rhombus is a square., (ii) Angles in a linear pair are complementary., (iii) A triangle is a figure consisting of three segments., (iv) A number with only two divisors is called a prime number., , 8., , Write the inverse of each of the following numerical statements., (i) If the sum of the angles measured in a figure is 180°, then the figure is a triangle., (ii) If the sum of the angular measures of two angles n is 900, then they are complementary s to each other., (iii) If the corresponding angles formed by a transversal of two lines are congruent, then the two lines are parallel., (iv) If the checksum of a number is divisible by 3 then the number is divisible , by 3., , 9. , , Write the antecedent (part given) and consequent (part to be proved) in the following statements., (i) If all sides of a triangle are congruent, then all angles are congruent. , (ii) The diagonals of a parallelogram divide in half., , 10*. Draw a labeled figure that contains information in each of the following statements and write the premise and theorem., (i) Two equilateral triangles are similar., (ii) If the angles in a linear pair are congruent, then each of them is straight angle., (iii) If the heights drawn on two sides of a triangle are congruent, then those two sides are congruent., , qqq, 12

Page 23:

2, , Parallel lines, , Let's learn., , • Properties of angles formed by • Tests of the parallelism of two lines, parallel lines and their transversals • Use of properties of parallel lines, Let's remember., Parallel lines : The lines , which are coplanar and do not intersect, , l, , are called parallel., m, Hold a stick over the horizontal parallel, bar of a window as shown in the figure., How many angles are formed?, n, , · Make yourself Remember the pairs of angles formed by two lines and their transversals? be formed? In Figure 2.1, line n is a transversal of line l and line m. A total of 8 angles are formed here. Pairs of angles formed from these angles are , angles, , Ðd, Ðh, (ii) Ða,, (iii) Ðc,, (iv) Ðb,, (i), , h e, g f, , l, m, , Fig . 2.1 , , follows:, pairs of correspondents, , d a, c b, , pairs of alternate interior angles, (i) Ðc, Ðe (ii) Ðb, Ðh, pairs of alternate exterior angles, (i, (1) If two lines intersect, the pairs of opposite angles formed are congruent., (2) The angles in a linear pair are complementary., , 13

Page 24:

(3), , If a pair of corresponding angles is congruent, then all remaining pairs of corresponding angles are congruent. , , (4), , If a pair of reciprocal angles are congruent, then all the remaining angles are congruent., , (5), , If a pair of interior angles on one side of the transversal is complementary, then the other pair of interior angles are also complementary., , Let's learn., Properties of parallel lines, Activity , To check the properties of the angles formed by a transversal of two parallel lines., Take a piece of thick colored paper. On it draw a pair of parallel lines and a transverse line. Glue straight toothpicks to the lines. Eight angles are formed. Cut pieces of colored paper to fit just the corners of Ð1 and Ð2 as shown in the illustration. Place the pieces near different pairs of corresponding angles, alternate angles and interior angles and check their properties., 1, , 2, , 1, , 2, , 14

Page 25:

Let's learn., We check the properties of the angles formed by a transversal of two parallel lines. Now let's prove the properties using Euclid's famous fifth postulate given below. If the sum of two interior angles formed on one side of a transverse line of two lines is less than two right angles, then the lines formed in that direction intersect. , Interior Angle Theorem, Theorem : When two parallel lines are intersected by a transversal, the interior angles on both sides of the transversal complement each other. Given : Line l || line m and line n is its transversal. Therefore, as shown in the figure, Ða,Ðb are interior angles formed on one side and Ðc,Ðd are interior angles formed on the other side of the transversal. For the proof: Ða + Ðb = 180°, Ðd + Ðc = 180 °, , n, d a, c b, , m, l, , Fig. 2.2, , Proof: There are three possibilities for the sum of the dimensions of Ða and , , Ðb., , (i) Ða + Ðb < 180° (ii ) Ða + Ðb > 180° (iii) Ða + Ðb = 180°, Suppose the possibility (i) Ða + Ðb < 180° is true. Then, according to Euclid's postulate, when the line l and the line m are formed, they intersect on the side of the transverse line where Ða and Ðb are formed. the l line and the m line are parallel lines. ..........given, \ Ða + Ðb < 180° impossible ..........(I), Now suppose that \, , Ða + Ðb > 180° true., , Ða + Ðb >180°, but Ða + Ðd = 180°, , and Ðc + Ðb = 180° . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 Angles in linear pairs, \Ða + Ðd + Ðb + Ðc = 180° +180° = 360°, \Ðc + Ðd = 360° - (Ða + Ðb), If Ða + Ðb >180° then [360° - ( Ða + Ðb)] < 180°, \Ðc + Ðd < 180°, 15

Page 26:

\ In this case, the l line and the m line intersect on the same side of the transverse line where Ðc and Ðd are formed. , , \Ðc + Ðd < 180 is impossible ° is impossible...... (II), , \ the remaining possibility,, , , Ða + Ðb = 180° is true......of (I) and (II), , , \ Ða + Ðb = 180° Likewise, Ðc + Ðd = 180°, , , Note that in this proof, due to contradictions, the possibilities Ða + Ðb >180° and Ða + Ðb <180 ° were negated., Therefore, this proof is an example of an indirect proof., Theorems about corresponding angles and alternate angles, Theorem : The corresponding angles formed by a transversal of two parallel lines are equal, n, : line l | 🇧🇷 Line m, given line n is a transversal., , a, c, b, , Ða = Ðb, : Ða + Ðc = 180° ........(I) angles in the linear pair, Fig. 2.3 , Ðb + Ðc = 180° ..............(II) Property of the interior angles of parallel lines, Ða + Ðc = Ðb + Ðc .......de (I ) and (II ), \ Ða = Ðb, , l, m, , Prove :, Prove, , Theorem : The alternate angles formed by a transversal of two parallel lines are equal, measures., n, Given, : line l | 🇧🇷 Line m, line n is a transversal., , d c, b, , proof :, , Ðd = Ðb, , proof, , Ðd + Ðc = 180°............... . (I) Angles in the pair of lines, , :, , Fig. 2.4, , Ðc + Ðb = 180° ..............(II) Property of the interior angles of parallels , Ðd + Ðc = Ðc + Ðb.......... ......of (I) and (II), \ Ðd = Ðb, , 16, , l, m

Page 27:

Exercise set 2.1, 1st, , In Figure 2.5 line RP || MS line and DK line, is its transversal. ÐDHP = 85°, Find the dimensions of the following angles., (i) ÐRHD (ii) ÐPHG, (iii) ÐHGS (iv) ÐMGK, , R, , H, , D, 85° P, , G, , M , , S, , K, , Fig. 2.5, , p, , 2., , In Fig. 2.6 line p || Line q and line l and line m are transversal., The measures of some angles are shown., Find, therefore, the measures of, Ða, Ðb, Ðc, Ðd., , n, , a, b, , 110°, , b , l, , 115°, , c, , d, , m, , Fig. 2.6, , p, 45°, , c, , a, , q, , 3., , l, m, , In Fig. 2.7, Line 1 || line m line n || line p Find Ða, Ðb, Ðc from the given angle measure., , Fig. 2.7, 4*. In Figure 2.8, the sides of ÐPQR and ÐXYZ are parallel to each other. Prove that,, ÐPQR @ ÐXYZ, , P, , X, , Y, Q, , R, , Fig. 2.8, , 17, , Z

Page 28:

5., In Figure 2.9, line AB || The CD line and the PQ line are transversal. The measure of one of the angles is given. Therefore, find the measures for the following angles. , , (i) ÐART, , (iii) ÐDTQ, , R, 105°, , A, T, , C, , (ii) ÐCTQ, , B, D, , Q, , (iv) ÐPRB, , Fig. 2.9, , Let's learn., Using properties of parallel lines, Let's prove a property of a triangle using the properties of the angles formed by a transversal of parallel lines., Theorem : The sum of the measures of all the angles of a triangle is 180° ., Given :, , D ABC is an arbitrary triangle., , Proof : ÐABC + ÐACB + ÐBAC = 180°., , A, , Construction : Draw a line parallel to sec BC and passing through, , , passing through A. On the line take the points P and Q, , , , such that P - A - Q., , B, , Fig. 2.10, , Proof : line PQ || The line BC and the section AB is a transversal., , \ ÐABC = ÐPAB.....alternative angles.....(I), line PQ || The line BC and the segment AC is a transversal., \ ÐACB = ÐQAC.....alternative angles.....(II), \ From I and II ,, ÐABC + ÐACB = ÐPAB + ÐQAC . 🇧🇷 🇧🇷 (III), , P, , B, , Adding ÐBAC to both sides of (III)., , A, , C, , Q, , Fig. 2.11, , C, , ÐABC + ÐACB + ÐBAC = ÐPAB + ÐQAC + ÐBAC, , , , , , = ÐPAB +, , ÐBAC + ÐQAC, , , , = ÐPAC +, , ÐQAC ...(Q ÐPAB + ÐBAC = ÐPAC), , = 180°, , ...angle in linear pair , , That is, the sum of the measures of the three angles of a triangle is 180°., , 18

Page 29:

Let's discuss., in Fig. 2.12, How will you decide whether line l and line m are parallel or not?, , l, , Fig. 2.12, , m, , Let's learn., Tests for parallel lines, Whether or not two given lines are parallel can be decided by examining the angles formed by a transversal of the lines., (1) If the interior angles on the same side of a transversal are complementary, then the lines are parallel., (2) If any of the pairs of alternate angles are congruent, then the lines are parallel., (3) If any of the pairs of corresponding angles are congruent, then the lines are parallel lines., Interior angle test, theorem: If the interior angles form a transversal of two If different lines are formed, then the two lines are parallel., Given, , : The line XY is a transversal of the line AB and, Line CD ., , A, , ÐBPQ + ÐPQD = 180°, , Proof : Line AB || Line CD, proof, , means line AB and CD intersect at point T, so D is formed PQT., , Fig. 2.13, , X, , A, , line AB and line CD are not parallel,, , B, D , , Y, , Suppose the statement to be proved is false. That is, we assume,, , X, , Q, , C, , : We perform an indirect proof., , , , P, , P, Q, , C, Y, , B T, D, , Fig 2.14, , , \ ÐTPQ + ÐPQT + ÐPTQ = 180° ..........sum of the angles of a triangle, but ÐTPQ + ÐPQT = 180° ..........given , , That is, the sum of the two angles of the triangle is 180°., , But the sum of the three angles of the triangle is 180°., \, , ÐPTQ = 0°., , 19

Page 30:

\ PT line and QT line means that AB line and CD line are not distinct lines., But we get that AB line and CD line are distinct lines., , \ we arrive at a contradiction., \ our assumption is wrong. Therefore, line AB and line CD are parallel. Thus it is proved that if the interior angles formed by a transversal are complementary, then the lines are parallel. This property is called the test for interior angles of parallel lines., Test for alternate angles, Theorem : If a pair of alternate angles formed by a transversal of two lines are congruent, then the two lines are parallel., Given, : The line n is a transversal of the line l and the lines m, n, , Ða and Ðb are a congruent pair of alternate angles., l, That is, Ða = Ðb, ac, To prove: line l || Line m, b, m, proof, : Ða + Ðc = 180° . ....Angles in the linear pair, Ða = Ðb .......... given, , , \ Ðb + Ðc = 180°, , , Fig. 2.15, But Ðb and Ðc are interior angles on the same side of the transversal. , , \ line l || Line m ....... Test of the interior angle, , This property is called the test of the alternating angle of parallel lines., Test of the corresponding angle, Theorem : If a pair of corresponding angles formed by a transversal of two lines are congruent, then the two lines are parallel. , Given, , : Line n is a transversal of line l and line m., , , Ða and Ðb is a congruent pair of corresponding angles., , That is, Ða = Ðb, Proof: , , n, , line l | | Line m, , Аa + Аc = 180° ..........angles in pair of lines, , Аa = Аb .......... given, \ Аb + Аc = 180° , , proof , , a, c, , :, , This is a pair of interior angles on the same side of the transversal is congruent., , , \ Line l || Line m .......... internal angle test, , b, , Fig. 2.16, , This property is called the corresponding angle test of parallel lines., , 20, , l, , m

Page 31:

Corollary I: If a line is perpendicular to two lines in a plane, then the two lines are parallel., n, given, : line n ^ line l and line n ^ line m, proof: line l | 🇧🇷 line m, proof, : line n ^ line l and line n ^ line m ... given, \ Ða = Ðc = 90°, , Ða and Ðc are corresponding, , angles formed by the transversal n of line l and line m . , , \ line l || Lines m, , , a, , l, , c, , m, , Fig. 2.17, , ....corresponding angle test, , Corollary II: If two lines in a plane are parallel to a third line in the plane, then these two lines are parallel to each other. Write the proof of the corollary., , Exercise 2.2, l, , 1. In Figure 2.18, y = 108° and x = 71°, are the lines m and n parallel? Justify ?, , m, , x, y, , Fig. 2.18, , n, , n, a, , 2. In Fig. 2.19, if Ða @ Ðb then prove this line l || line m., , l, m, , b, l, , Figure 2.19, 3. In figure 2.20, if Ða @ Ðb and Ðx @ Ðy, then prove that the line l || Line n., , b, , m, , K, , a, x, , D, A, , 50°, C, , Fig. 2.21, , y, , E, 100°, , B, , n, , Fig. 2.20, , 4. In Fig. 2.21, if radius BA || Radius DE,, ÐC = 50° and ÐD = 100°. Find the measure of ÐABC., (Hint: draw a line through point C and parallel to line AB), , 21

Page 32:

In figure 2.22 radius AE || Ray BD, Ray AF is the bisector of ÐEAB and Ray BC is the bisector of ÐABD. Prove that the line AF || Line BC., , 5., F, E, , xx, , B, y y, , A, C, , D, , Fig. 2.22, 6. A transversal EF of the line AB and the line CD intersects the lines at the point P , , A and Q. Radius PR and radius QS are parallel bisectors of ÐBPQ and , , C, , ÐPQC, respectively. Prove that the line AB || Line CD., , P, , E, R, , S, F, , Q, , B, D, , Fig. 2.23, , Exercise set 2. Mark the correct alternative and fill in the gaps in the following statements ., (i) If a transversal intersects two parallel lines, then the sum of the interior angles on the same side of the transversal is ...... ......, (B) 90°, (C) 180° , ( D) 360°, , (A) 0°, (ii) The number of angles formed by a transversal of two lines is ... ........., , (A) 2, ( B ) 4, (C) 8, (D) 16, (iii) A transversal intersects two parallel lines. If the measure of one of the angles is 40°, then the measure of the corresponding angle is ............., (B) 140°, (C) 50°, (D) 180° ° , , (A) 40°, 1st, , (iv), , (v), , , , In D ABC, ÐA = 76°, ÐB = 48°, \ ÐC = ........ . .. ..., (B) 56°, (C) 124°, (D) 28°, (A) 66°, Two parallel lines are intersected by a transversal. If the measure of one of the alternate interior angles is 75°, then the measure of the other angle is .............., (B) 15°, (C) 75°, (D ) 45°, (A) 105°, , 2*. Radius PQ and radius PR are perpendicular to each other. Points B and A are inside and outside ÐQPR, respectively. Radius PB and radius PA are perpendicular to each other., , Draw a figure showing all these rays and write (i) a pair of complementary angles, (ii) a pair of additional angles, (iii) a pair of congruent angles. , , 22

Page 33:

3. Prove that if a line is perpendicular to one of the two parallel lines, then it is also perpendicular to the other line. 4. Figure 2.24 shows the dimensions of some angles. Using measures, find the measures of Ðx and Ðy and therefore show that the line l || The line m., Q, , A, , B, , x, , E, , 50°, , y, , m, , Fig. 2.24, , is its transversal. If y : z = 3 : 7 then find, , D, , z, , l, , x, , 5th row AB || CD line || EF line and QP line, , y, C, , 130°, , the measure of Ðx. (See Figure 2.25.), , F, P, , Fig. 2.25, , p, , 6. In Figure 2.26, if line q || line r,, line p is your, transversal and if, a = 80° find the values of f and g., , b a, c d, , q, , f e, gh, , r, , Fig. 2.26, A , , P, , B, , 7. In Figure 2.27, if the line AB || CF line and BC line || Line ED then proves that ÐABC = ÐFDE., , C, , E, , D, F, , Fig. 2.27, , P, Q, , A, , 8. In Figure 2.28 the line PS is a transversal, , Y , , X, , the parallel line AB and the line CD. If radius, QX, radius QY, radius RX, radius RY are bisectors, , C, , R, , bisectors, then prove that ¨ QXRY is a, , B, D, , S, , rectangle., Fig. 2.28, , 23, , qqq

Page 34:

3, , triangles, , let's learn., , • median of a triangle, • property of the median over the hypotenuse, , • set of distant interior angles, •, •, •, , of a triangle, congruence of triangles, set of an isosceles triangle, property of angles 30°- 60°- 90°, triangle, , •, •, •, , right triangle, bisector theorem, bisector theorem, similar triangles, , Activity :, Draw a triangle of each measure on thick paper. Take a point T on the QR radius as in Fig. 3.1. Cut two pieces of thick paper that fit perfectly at the corners of ÐP and ÐQ. Make sure the same two pieces fit snugly into the corner of ÐPRT as shown., , P, , Q, , Fig. 3.1, , R, , T, , Let's learn., Theorem of distant interior angles of a triangle, Theorem: The measure of an exterior angle of a triangle is equal to the sum of its distant interior angles., P, Given: ÐPRS is an exterior angle of D PQR., Fig. 3.2, To prove: ÐPRS = ÐPQR + ÐQPR, Proof, : The sum of all angles of a triangle is 180°., \ ÐPQR + ÐQPR + ÐPRQ = 180°..... ..(I), , Q , R S, ÐPRQ + ÐPRS = 180°.....angles in the linear pair.. ....(II), \ of (I) and (II), ÐPQR + ÐQPR + ÐPRQ = ÐPRQ + ÐPRS, \ ÐPQR + ÐQPR = ÐPRS .......If you eliminate ÐPRQ from both sides, \ the measure of an exterior angle of a triangle equals the sum of its distant interior angles., , 24

Page 35:

Use your mind power!, , Can we give an alternative proof of the theorem by drawing a line through point R and parallel to sec PQ in Figure 3.2?, , Let's learn., property of an exterior angle of the triangle, The sum of two numbers positive a and b, that is, (a + b) is greater than a and greater than b as well. That is, a + b > a, a + b > b, P, This inequality gives us a property related to the exterior angle of a triangle. If ÐPRS is an exterior angle of D PQR then ÐPRS > ÐP , ÐPRS > ÐQ, S, R, Q, Fig. 3.3, \an exterior angle of a triangle is greater than its distant interior angle., Worked examples, Ex. ( 1 ) The angular dimensions of a triangle are in the ratio 5 : 6 : 7. Find the dimensions., Solution : The dimensions of the angles of a triangle are 5x, 6x, 7x., \ 5x + 6x + 7x = 180° , 18x = 180°, , , , x = 10°, , , , 5x = 5 ´ 10 = 50°, 6x = 6 ´ 10 = 60°, 7x = 7 ´ 10 = 70°, , \ the angular dimensions of do triangle are 50°, 60° and 70° ., Ex. (2) Consider Figure 3.4 and determine the dimensions of ÐPRS and ÐRTS., Solution: ÐPRS is an exterior angle of D PQR., So from the set of distant interior angles,, , P, 30°, , , ÐPRS = ÐPQR + ÐQPR, = 40°, , , , T, , + 30°, , = 70°, In D RTS, , Q, , \, , + ÐRTS + , , R, , 20°, , Fig 3.4 , , , ÐTRS + ÐRTS + ÐTSR =, , , 40°, , ........ sum of all angles of a triangle, , = 180°, , \ ÐRTS + 90° = 180°, \ ÐRTS =, , , , 25, , p

Page 37:

Ex (4) In Figure 3.7, the bisectors of , , ÐB and ÐC of D ABC intersect at P., A, ÐBAC., , 1, Prove that ÐBPC = 90 +, 2, , P, , Complete Complete the proof by filling in the blanks., Proof: In D ABC,, , Fig. 3.7, ...... Sum of the angles of a triangle, , ÐBAC + ÐABC + ÐACB =, , , , \, , 1, 2 , , ÐBAC +, , C, , B, , 1, 2, , ÐABC +, , 1, 2, , ÐACB =, , 1, 2, , ´, ....Multiply each term by 1, , 2, , \ , , 1, 2, , ÐBAC + ÐPBC + ÐPCB = 90° , , \ ÐPBC + ÐPCB = 90° In D BPC, , 1, 2, , ÐBAC ......(I), , ÐBPC + ÐPBC + ÐPCB = 180° ......sum of angular dimensions of a triangle, \ÐBPC +, , = 180° ......of (I), , 1, ÐBAC), 2, , , = 180 ° - 90° + 1 ÐBAC, 2, = 90° + 1 ÐBAC , 2, , , , \ ÐBPC = 180° - (90° -, , Exercise Theorem 3.1, 1., , In Figure 3.8, ÐACD is an exterior angle of D ABC., , A, , ÐB = 40°, ÐA = 70° Find the measure of ÐACD., 2º, , in D PQR, ÐP = 70°,, , 3º, , The measures are the angles of a triangle x°, (x-20)°, (x-40) °., Find the Mate for each angle., , 4., , ÐQ = 65° then find ÐR., , B, , Fig. 3.8, , C, , D, , An angle of a triangle is twice the angle its smallest angle and the other is three times the size of the smallest angle. Find the measures of the three angles., , 27

Page 38:

T, , 5th, , 100°, , Figure 3.9 gives the dimensions of some angles. Using measurements, find the, , E, y, z, , x, N, , values of x, y, z., , M, , 140°, R, , Fig. 3.9, D, , 6. , , B , , In Figure 3.10 line AB || line of. Find the measures of ÐDRE and ÐARE using the given measures of some angles., , R, 40°, , 70°, , A, , 7°, , 8°, , In D ABC, bisectors of , of ÐAOB. , , And, , Fig. 3.10, , ÐA and ÐB intersect at point O. When ÐC = 70°. Find the measure, , in Figure 3.11, line AB || Line CD and line PQ is the transversal. PT ray and QT ray bisect ÐBPQ and ÐPQD, respectively., Prove that mÐPTQ = 90°., , A, , B, , P, T, , C Q, , D, , Fig. 3.11, , 9., , a, , Using the information in Figure 3.12, determine the dimensions of Ða, Ðb, and Ðc., , b, 70°, , 10. In Figure 3.13 line DE || The line GF , the ray EG and the ray FG bisect the angles of ÐDEF and ÐDFM respectively. Prove that,, (i) ÐDEG =, , 1, 2, , 100°, , c, , Fig. 3.12, , D, , ÐEDF (ii) EF = FG., , E, , 28, , G , , F, , Fig. 3.13, , M

Page 39:

Let's learn., Congruence of triangles. We know that when a segment placed on top of another corresponds exactly, the two segments are congruent. If an angle placed on top of another corresponds exactly, then the two angles are congruent. When a triangle placed on top of another triangle fits perfectly, we say that the two triangles are congruent. If D ABC and D PQR are congruent, it is written as D ABC @ D PQR., A1, , A3, , A2, , C1, , B1, , B2, , C2, , B3, , Activity : Draw D, , A , , C, , B, A4, , A, , B, , C3, , C, , C4, , B4, , Fig. 3.14, ABC of any size on a sheet of cardboard and cut out., , Place on a sheet of map. Make a copy of it by drawing the border. Name it D A1B1C1, . Now slide the D ABC, which is the cutout of a triangle, some distance away and make another copy of it. Name D A2B2C2. Then rotate the cutout of triangle ABC a little bit as shown in the figure and make another copy of it. Name the copy D A3B3C3 . Then turn the triangle ABC over, place it on another sheet of cards and make a new copy. Name this copy D A4B4C4 . Did you notice that D A1B1C1 , D A2B2C2 , D A3B3C3 and D A4B4C4 are coincident with D ABC, respectively? Since each of them corresponds exactly to D ABC, let's check D A3B3C3. If we put ÐA in ÐA3 , ÐB in ÐB3 , and ÐC in ÐC3 , then only they match and we can say that D ABC @ D A3B3C3 . We also have AB = A3B3, BC = B3C3, CA = C3A3. , Note that when examining the congruence of two triangles, we must write their angles and sides in a specific order, that is, with a specific one-to-one correspondence., If D ABC @ D PQR, we get the following six equations : , ÐA = ÐP, ÐB = ÐQ, ÐC = ÐR . 🇧🇷 (I) and AB = PQ, BC = QR, CA = RP . 🇧🇷 🇧🇷 🇧🇷 (II), That is, given a one-to-one relationship between the angles and sides of two triangles, we obtain three pairs of congruent angles and three pairs of congruent sides., , 29

Page 40:

Assuming that six equations above are valid for congruent triangles. For that, let's see that three specific equations are true, so all six equations become true and therefore two triangles, congruent. (1) In a correspondence, if two angles of DABC are equal to two angles of DPQR and the sides are included. If the respective pairs of angles are also equal, then the two triangles are congruent., A, , P, R , , CQ, , B, , This property is called the angle-side-angle test, in short we write A-S-A test ., , Fig. 3.15, , If in a correspondence two sides of D ABC are equal to two sides of D PQR and those of the respective Angles included by pairs of sides are also equal, then the two triangles are equally congruent., , (2), , A, , P, , C, , This property is called the side-angle-side test, which we call the S-A-S test for short writing., R, , Q, , B, , Fig. 3.16, , (3) If in a correspondence three sides of D ABC are equal to three sides of D PQR , then the two triangles are congruent., A, , Q, , B, , R, , C, This property is called side-side-side test, which we will call the S-S-S test for short., , P, , Fig. 3.17, , (4) If in D ABC un d D PQR ÐB and ÐQ are right angles, the hypotenuses are equal and AB = PQ, then the two triangles are congruent., P, , A, , This property is called the test on the side of the hypotenuse., B, , C, , Q, , R, , Fig. 3.18, , 30

Page 41:

To remember! We construct triangles using the information provided about the parts of the triangles. (For example, two angles and the included side, three sides, two sides and one included angle). Our experience is that the triangle constructed using any of this information is unique. , between two triangles, if these three parts of one triangle are congruent with the corresponding three parts of the other triangle, then the two triangles are congruent. So we learn that in this correspondence its three angles and three sides are congruent. If two triangles are congruent, then their respective angles and respective sides are congruent. This property is useful for solving many geometry problems., , Exercise Theorem 3.2, 1., , A pair of triangles is shown in each of the examples below. Equal parts of triangles in each pair are marked with the same signs. Look at the figures and give the test that the triangles in each pair are congruent., , (i), P, , A, , B, , (ii), , C, , Q, , R, Y , , Z , , N, , M, , Through . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 test, by . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 test, , D ABC @ D PQR, , D XYZ @ D LMN, , (iii), , (iv), , L, , M, , T, , S, , P, , Q, , L, , X, , R, , T, , U, N, , R, , Through . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 test, by . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 Test, , D PRQ @ D STU, , D LMN @ D PTR, Fig. 3.19, , 31, , P

Page 42:

2., , Consider the information presented in the pairs of triangles below. Test that the two triangles are congruent. Write the remaining congruent parts of the triangles., , (i), , (ii), , A, , R, , P, , P, , T, C Q, , B, , Fig. 3.20, , R, , 3.21 , From the information shown in the figure,,, In D PTQ and D STR, seg PT @ seg ST, ÐPTQ @ ÐSTR....vertically opposite angles, seg TQ @ seg TR, \ D PTQ @ D STR . ......, test, \ÐTPQ @, ..... corresponding, congruent angles, @ ÐTRS, and, triangles., , From the information shown in the figure, in D ABC and D PQR, ÐABC @ ÐPQR , sec BC @ sec QR, ÐACB @ ÐPRQ, \ D ABC @ D PQR ......., test, \ÐBAC @, .......corresponding, , angles of congruent triangles., sec AB @ , , @, , and, 3rd, , C, , 5th, , }, , corresponding, .... congruent sides, sec PR, triangles, , }, , seg PQ @, , , From Enter the in Provide the information presented in the figure and indicate the test that guarantees the congruence of D ABC and D PQR. Write the remaining congruent parts of the triangles. 4., , corresponding sides of, congruent triangles., , as shown in the figure below, in, D LMN and D PNM, LM = PN, LN = PM. Write the test that guarantees the congruence of the two triangles. Write their remaining congruent parts @ sec CD., Prove that, D ABD @ D CBD, , Fig. 3.23, , N, , C, , A, , D, , Fig. 3.24, , Note: Corresponding sides of congruent triangles we write c.s.c.t. and corresponding angles of congruent triangles we write c.a.c.t., , 32

Page 43:

6., , In Figure 3.25, ÐP @ ÐR, sec PQ @ sec RQ, Prove that,, D PQT @ D RQS, , Q, S, , T, , Fig. 3.25, , P, , R, , Let's learn ., Isosceles Triangle Theorem, Theorem : If two sides of a triangle are congruent, then their opposite angles are congruent., : In D ABC, Side AB @ Side AC, Given, A, Proof : ÐABC @ ÐACB, Construction : Draw the bisector of ÐBAC, , which intersects the side BC at point D. Proof, : In D ABD and D ACD, B, C, D, sec AB @ sec AC ... .... given, ÐBAD @ ÐCAD ........construction, , Fig. 3.26, seg AD @ seg AD ....... common side, \D ABD @ D ACD ......, \ÐABD @, ... .... (c.a.c.t.), , \ÐABC @ ÐACB , , Q B-D-C , , Corollary : If all sides of a triangle are congruent, then all angles are congruent., (Write the proof of this corollary.), Opposite of isosceles Triangle Theorem, Theorem : If two angles of a triangle are congruent, so their opposite sides are congruent., Data: In D PQR, ÐPQR @ ÐPRQ, P, Proof: Side PQ @ side PR, Construction: Draw the gate bisector of ÐP, , , Intersection side QR at point M, , Proof: In D PQM and D PRM, , ÐPQM, , @, ........ given, , ÐQPM @ ÐRPM..... ., sec PM @, ....... common side , \ D PQM @ D PRM . ....., test, , \sec PQ @ sec PR..... c.s.c.t., 33, , Q, , M, , Fig. 3.27, , R

Page 44:

Corollary :, , , , , If three angles of a triangle are congruent, then its three sides are also congruent., (Write the proof of this corollary yourself.), The above two theorems are also inverses of each other., Likewise way, the corollaries of the theorems are opposite to each other., , Use your brain!, (1), (2), , Can the isosceles triangle theorem be proved by doing another construction?, Can the isosceles triangle theorem be proved without doing any construction ?, , let's learn., property of the triangle 30° - 60° - 90°, activity I, , A, , Each student in the group must draw a, , 60°, , right triangle, one of the angles , measuring 30°. The choice of side lengths should be yours. Each should measure, , 30°, , B, , the length of the hypotenuse and the length of, , C, , Fig. 3.28, , the side opposite the 30° angle., , One of the students in the group should fill in the table below., triangle number, , 1, , 2, , 3, , 4, , length of the side opposite the angle of 30°, length of the hypotenuse, Did you notice a property of the sides of a right triangle? with one of the angles measuring 30°?, Activity II, The angular measures of a square in its compass are 30°, 60° and 90°., Check the property of the sides of the square., Let's prove an important property that resulted from these activities ., , 34

Page 45:

Theorem: If the acute angles of a right triangle measure 30° and 60°, then the length of the side opposite the 30° angle is half the length of the hypotenuse., (Fill in the blanks and complete the proof.), A , Given, , : In D ABC, , 60°, , , ÐB = 90°, ÐC = 30°, ÐA = 60°, To prove : AB =, , 1, AC, 2, , 30°, , B , , Construction : Take a point D on the extended segment AB such that AB = BD. Draw Mon DC., , Fig. 3.29, , C, , A, , 60°, D ABC and D DBC, sec AB @ sec DB ..........., 30°, B, C , ÐABC @ ÐDBC ..... ..., sec BC @ sec BC ............, D, \ D ABC @ D DBC ....., Fig. 3.30, \ ÐBAC @ ÐBDC ........ (c.a.c.t.), In D ABC, ÐBAC = 60° \ ÐBDC = 60°, ÐDAC = ÐADC = ÐACD = 60° ...Q sum of angles of D ADC is 180°, , \ D ADC is an equilateral triangle., , \ AC = AD = DC ........ Corollary of the inverse isosceles triangle theorem, , proof, , :, , 1, AD.. ... . 🇧🇷 .bau, 2, \AB = 1 AC ........ Q AD = AC, , 2, , , , But AB =, , Activity, Using Figure 3.29 above, fill in the blanks and complete the proof of the following theorem. Theorem: If the acute angles of a right triangle are 30° and 60°, then the length of the side opposite the 60° angle is, Proof: In the theorem above, we prove AB =, AB2 + BC2 =, , .. .. .......... Pythagorean theorem, , 1, 2, 2, 4 AC + BC =, , \ BC2 = AC2 -, , 1, AC , 2, , 3, 2 ´ hypotenuse, , 1 , 2, 4 AC, , \ BC2 =, \ BC = 23 AU, 35

Page 46:

Exercise: Complete the proof of the theorem., Theorem: If the angular dimensions of a triangle are 45°, 45°, 90°, then the length of each side containing the right angle is 1 ´ hypotenuse., A, Proof: In D ABC, ÐB = 90° and, , 2, , ÐA = ÐC = 45°, , 45°, , \ BC = AB, By the Pythagorean theorem, , \ 2AB2 =, \ AB2 =, \ AB =, , 45 ° , , B, , AB2 + BC2 =, AB2 +, = AC2 ... Q (BC = AB), , C, , Fig. 3.31, , 1, AC, 2, , This property is called the 45°- 45 ° - 90° Theorem., , Remember !, (1) If the acute angles of a right triangle are 30°, 60° , then the length of the side, , , opposite the 30° angle, is half the hypotenuse, and the side length opposite the 60° angle, is, , 3, the hypotenuse . This property is called the 30°-60°-90° theorem., 2, , (2) If the acute angles of a right triangle are 45°, 45°, then the length of each side, hypotenuse, containing the right angle is , ., 2, , This property is called the 45°-45°-90° set., , Remember, Median of a triangle, The segment joining a vertex and the midpoint of the opposite side is called a , median of the triangle., , A, , In Figure 3.32, the point D is the midpoint of the side BC., , \seg AD is a median of D ABC., , B, , D, , Fig. 3.32, , 36, , C

Page 47:

Activity I, , : Draw a triangle ABC. Draw the medians, , AD, BE, and CF of the triangle. Their point of agreement , , B, , is G, which is called the barycenter of the triangle , D, . Compare the lengths of AG and GD, , F, , with a divisor. Make sure the length of AG is twice the length of GD. Likewise, check that the length of BG is twice the length of GE and the length of CG is twice the length of GF. Therefore, observe the following, , C, , G, E, , A, , Fig. 3.33, , property of the medians of a triangle. , Activity II : Draw a triangle ABC on a map, blackboard. Draw your centerlines and label your G-spot. Cut out the triangle. Now take a pencil. Try to balance the triangle on the flat end of the pencil. The triangle is in equilibrium only when the G point is on the flat point of the pencil. This activity demonstrates an important property of the center of gravity (point of convergence of the medians) of the triangle., , Fig. 3.34, , Let's learn., Property of the median plotted on the hypotenuse of the right triangle, activity: In Figure 3.35, D ABC is a right triangle. Segment BD is the median at, B, Hypotenuse., Measure the lengths of the following segments., AD =........., , DC =............, , BD = ............, , Check by measurements that BD =, , 1, AC ., 2, , A, , D, , Fig. 3.35, Now we want the property Prove the length of the median is half the length of the hypotenuse., , 37, , c

Page 48:

Theorem : In a right triangle, the length of the centerline of the hypotenuse is half the length of the hypotenuse., Given, , : In D ABC,, , E, , A, , ÐB = 90°, sec BD is the median. , , Proof : BD = 1 AC, , D, , B, , 2, , Construction : Take point E on radius BD such that B - D - E, , , C, , Fig. 3.36, , and l(BD) = l(DE). Draw sec EC., , : (main steps are given. Write the ratios between steps and complete the proof.), , D ADB @ D CDE .......... by S-A-S test, , proof , , , , || line EC ..........by the alternate angles test, D ABC @ D ECB ..........by the S-A-S test, , , , BD =, , line AB, , 1, AC, 2, , Remember !, In a right triangle, the length of the center line on its hypotenuse is half the length of the hypotenuse., , Exercise Theorem 3.3, A, , 1., , Find the values of x and y in Using the information, , x, , from Figure 3.37., , find the measure of ÐABD and mÐACD., , 50°, , B 60°, , y C, , Fig. 3.37, D, , 2nd, , Die Length of the hypotenuse of a right triangle is 15. Find the length of the median of its hypotenuse., , 3., , In D PQR, ÐQ = 90°, PQ = 12, QR = 5 and QS is a median. Find l(QS)., , 4., , In Figure 3.38,, , P, , point G is the point , , meeting the medians of, , D PQR ., , G, , when GT = 2.5, find the lengths of PG and PT., Q, , T, , Fig. 3.38, , 38, , R

Page 49:

Let's remember., Activity: Draw a segment AB of suitable length. Denote its center M. Draw a line l through point M and perpendicular to sec AB. Did you notice that line l is the perpendicular bisector of seg AB? Now take a point P line l somewhere. Compare PA and PB distance with a divider. What did you find? You A must have noticed that PA = PB. This observation shows that every point on the perpendicular bisector of a segment is equidistant from its endpoints. Now, using a compass, pick any two points, such as C and D, that are equidistant from A and B. Are the points on the straight line l? What did you notice while watching? Any point equidistant from the endpoints of a segment lies on the perpendicular bisector of the segment. These two properties are two parts of the perpendicular bisector theorem. Let's prove them now., , D, P, M, , B, , C, , l, Fig. 3.39, , Let's learn., Perpendicular Bisector, Part I, , Given, , : Any point on the perpendicular bisector of a , The segment is equidistant from the ends of the segment., , l, , : the line l is the perpendicular bisector of the segment AB in the point M., , the point P is any point in l,, Proof: PA = PB, , A , , Construction : Draw in AP and in BP., Proof, , P, , M, , : In D PMA and D PMB, , seg PM @ seg PM ....... common side, , ÐPMA @ ÐPMB . ........each a right angle, sec AM @ sec BM .......given, Fig. 3.40, , 39, , B

Page 50:

\ D PMA @ D PMB ...... test S-A-S, \ sec PA @ sec PB .......c.s.c.t., \ l (PA) = l (PB), , , , So each point on perpendiculars The bisector of a segment is equidistant from the ends of the segment., , Part II : Every point equidistant from the ends of a segment lies on the bisector of the segment. Given: Point P is any point equidistant from the endpoints of segment AB. That is, PA = PB., Proof: Point P lies on the bisector of segment AB., Construction: Take the midpoint M of segment AB and draw the line PM., Proof, , P, , :In D PAM and D PBM, , Mon PA @ Mon PB ........., Mon AM @ Mon BM ......., Mon PM @, , M, , A, , B, , .... . .. common Page, , \D PAM @ D PBM ......, test., \ ÐPMA @ ÐPMB.....c.a.c.t., Fig. 3.41, But ÐPMA +, = 180°, ÐPMA + ÐPMA = 180° ........ (Q ÐPMA = ÐPMA), 2 ÐPMA =, , \ ÐPMA = 90°, \sec PM ^ sec AB , . .....(1), But point M is the center of Segment AB., , ......Construction .... (2), , \ Line PM is the bisector of segment AB. The point P is therefore on the perpendicular bisector of the section AB. Bisector Theorem, Part I, , : Every point on the bisector is equidistant from the sides of the angle., , :, Given, , , To prove :, Proof, , Radius QS is the bisector of ÐPQR ., point A is a point arbitrary on ray QS, sec AC ^ ray QR, sec AB ^ ray QP, sec AB @ sec AC, , B, , S, , A , Q, , : Write the proof using the triangle congruence test., , 40 , , P, , C, , Fig. 3.42, , R

Page 51:

Part II, , : Every point equidistant from the sides of an angle lies on the bisector., , Given, , : A is a point inside ÐPQR., sec AC ^ radius QR, and AB = AC, , P, , sec AB ^ Ray QP, , B, , Proof : Ray QA is the bisector of ÐPQR., That is ÐBQA = ÐCQA, , Proof, , D, A, , Q, , R, , C, , Fig 3.43, : Write a proof using the right triangle congruence test., , Recalling., , X, , Activity, As shown in the figure, draw D XYZ such that XZ > side XY, Find out which of ÐZ and ÐY is greater., , Y, , Z , , Fig. 3.44, , Let's learn., Properties of inequalities of sides and angles of a triangle, Theorem : If two sides of a triangle are not equal, then the angle opposite the larger side , is greater than the angle opposite the smaller side. , Given, , : In D XYZ, Side XZ > Side XY, , Proof : ÐXYZ > ÐXZY, Construction : Take point P on side XZ such that, , XY = XP, Draw Sec YP., : In D XYP, Proof , XY = XP .........Construction, , X, P, Y, , Fig. 3.45, , Z, , \ ÐXYP = ÐXPY.....Isosceles Triangle Theorem.....(I ) , ÐXPY is an exterior angle of D YPZ., \ ÐXPY > ÐPZY ........ .exterior angle theorem, \ ÐXYP > ÐPZY ..........of (I), \ ÐXYP + ÐPYZ > ÐPZY ........If a > b and c > 0, then a + c > b, \ÐXYZ > ÐPZY, then ÐXYZ > ÐXZY, 41

Page 52:

Theorem: If two angles of a triangle are not equal, then the side opposite the larger angle is greater than the side opposite the smaller angle. The theorem can be proved by indirect proof. Complete the following proof by filling in the blanks., Data, , : In DABC,, , ÐB > ÐC, , A, , Prove : AC > AB, Proof, , : There are, , only , , three , , Possibilities, , B, , wrt the side lengths AB e, side AC of, , D ABC., , (i) AC < AB , , C, , (iii), , (ii), , (i) Let's take AC < AB . If, , Fig. 3.46, , (ii) If AC = AB, then ÐB = ÐC, , are two sides of a triangle, , unequal then the opposite angle, , but, , longer side is, , This also creates a contradiction. , , ., , \ÐC >, But ÐC < ÐB ......... (given), <, , ...... (given), , \ , =, is false, \AC > AB is the only thing left, , This creates a contradiction., , \, , >, , possibility., , \ AC > AB, , is false., , Let's remember., As shown in the image on the side , there is a store on A. Sameer stood next to C. To get to the store, he chose path C, , ®A, , ® B ® A because he knew that path C ® A was shorter than path C ® B ® A . So what is the quality of a triangle, instead of C, , A, , B, , he realized ?, The sum of two sides of a triangle is greater than its third side., Now let's prove the property, , C , 42

Page 53:

Theorem : The sum of any two sides of a triangle is greater than the third side., Given, , D ABC is any triangle., , :, , D, , Proof : AB + AC > BC, AB + BC > AC , AC + BC > AB, , A, , Construction : Take a point D on radius BA such that AD = AC., Prove, , : In D ACD, AC = AD ..... construction, B, C, \ ÐACD = ÐADC ...... c.a.c.t., Fig. 3.47, \ ÐACD + ÐACB > ÐADC, \ ÐBCD > ÐADC, \ side BD > side BC ......... the side opposite the largest angle is greater , \ BA + AD > BC ...... ........Q BD = BA + AD, , , , , , BA + AC > BC ........ Q AD = AC , We can also prove that AB + BC > AC and BC + AC > AB., , Exercise Theorem 3.4, 1., , In Figure 3.48, point A is on the bisector of ÐXYZ., If AX = 2 cm, find AZ ., , X, A, Y, , Z, , Fig . 3.48, , T, , 2nd, S, , P, R, , In Fig. 3.49, ÐRST = 56°, sec PT ^ ST radius,, sec PR ^ SR radius and sec PR @ sec PT, Find the measure of ÐRSP., Justify your answer., , Fig. 3.49, 3., , In D PQR, PQ=10cm, QR=12cm, PR=8cm. Find the largest and smallest angles of the triangle., , 4th, , In D FAN, ÐF = 80°, ÐA = 40° . Find the longest and shortest sides of the triangle. Give the reason., , 5., , Prove that an equilateral triangle is equiangular., , 43

Page 54:

6., , 7., , Prove that if the bisector of ÐBAC of , is an isosceles triangle., , D ABC is perpendicular to the side BC, then D ABC, P, , In Figure 3.50, if sec PR @ sec PQ , , show that sec PS > sec PQ., , Q, , R, , Fig. 3.50, , S, , A, , 8., , E, , In Figure 3.51, in D ABC, seg AD and seg BE are heights , and AE = BD., Prove that seg AD @ seg BE, , B, , D, , Fig. 3.51, , C, , Let's learn., Similar triangles, Consider the following figures., , The pairs of figures The parts shown in each part have the same shape, but their sizes are different. This means that they are not overlapping. These similar looking figures are called similar figures., , 44

Page 55:

We find similarity in a photograph and its enlargement, we also find similarity between a road map and streets., The proportionality of all sides is an important property of the similarity of two figures., But the angles in the figures must be of equal measure. If the angle between these streets on the map is not the same, then the map is misleading., , ICT tools or links, Take a picture on a mobile phone or computer. Remember what you are doing to make it smaller or larger. Also remember what you do to see part of the photo in detail., Now let's learn about the properties of similar triangles through an activity., Activity: on a map sheet draw a triangle with sides of 4 cm, 3 cm and 2 cm. Stop this. Make 13 more copies of the triangle and cut them out of the card stock. Note that all these triangle parts are congruent. Arrange them as shown in the following figure, forming three triangles., , P, , D, 4, A, 4, B, , 3, , 2, , 4, , 2, C, , E, , Fig. 3.52, 1 triangle, , 3, , 4, 4, , 2, 3, , Fig. 3.53, , 2, , ´, , 4, Q, , F, , 2, , 4 triangles, , 3, , 2, 3, , 3, , Fig. 3.54, , R, , 9 triangles, , D ABC and D DEF are similar in the correspondence ABC « DEF., ÐA @ ÐD, ÐB @ ÐE, ÐC @ ÐF, and, , AB, =, DE, , 4 1, = ;, 8 2, , BC, =, EF, , 3 1, = ;, 6 2, , AC =, DF, , 2 1, = ,, 4 2, , , .. .. ...the sides correspondents are proportional., , D DEF and D PQR. Their angles are congruent and their sides proportional in the correspondence DEF « PQR ?, Consider similar, , 45

Page 57:

Remember !, y•y If the corresponding angles of two triangles are equal, then the two triangles are similar., y•y If two triangles are similar, then their corresponding sides are proportional and their corresponding angles are congruent., , Eg ., , P, , Some information is shown, , in D ABC and, , O, , D PQR in, , A, , Figure 3.57. Observe it., So, find the lengths of sides AC and PQ., , 7.5, , O, , 3, 4, , C, , Solution: The sum of all angles of a triangle is 180°., , B , , R, , 6, , Q, , Fig. 3.57, , Given that,, , АA = АP and АB = АQ, , \ АC = АR, , \ D ABC and D PQR are equiangular triangles., \ No sides are proportional., AB, BC AC, \ , =, =, PQ, QR PR, 3, 4, AC, \ , =, =, PQ, 6, 7.5, , , , \ 4 ´ PQ = 18 , 18 , \ PQ =, = 4,5, 4, Analog 6 ´ AC = 7.5 ´ 4, , \ AC =, , 7.5 ´ 4, 30, =, =5, 6, 6, , Set of exercises 3.5, 1st, 2nd ., 3., , If D XYZ ~ D LMN, write the corresponding angles of the two triangles and also the ratios of the corresponding sides., In D XYZ, XY = 4 cm, YZ = 6 cm, XZ = 5 cm, If D XYZ, PQ = 8 cm, then find the lengths of the remaining sides of D PQR., , ~ D PQR and, , Draw a sketch of a pair of similar triangles. Sort them. Show their corresponding angles using the same signs. Show the lengths of the corresponding sides by numbers in proportion., , 47

Page 58:

Remember: when creating a map of a location, you need to show the distances between different points on roads with an appropriate scale. For example 1cm=100m, 1cm=50m etc. Have you ever thought about the properties of the triangle? Remember that the side opposite the larger angle is larger., Project:, Make a map of the street around your school or house, at a distance of about 500 meters., How do you measure the distance between two points on a Street? , When walking, count how many steps cover a distance of about two meters. Suppose your three steps cover a distance of 2 meters. Considering this proportion, 90 steps means 60 meters. This allows you to judge the distances between different points on the roads and also the lengths of the roads. You also need to evaluate the measurements of the angles where two roads meet. Choose an appropriate scale for the length of the streets and prepare a map. Try to map stores, buildings, bus stop, rickshaw stop, etc. See below an example of a map with legend., , 6, 2, , 1, 7, , 2, 1, 4, , 6, 3, , 5 , , 4, , 6, , 3, Key: 1st bookstore, 5th medical supply store, , 2nd bus stop, 6th restaurant, , 3rd stationery store, 7th bicycle store, , 48, , 4th bank

Page 59:

Problem set 3, 1., , Choose the alternative correct answer to the following questions., (i) If two sides of a triangle are 5 cm and 1.5 cm long, the length of its third side cannot be, ..... .., , , , (A) 3.7 cm, , (B) 4.1 cm, , (C) 3.8 cm, (D) 3.4 cm, , (ii) In D PQR, If ÐR > ÐQ then . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 ., , , (A) QR > PR, , (B) PQ > PR, , (C) PQ < PR, , (D) QR < PR, , (iii) In D TPQ, АT = 65°, АP = 95° Which of the following statements is true?, , 2., , (A) PQ < TP, , (B) PQ < TQ, , (C) TQ < TP < PQ, , (D) PQ < TP < TQ , , D ABC is isosceles with AB = AC. Seg BD and Seg CE are medians. Show that BD = CE., P, , 3., , In D PQR, if PQ > PR and bisectors of , intersect in S. Show that SQ > SR., , ÐQ and ÐR, , S, Q, , 4 ., , R, , Fig. 3.58, A, , In Fig. 3.59 the points D and E lie on the side BC of D ABC such that BD = CE and AD = AE. Show that D ABD @ D ACE. , B, , D, , Fig. 3.59, , 5., , E, , C, , P, , In Figure 3.60, point S is any point on the QR side of D PQR. Prove: PQ + QR + RP > 2PS, , Q, , 49, , S, , Fig. 3.60, , R

Page 60:

In Figure 3.61 the bisector of, , 6., , ÐBAC, , A, , intersects the side BC at point D. Prove that AB > BD, B, , C, , D, , Fig. 3.61, S, , In Figure 3.62, seg PT is the bisector of ÐQPR., , 7. , , A line parallel to seg PT and passing through R, , P intersects the radius QP at point S. Prove that PS = PR ., , Q, , T , , R, , Fig. 3.62, C, , 8. In Fig. 3.63, Seg AD ^ Seg BC., , Seg AE is the bisector of ÐCAB and C - E - D., , D, , Prove that, , ÐDAE =, , E, 1, (Ð C 2, , ÐB ), A, , B, , Fig. 3.63, , Use your wits!, We learned that if two triangles are equiangular, their sides are the same length. What do you think when two quadrilaterals are equiangular? Are your sides proportional? Draw different figures and check., Check the same for other polygons., , qqq, 50

Page 61:

4, , Constructions of triangles, , Let's study., To construct a triangle given the following information., • base, an angle adjacent to the base and the sum of the lengths of the two remaining sides., • base, an angle close to the base and the difference in length of the two remaining sides., • Perimeter and angles close to the base., , Let's remember., In the previous pattern, we learned the following triangle constructions., * To construct a triangle if its three sides are given., * Construct a triangle, given its base and two adjacent angles., * Construct a triangle, given two sides and the included angle., * Construct a right triangle, given its hypotenuse and one side., Perpendicular bisector theorem, • Each point on the perpendicular bisector of a segment is equidistant from its endpoints., , Q, , • Each point that is separated from the endpoints of , , a segment is equidistant lies on the bisector of , the segment., , A, , T, l, , B , , Fig. 4.1, , Let's win l., Constructions of triangles, To construct a triangle, three conditions are necessary. Three sides and three angles of a triangle are two parts and given some additional information about them, then we can build a triangle from them. is the intersection of the two lines., , 51

Page 62:

Construction I, To construct a triangle, given its base, an angle adjacent to the base, and the sum of the lengths of the remaining sides., Ex. Construct D ABC with BC = 6.3 cm, ÐB = 75° and AB + AC = 9 cm. Solution: First, let's draw an approximate figure of the expected triangle. Explanation: As shown in the close-up figure, we first draw the segment BC = 6.3 cm long. On, the ray makes an angle of 75° with the sec BC,, mark the point D such that BD = AB + AC = 9 cm, Now we need to locate the point A on the ray BD, BA + AD = BA + AC = 9 , , A, , 6.3 cm, , C, , gross figure 4.2, , 9 cm, , B, , D, , 75°, , \ AD = AC, \ point A is vertical, , A, , Bisector of section CD., , B, , \ the point of intersection of the ray, , 6.3 cm, , approximate figure 4.3, , , BD and the perpendicular bisector of section CD is the point A., , P, , steps of construction, (1) Draw the segment BC of length 6.3 cm., (2) Draw the radius BP such that mÐ PBC = 75°., (3) Mark the point D on the radius BP such that d(B,D ) = 9 cm, (4) Draw the sec DC., (5) Construct the perpendicular bisector of sec DC ., (6) Call the intersection of the radius BP and the perpendicular bisector of CD as A., (7) draw sec B.C. , , D, , A, , D ABC is the desired triangle., 75°, , B, , 6.3 cm, , Fair fig. 4.4, , 52, , C, , C

Page 63:

Exercise set 4.1, 1st, 2nd, 3rd, 4th, , building D, building D, building D, building D, is 10 cm, , PQR, where QR = 4.2 cm, mÐQ = 40° and PQ + PR = 8.5 cm, XYZ, where YZ = 6cm, XY + XZ = 9cm. ÐXYZ = 50°, ABC, where BC = 6.2 cm, ÐACB = 50°, AB + AC = 9.8 cm, ABC, where BC = 3.2 cm, ÐACB = 45° and perimeter of D ABC, , Construction II, To construct a triangle, given its base, the angle adjacent to the base, and the difference between the remaining sides. Ex. (1) Construct D ABC such that BC = 7.5 cm, ÐABC = 40°, AB - AC = 3 cm. , Solution : Let's draw an approximate number., A, Explanation : AB - AC = 3 cm \ AB > AC, Draw Mon BC. We can plot the radius BL, 40°, such that Ð LBC = 40°. We need to locate the point C, B, 7.5 cm, A on the BL radius. Take the point D on the radius BL, approximately as in Figure 4.5, A where BD = 3 cm. Now B-D-A and BD = AB - AD = 3. Given that AB - AC = 3, \ AD = AC, D , , \ point A lies on the perpendicular bisector of sec DC., , B, , 40°, , C, , \ Point A is the intersection of radius BL and , Approximate figure 4.6, , the perpendicular bisector of sec DC ., , A, , Construction steps, (1) Draw segment BC with length 7.5 cm, (2 ) Draw the ray BL such that Ð LBC = 40°, (3) Take the point D on the ray BL such that, BD = 3 cm., (4) Construct the perpendicular bisector of the segment CD., (5) Name the point of intersection of radius BL and the perpendicular bisector of sec CD as A., (6) draw sec AC., D ABC is the required triangle., , L, , D, 40°, , B, , C, , 7,5 cm, , Figure just 4.7, , 53

Page 64:

Ex. 2 Build D ABC, with sides BC = 7 cm, ÐB = 40° and AC - AB = 3 cm. Solution : Let's draw an approximate figure., A, sec BC = 7 cm. CA > AB. We can plot the ray BT so that 40°, Ð TBC = 40° , C, B, 7 cm, point A is on the ray BT. Locate point D, approximated from Figure 4.8, opposite radius BT such that T, A, BD = 3 cm. Now AD = AB + BD = AB + 3 = AC, (Q AC - AB = 3 cm.), , \ AD = AC, \ Point A is perpendicular, , 7 cm, , B, S, , bisector of Segment CD., , C, , D, , Figure 4.9, , Construction steps, (1) Draw BC of length 7 cm., , T, , (2) Draw the radius BT such that, Ð TBC = 40°, , (3) Take point D on the opposite side, radius BS from radius BT, such that, , A, , BD = 3 cm., (4) Construct the perpendicular, , B, , bisector of the line segment DC., , m , 3c, , (5) Call the intersection point, , of DC as A., (6) Draw the segment AC ., , 7 cm, , D, , of the radius BT and, the perpendicular bisector, , 40 °, , S , , Just Fig. 4.10, , D ABC is the required triangle., , Exercise Theorem 4.2, 1. Construct D XYZ such that YZ = 7.4 cm, ÐXYZ = 45° and XY - XZ = 2.7 cm ., , 2 . Construct D PQR so that QR = 6.5 cm, ÐPQR = 40° and PQ - PR = 2.5 cm., , 3. Construct D ABC so that BC = 6 cm, ÐABC = 100 ° and AC - AB = 2.5 cm., , 54, , C

Page 65:

Construction III, Construct a triangle given its perimeter, base and the angles that form the base., Ex. Construct D ABC such that AB + BC + CA = 11.3 cm, ÐB = 70°, Solution: Draw an approximate number. , , P, , 35°, , ÐC = 60°., , A, , B, , 70°, , 60°, , 11.3 cm, , C, , 30° , , Q, , Rough Fig. 4.11, Explanation : As shown in the figure, the points P and Q are taken on the line BC so that,, PB = AB, CQ = AC, , \PQ = PB + BC + CQ = AB + BC + AC = 11, 3 cm., Now in DPBA, PB = BA, \ÐAPB = ÐPAB and ÐAPB + ÐPAB = Exterior angleABC = 70°, ......Remote Interior Angles Theorem, , \ ÐAPB = ÐPAB = 35° In the same way ÐCQA = ÐCAQ = 30° , Now we can draw D PAQ since its two angles and the included side are known., Since BA = BP , the point B lies on the perpendicular bisector of sec AP., , According to CA = CQ, that is, point C is on the perpendicular bisector of sec AQ, , \ obtained by constructing the perpendicular bisector of sec AP and AQ n we find the points B and C where the perpendicular bisector i intersection line PQ., Construction steps, (1 ) Draw the 11.3 cm long segment PQ., , (5) Draw the perpendicular bisector of segment AP and segment AQ. Label the points B and C, respectively, where the perpendicular bisectors intersect the line PQ. (6) Draw segment AB and segment AC. 3) Draw another ray with an angle of 30° at point Q. (4) Call the intersection of the two rays A, , D ABC is the desired triangle, 55

Page 66:

A, , P, , , 35°, , B, , 70°, , 60°, , C, , 30°, , 11.3 cm, end Fig. 4.12, , Q, , exercise set 4.3, 1st construction D PQR , where ÐQ = 70°, ÐR = 80° and PQ + QR + PR = 9.5 cm., 2. Construct D XYZ, where ÐY = 58° , ÐX = 46° and the perimeter of the triangle is 10.5 cm . Construct D LMN, where ÐM = 60°, ÐN = 80° and LM + MN + NL = 11 cm., , Problem set 4, 1st, 2nd, 3rd, , Construct D XYZ such that XY + XZ = 10.3 cm , YZ = 4.9 cm, ÐXYZ = 45°., Construct D ABC, where ÐB = 70°, ÐC = 60°, AB + BC + AC = 11.2 cm., The perimeter of a triangle is 14 , 4 cm and the ratio of the lengths of the sides is 2 : 3 : 4., Construct the triangle., , 4., , Construct D PQR, where PQ - PR = 2.4 cm, QR = 6.4 cm and, , ÐPQR = 55°. , , ICT tools or links, Make constructions of the above types with Geogebra software and enjoy the constructions. The third type of construction given above is represented in Geogebra by a different method. Also study this method., , qqq, , 56

Page 67:

5, , squares, , let's learn., , • parallelogram, • parallelogram tests, • rhombus, , • rectangles, • square, • trapezoid, , • center point theorem, , let's remember., B, , 1., , write the following pairs considering C, , A, , pairs of adjacent sides:, (1) ... , ..., (2) ... , ..., (3) ... , ... , ( 4) ... , ..., , �ABCD, , pairs of adjacent angles :, (1) ... , ..., (2) ... , ..., (3) .. . , ..., (4) ... , ..., , pairs of opposite sides (1) ..... , ..... (2) ..... , ..... , pairs of opposite angles (1 ) ..... , ..... (2) ..... , ....., , D, , Fig. 5.1, , Let's recall the types of quadrilaterals and their properties ., I am a quadrilateral , , My two opposite pairs of sides are parallel., , All my angles are right angles., , My sides are the same length., , My properties, , My properties, , My properties, , Opposite sides coincide Opposite sides ..... , Opposite angles ......, Diagonals ....., Diagonals ....., , Opposite sides .. .. ., Diagonals ....., , My only pair of opposite sides are parallel., , 57, , All my angles are equal and all sides are equal., , My properties, , diagonals .... .

Page 68:

You know different types of quadrilaterals and their properties. So you learned, through various activities like measuring sides and angles, through paper folding method, etc., now let's examine these properties by giving their logical proofs., A logically proved property is called a proof. You will learn how a rectangle, a rhombus and a square are parallelograms. Let's start our study with the parallelogram., , , Let's learn., , parallelogram, A quadrilateral where both pairs of opposite sides are parallel is called a parallelogram., To prove theorems or solve problems, we need the figure a drawing, parallelogram often. Let's see how to draw a parallelogram. Suppose we need to draw a parallelogram ABCD. Method I :, � Let's draw segments AB and segments BC of any length and with any angle A of any other measure., � Now we want seg AD and seg BC, parallel to each other. Therefore, draw a parallel line B to separate BC through point A, Fig. 5.2, � In the same way, we will draw a line parallel to AB through point C. These lines will intersect at point D. The quadrilateral ABCD constructed in this way will be a parallelogram., Method II:, � Let's draw the section AB and the section BC of any length and any angle between them., • Draw a circle arc with center, A and radius BC., • Draw In the same way, draw an arc with center C and radius AB that intersects the previously drawn arc. • Name the intersection of two arcs D. Draw section AD and section CD. The quadrilateral formed in this way is a parallelogram ABCD. , D, , C, , A, , B, , 58, , D, , Fig. 5.3, , C

Page 69:

In the second method, we draw �ABCD where opposite sides are equal. Let's prove that a quadrilateral whose opposite sides are equal is a parallelogram. Task I Draw five parallelograms using different measurements for lengths and angles. For proofs on parallelograms we use congruent triangles. To understand how they are used, let's do the following activity., Activity II, , • Draw a parallelogram ABCD on a map sheet , ,. Draw the diagonal AC. Write the names of the vertices inside the triangle as shown in the figure. Then cut out., , B, B, , C C, , C, , I, , • Fold the square diagonally, AC and see if DADC and DCBA, , A, , match or not., , A A, , II , , D, D, , Fig. 5.4, D, , • Cut �ABCD along the diagonals AC and separate DADC and DCBA. Twist and rotate the DCBA to ensure it matches the DADC exactly., , B, B, , What do you think? Which DCBA pages correspond to which DADC pages? Which DCBD angles correspond to which DADC angles?, , A, , C, , Fig. 5.5, B, , C, A, , D, , Side DC corresponds to side AB and Side AD corresponds to side CB. Similarly, ∠ B fits ∠ D., , A, C, , So we can see that opposite sides and angles of a parallelogram are congruent., , Let's prove these properties of a parallelogram., , 59, , Fig. 5.6

Page 70:

Theorem 1. Opposite sides and opposite angles of a parallelogram are congruent., , Given, : �ABCD is a parallelogram., D, C, , Means side AB || DC side,,°´, AD side || side BC., ´°, to prove: sec AD ≅ sec BC; sec DC ≅ sec AB, A, B, , ∠ADC ≅ ∠CBA, and ∠DAB ≅ ∠BCD., Fig. 5.7, , construction: diagonal AC., , || draw the segment AB and the diagonal AC is a transversal., ∴ ∠DCA ≅ ∠BAC ................(1), and ∠DAC ≅ ∠BCA ..... .. .......(2), ..... alternate angle, now , in ∆ADC and ∆CBA,, , ∠DAC ≅ ∠BCA, .......... of ( 2) , ∠DCA ≅ ∠BAC, .......... of (1), , .......... common side, AC sec ≅ AC sec, ∴ ∆ADC ≅ ∆CBA , . 🇧🇷 ........ ASA test, ∴ AD side ≅ CB side, .......... c.s.c.t., and DC side ≅ AB side .......... c.s.c.t.,, . ......... c.a.c.t., So, ∠ADC ≅ ∠CBA, Similarly, we can prove ∠DAB ≅ ∠BCD., Proof: sec DC, , }, , Use your brain power!, In the theorem above , Does a ∠ Prove DAB ≅ ∠BCD require a design change? If so, how will you write the proof that makes the change? To learn about another property of a parallelogram, let's do the following activity., Activity: Draw a PQRS parallelogram. Loops, PR and QS diagonals. Denote the intersection of the diagonals with the letter O. Compare the two parts of each diagonal with a divider. What do you find ?, , 60, , P, , S, , X, , O, Q, , X, , Fig. 5.8, , R

Page 71:

Theorem : The diagonals of a parallelogram divide in half., P, , S, , X, , Q, , Given, , O, X, , Fig. 5.9, , R, , :, , �PQRS is a parallelogram. Diagonals PR, , and QS intersect at point O., Proof: sec PO ≅ sec RO,, sec SO ≅ sec QO., , Proof: At ∆POS and ∆ROQ, ∠OPS ≅ ∠ORQ ..... . .... alternative angle, , side PS ≅ side RQ ......... opposite sides of the parallelogram, ∠PSO ≅ ∠RQO .......... alternative angle, , ∴∆POS ≅ ∆ROQ ....... Test ASA, , ∴sec PO ≅ sec RO .............., , ..... corresponding sides of e seg SO ≅ sec QO . 🇧🇷 ........, Congruent triangles, , }, , Remember !, , •, •, •, •, , Adjacent angles of a parallelogram complement each other., Opposite sides of a parallelogram are congruent., Opposites angles of a Parallelograms are congruent., Diagonals of a parallelogram divide in half., , Worked examples, Ex. (1) PQRS is a parallelogram. PQ = 3.5, PS = 5.3, , ∠Q = 50° then determine the lengths of, , remaining sides and measures of remaining angles., , P, , 5.3, , S, , �PQRS is a parallelogram., \∠ Q + ∠ P = 180° ........ The interior angles are, 50°, supplementary., \ 50° + ∠ P = 180°, Q, R, Fig. 5.10, \ ∠ P = 180° - 50° = 130°, Now , ∠ P = ∠ R and ∠ Q = ∠ S ........opposite angle of a parallelogram., \ ∠ R = 130° and ∠ S = 50°, 3.5, , Solution :, , Similarly PS = QR and PQ = SR........opposite sides of a parallelogram., , , \ QR = 5.3 and SR = 3.5, , 61

Page 72:

Example (2), , �ABCD is a parallelogram. If Ð A = (4x +13)° and Ð D = (5x -22)° then find the dimensions of Ð B and Ð C., , Solution: Adjacent angles of a parallelogram are complementary., Ð A and Ð D are adjacent angles., D, C, 5x - 22, \ (4x +13)°+ (5x - 22)° = 180, \ 9x - 9 = 180, 4x +13, A, \ 9x = 189, Fig. 5.11 , \x = 21, \ Ð A = 4x +13 = 4 ´ 21 + 13 = 84+13 = 97° \ Ð C = 97°, Ð D = 5x - 22 = 5 ´ 21 - 22 = 105 - 22 = 83° \ Ð B = 83°, , B, , Exercise 5.1, 1. The diagonals of a parallelogram WXYZ intersect at point O. If ∠XYZ = 135°, what is the measure of ∠XWZ and ∠YZW ?, If l ( OY)= 5 cm then l(WY)= ?, 2nd, In a parallelogram ABCD, If ∠A = (3x + 12)°, ∠B = (2x - 32)° then find the value , of x and then find the dimensions of ∠C and ∠D., 3.Perimeter of a parallelogram is 150 cm. One of its sides is 25 cm longer than the other side. Find the lengths of all sides., 4. If the ratio of the measures of two adjacent angles of a parallelogram is 1 : 2, find the measures of all the angles of the parallelogram., 5*. The diagonals of a parallelogram intersect at point O. If AO = 5, BO = 12 and AB = 13, it turns out that �ABCD is a rhombus., 6. In Figure 5.12, �PQRS and �ABCR are two parallelograms . , If ∠P = 110°, then determine the measures of all angles of �ABCR., , Q, , P, , 110°, , S, , B, , A, , C, R, , Fig. 5.12, 7 In Fig. 5.13, �ABCD is a parallelogram. Point E is on radius AB such that BE = AB then prove that line ED intersects Seg BC at point F., , B, , A, F, D, , 62, , C, , Fig. 5.13, , And

Page 73:

Remember, Tests for parallel lines, 1., , If a transverse line intersects two lines and a pair of corresponding angles are congruent, those lines are parallel., , 2., , If a transverse line intersects two lines and a pair of alternate angles is then congruent, these two lines are parallel., , 3., , If a transversal line intersects two lines and a pair of interior angles are complementary, then these two lines are parallel., , Let's learn., Tests for parallelogram, Suppose that in ��PQRS are PS = QR and PQ = SR, S, P, and we have to prove that ��PQRS is a parallelogram. To prove this, which pairs of sides of ��PQRS should be shown in parallel?, R, Q, What test can we use to show the sides, Fig. 5.14, in parallel? Which line is suitable as a transversal to obtain the necessary angles to apply the test?, Theorem: If pairs of opposite sides of a quadrilateral are congruent, then this quadrilateral is a parallelogram., P, S, : In �PQRS , Given, Side PS @ Side QR, Side PQ @ Side SR, Proof: �PQRS is a parallelogram., Construction: Draw diagonal PR., Proof, : In D SPR and D QRP, Side PS @ Side QR ... .... . Dice, R, Q, Fig. 5.15, SR side @ QP side ........ given, PR side @ RP side ........ common side, \D SPR @ D QRP .. .... sss test, , \ ∠ SPR @ ∠ QRP ....... c.a.c.t., Likewise, ∠PRS @ ∠RPQ ..... c.a.c.t., ∠SPR and ∠QRP are alternative angles formed by the transversal PR of , , seg PS and seg QR ., , 63

Page 74:

\ PS page || QR side ......(I) Alternating angle test for parallel lines., Likewise, , ∠PRS and ∠RPQ are the alternating angles formed by the transversal PR of, , sec PQ and sec SR., , \ PQ side | 🇧🇷 Page SR ......(II) .....alternative angle test, \ of (I) and (II) �PQRS is a parallelogram., Page 56 provides two methods for drawing a parallelogram. In the second method, we draw a quadrilateral whose opposite sides are equal. Now do you understand why such a quadrilateral is a parallelogram?, Theorem : If both pairs of opposite angles of a quadrilateral are congruent, then it is a parallelogram., H, , G, , E, , Fig. 5.16, , Proof : Let , , F, , �EFGH ∠ E @ ∠ G, and ∠......... @ ∠........, To prove: �EFGH is a .. ... ..... ....., , given, , : In, , ∠ E = ∠ G = x and ∠ H = ∠ F = y, , the sum of all angles of a quadrilateral is .. .... ....... , , \ ∠ E +, , ∠ G + ∠ H + ∠ F = ........., , , \ x + y + ..... .... . + .......... = .........., , \ x +, , y = ....., , \ x + y = 180°, , \ ∠ G + ∠ H = .........., , ∠ G and ∠ H are interior angles formed by the transversal HG of Seg HE and Seg GF., , \ side HE, , || Side GF .......... (I) Internal angle test for parallel lines., Likewise ∠ G + ∠ F = .........., , \ side... .. . 🇧🇷 Side .......... .......... (II) Internal angle test for parallel lines., , \ Off (I) and (II) �EFGH is on ... . .. .............. ., , 64

Page 75:

If the diagonals of a quadrilateral divide, then it is a parallelogram., Diagonals of �ABCD divide at point E. This means seg AE @ seg CE, A, B and seg BE @ seg DE, �ABCD is a parallelogram ., E, Find the answers to the following questions, C, D, and write the proof yourself., Fig. 5.17, 1. Which pair of alternating angles should be shown, congruent with the proof of sec AB || sec dc?, Which transversal forms a pair of alternating angles?, 2. Which triangles contain the alternating angles formed by the transversal?, 3. Which test allows us to say that the two triangles are congruent?, 4. Similarly, you can prove that seg AD || sec BC?, The above three theorems are useful to prove that a given quadrilateral is a parallelogram. Therefore, they are called tests of a parallelogram., Another theorem useful as a test for a parallelogram is given below., , theorem :, given, :, , , to prove :, proof, :, , theorem : a A quadrilateral is a parallelogram if any pair of its opposite sides are parallel and, C, D, congruent Given: In �ABCD, seg CB @ seg DA and seg CB || sec DA, To prove: �ABCD is a parallelogram., B, A, Construction: Draw the diagonal BD., Fig. 5.18, Write full proof which will be provided shortly., D CBD @ D ADB ...... . Test SAS, , \ ∠CDB @ ∠ABD ..... c.a.c.t., , \seg CD || sec BA ..... Test of alternate angles for parallel lines, , , Remember !, •, •, •, •, , A quadrilateral is a parallelogram if its pairs of opposite angles are congruent., A quadrilateral is a parallelogram if its opposite pairs of sides are congruent., A quadrilateral is a parallelogram if its diagonals divide in half., A quadrilateral is a parallelogram if one pair of its opposite sides are parallel and congruent., These theorems are called tests of parallelogram., , Remember: The lines in a notebook are parallel. How can we draw a parallelogram with these lines?, , 65

Page 76:

Worked Examples �PQRS is a parallelogram. M is the midpoint of side PQ and N is the midpoint, eg. (1), on the RS side. Prove that �PMNS and �MQRN are parallelograms., P, S, Given, : � PQRS is a parallelogram., M and N are the midpoints of side PQ and side RS, respectively., M, N, Proof: � PMNS is a parallelogram., �MQRN is a parallelogram., , R, Q, Fig. 5.19, proof, : side PQ || SR Page, \ PM Page || SN side ...... ( P-M-Q; S-N-R) ......(I), , , PQ side @ SR side., 1, 2, , \ PQ side =, , , 1, SR side, 2, , \ PM side @ SN side ..... ( M and N are midpoints.)......(II), , \ of (I) and (II), �PMNQ is a parallelogram , , , We can also prove that �MQRN is a parallelogram., Example (2) Points D and E are the midpoints of side AB and side AC of D ABC, respectively. Point F is on radius ED, so ED = DF. Prove that �AFBE is a parallelogram., For this example, write 'given' and 'to be proved' and complete the proof below., Given: --------------- --- To prove: ---------------------A, , Proof: Mon AB and Mon EF are, Mon AD, Mon, , of, , �AFBE ., , @seg DB......., @seg, , , \ diagonals of, , F, , D, , E, , .......construction., , �AFBE, , each other, , , \ �AFBE is a parallelogram ...through, , Test., , B, , C, , Fig. 5.20, , Ex (3) Prove that every rhombus is a parallelogram., Given, , :, , Proof :, Prove , , B, , �ABCD is a rhombus., , C, , �ABCD is a parallelogram., , : sec AB @ sec BC @ sec CD @ sec DA (given), , \side AB @ side CD and side BC @ side AD, , , A, , D, , Fig. 5.21, \ �ABCD is a parallelogram..... opposite side test for parallelogram, , , 66

Page 77:

Exercise Theorem 5.2, A, , 1. In Figure 5.22 �ABCD is a parallelogram, P and Q are midpoints of sides AB and DC, respectively, so prove that �APCQ, is a parallelogram., , D, , P, , Q , , B, , C, , Fig. 5.22, , 2. Prove that every rectangle is a parallelogram using the opposite angles test for parallelograms., D, , 3. In Figure 5.23, G is the point of coincidence of the medians by D DEF. Take the point H on the radius DG such that D-G-H and DG = GH, then prove that �GEHF is a parallelogram., , G, F, , E, H, , A, , 4. Prove that the quadrilateral is formed by the intersection of the bisectors of all angles of a parallelogram is a rectangle., (Figure 5.24), , Fig. 5.23, B, , P, , Q, , S, R, , C, , D, , Fig. 5.24, 5. If in Figure 5.25 the points P, Q, R, S are on the sides of the parallelogram so that AP = BQ = CR = DS, then prove that ��PQRS is a parallelogram., , A, , P, , B , Q, , S, D, , R, , C, , Fig. 5.25, , Let's learn., Properties of rectangle, rhombus and square, rectangle, rhombus and square are also parallelograms. Therefore, the properties that opposite sides are equal, opposite angles are equal, and diagonals divide also apply to this type of quadrilateral. But there are some other properties of these quadrilaterals. Proofs of these properties will be given shortly. Considering the steps in the given proofs, write the proofs in detail., , 67

Page 78:

Theorem: The diagonals of a rectangle are congruent., , �ABCD is a rectangle., Proof: Diagonal AC @ diagonal BD, , Given, , A, , D, , :, , : Complete the proof by giving adequate reasons., B , D ADC @ D DAB ...... SAS test, \ diagonal AC @ diagonal BD..... c.s.c.t., , , proof, , Fig. 5.26, , C, , theorem : the diagonals of a square are congruent . , Write 'Given', 'To be proved' and 'Proof' of the theorem., Theorem : The diagonals of a rhombus are perpendicular bisectors., Given, : EFGH is a rhombus, To be proved: (i) Diagonal EG is the perpendicular , diagonal bisector HF., (ii) diagonal HF is the perpendicular, diagonal bisector EG., Proof, , @ seg EH, seg GF @ seg GH, , : (i) seg EF, , } data , , E, , F, , H, , Fig. 5.27, G, , Every point that is equidistant from the endpoints of a segment lies on the perpendicular bisector of the segment., \ Point E and point G lie on the perpendicular bisector of sec HF. , , One and only one line passes through two different points., \ Line EG is the bisector of the diagonal HF., , \ Diagonal EG is the perpendicular of the diagonal HF., , (ii) We can also prove that the diagonal HF is the perpendicular bisector , , of EG. It's. Write the proofs of the following statements ., � Diagonals of a square are perpendicular bisectors., � Diagonals of a rhombus bisect its opposite angles., � Diagonals of a square bisect its opposite angles., , Remember!, � Diagonals of a rectangle are congruent. , � The diagonals of a square are congruent., � The diagonals of a rhombus are perpendicular, bisectors each other., , 68, , � The diagonals of a rhombus divide pairs of opposite angles., � The diagonals of a square are perpendicular , bisectors of each other., � Diagonals of an opposite square bisector , Angle.

Page 79:

Exercise Theorem 5.3, 1. The diagonals of a rectangle ABCD intersect at point O. If AC = 8 cm, calculate the length of BO, and if ∠CAD = 35°, calculate the measure of ∠ACB., 2. rhombus PQRS if PQ = 7.5 then find the length of QR. if ∠QPS = 75° then find the measure of ∠PQR and ∠SRQ., 3. Find the diagonals of a square that intersects IJKL at point M the dimensions of ∠IMJ, ∠JIK, and ∠LJK ., 4. The diagonals of a rhombus are 20 cm and 21 cm, respectively, then determine the side of the rhombus and its perimeter., 5. Justify whether the following statements are "true" or "false" . , (i) Every parallelogram is a rhombus., (ii) Every rhombus is a rectangle., (iii) Every rectangle is a parallelogram., (iv) Every square is a rectangle., (v) Every square is a rhombus. , (vi) Every parallelogram is a rectangle., , Let's learn., trapezoid, If only one pair of opposite sides of a quadrilateral are parallel, then the quadrilateral is called a trapezoid., B, , A, , In the adjacent Shown only the side AB and the DC side of � ABCD are parallel to each other. So this is a trapeze. ÐA and ÐD, is a pair of adjacent angles, just like the pair of ÐB and ÐC. Therefore, both pairs of parallel lines are complementary. , , D, , C, , Fig. 5.28, , P, , When the non-parallel sides of a trapezoid are congruent, this quadrilateral is called an isocele trapezoid., S, , Q, , Fig. 5.29, , R, , The segment , which connects the midpoints of the non-parallel sides of a trapezoid, is called the bisector of the trapezoid., , 69

Page 82:

PQ =, , 1, PR ...... (construction), 2, , \ PQ =, , , 1, BC, 2, , a PR = BC, , inverse center theorem, theorem : When a line is drawn passes through the midpoint of one side of a triangle and is parallel to the other side, so it bisects the third side. , This theorem has “Given”, “To be Proved”, “Construction” given below. Try writing the test. Given, , : The point D is the midpoint of the side AB of D ABC. Line l, which passes through point D and is parallel to side BC, intersects side AC at point E., A, , Prove: AE = EC, which passes through point C. Name the point, , B, , Fig. 5.35, , from the intersection where this line e, , F, , E, , D, , Construction : Draw a line parallel to the seg AB, , l, , C, , line l will intersect as F., Proof, , : Use the construction and line l || sec BC that is given. Prove, , , D ADE, , @ D CFE and complete the proof. , , Ex (1) Points E and F are the midpoints of section AB and section AC, respectively, of, , D ABC. If, , EF = 5.6, then find the length of BC., Solution: On, , D ABC, points E and F are midpoints of, , A, , side AB and side AC, respectively., EF = , , 1 , BC . ......Midpoint Theorem, 2, , 5.6 =, , 1, BC, 2, , E, B, , \ BC = 5.6 ´ 2 = 11.2, , Ex (2) Prove that the quadrilateral passing through Connecting the formed , , F, C, , Fig. 5.36, , midpoints of the sides of a quadrilateral in order is a parallelogram., Given, : �ABCD is a quadrilateral., P, Q, R, S are midpoints of the sides , AB , BC, CD or AD., Proof : �PQRS is a parallelogram., Construction: draw the diagonal BD, , A, S, D, , 72, , P, , B, , Q, R, ,. 5.37,C

Page 83:

Proof, , : In , , D ABD is the midpoint of the side AD S and the midpoint of the side AB is P. , , \ by the midpoint theorem PS, , , || DB and PS =, , 1, BD ........... (1), 2, , In D DBC the points Q and R are the midpoints of the side BC and the side DC respectively., , \ QR || BD and QR =, , \PS, , \, , , 1, BD ...........after the midpoint theorem (2), 2, , || QR and PS = QR ................ of (1) and (2), , �PQRS is a parallelogram., Exercise Theorem 5.5, A, , 1. In Figure 5.38 are the points X, Y, Z are the midpoints of side AB, side BC and side AC of D ABC, respectively. AB = 5cm, AC = 9cm and BC = 11cm. Find the length of XY, YZ, XZ., , Z, , X, Y, , B, , Figure 5.38, L, , S, , 2. In Figure 5.39, �PQRS and �MNRL are rectangles. If point M is the midpoint of side PR, 1, then prove that (i) SL = LR, (ii) LN = 2 SQ., , C, R, N, , M, P, , Q, , Fig 5.39, A, , 3. In Figure 5.40, DABC is an equilateral triangle. Points F, D, and E are the midpoints of side AB, side BC, and side AC, respectively. Show that D FED is an equilateral triangle. Produces, , PM, 1, QT intercepts PR in M. Show that PR = 3 ., , [Hint: Draw DN, , || QM.], , E, , Fig. 5.40, M, T, , Q, , C, , D, , D, , N, R, , Fig. 5.41, , Problem Set 5, 1., , Check the correct option for an alternative answer and fill in the blanks., , (i) If all pairs of adjacent sides of a quadrilateral are congruent, then says.. ., (A) rectangle (B) parallelogram (C) trapezium, (D ) rhombus , , 73

Page 84:

(ii) If the diagonal of a square is 12.2 cm, then the perimeter of the square is......, (A) 24 cm (B) 24.2 cm (C) 48 cm (D) 48, 2 cm, ( iii ) If the opposite angles of a rhombus are (2x)° and (3x - 40)°, then the value of x..., (A) 100, , ° (B) 80 ° (C) 160 ° (D) 40 °, , 2. The adjacent sides of a rectangle measure 7 cm and are 24 cm long. Find the length of its diagonals., 3. If the diagonal of a square is 13 cm, then find its side., 4. The ratio of two adjacent sides of a parallelogram is 3: 4, and its perimeter is 112 cm. Determine the length 5. The diagonals PR and QS of a rhombus PQRS are 20 cm and 48 cm, respectively. Find the length of the side PQ., , ÐQMR = 50° then, , 6. Diagonals of a rectangle PQRS intersect at point M. If, find the measure of ÐMPS., 7., , In Figure 5.42 below, if , , B, , then prove that,, sec BC, , Q, , || seg QR and seg BC @ seg QR., , C, , 8*. In Figure 5.43 �ABCD is a trapezoid., AB, , of segment AD and segment BC., 1, PQ = (AB + DC)., 2, , 9., , In adjacent Figure 5.44, trapezoid . AB, , �ABCD is a, , or prove that MN, , || AB., 74, , Q, , Fig. 5.43, , M, A, , C, , C, , D, , || direct current. The points M and N are, , midpoints of the diagonals AC and DB, , B, , P, D, , || AB e, , R, , Fig. 5.42, , A, , || direct current. Points P and Q are midpoints, , So prove that PQ, , P, , A, , || Mon PQ , Mon AB @ Mon PQ,, Mon AC || Mon PR, Mon AC @ Mon PR, Mon AB, , N, , Fig. 5.44, , B

Page 85:

Activity, To check the different properties of quadrilaterals, Material: A piece of plywood measuring about 15 cm × 10 cm, 15 thin screws, string, scissors., Note: On the plywood board, tighten five screws at a, •, • , •, •, •, horizontal row with a gap of 2 cm between two adjacent screws. In the same way, make two more, •, •, •, •, • rows of screws just below the first one. For the screw thread, make sure the vertical distance between two adjacent screws is also 2 cm., Fig. 5.45, . Use the screws to make different types of wire squares. Check the properties of the sides and angles of quadrilaterals., Additional information, You know the property that the intersection of the medians of a triangle divides the medians in the ratio of 2:1. The proof of this property is given below., A , , Given : seg AD and seg BE are the medians of D ABC that intersect at the point G such that, G-D-F and GD = DF, , G, B, , D, , proof : diagonals of �BGCF bisect, .... given and construction, F, \ �BGCF is a parallelogram., Fig. 5.46, \ BE segment || define FC, now point E is the midpoint of side AC of D AFC . .......... data, mon EB || sec FC, , , C, , line passing through the midpoint of one side and parallel to the other side bisects the third side., , \ point G is the midpoint of side AF., \ AG = GF, but GF = 2GD . ...... Construction, \ AG = 2 GD, \, , AG 2, GD = 1 ie AG : GD = 2 : 1, 75, , qqq

Page 86:

6, , circle, let's study., , • incircle, • circle, , • circle, • chord property of the circle, , let's remember., , C, , B, P, , Note the circle in the figure below with center P . With reference to this figure, complete the following table., , A, , --- segment PA, ------ chord, --diameter radius center, , D, Fig. 6.1, , -- central , angle, , ÐCPA, ---, , Let's learn., , circle, Let's describe this circle by a set of points., l The set of points in a plane, defined by a fixed point in the same distant plane is called a circle., Some terms related to a circle., l The fixed point is called the center of the circle., l The line segment connecting the center of the circle and a point on the circle is called the radius of that circle., l The distance of a point on the circle from the center of the circle is also called the radius of the circle. , l The line segment joining any two points on the circle is called the chord of the circle., l A chord passing through the center of a circle is called the diameter of the circle., A diameter is the largest chord of the circle. , Circles in a plane, congruent, concentric, intersecting circles, intersecting circles, circles, circles, at one point, at two points, , equal , , radii, , same center, different centers,, , different radii, Fig. 6.2, , different radii, only one point in common, , 76, , · different centers,, , different radii, two points in common

Page 87:

Let's Learn., , Properties of Chords, Activity I: Each student in the group will do this activity. Draw a circle in your notebook. Draw a rope from this circle. Draw perpendicular to the chord through the center of the circle. Measure the lengths of the two parts of the chord. The group leader prepares a table and the other students write their observations on it., Length, , Student, , l (AP), l (PB), , 1, , 2, , 3, , O, A, , P, , B, , Fig. 6.3, 4, , 5, , 6, , ...... cm, ...... cm, , Write the observed property., Let's write the proof of this property . , Theorem: A perpendicular drawn from the center of a circle to its chord divides the chord in half., Given, : seg AB is a chord of a circle with center O., seg OP ^ chord AB, Proof: sec AP @ sec BP , O , : Draw mon OA and mon OB, proof, in DOPA and D OPB, B, A, P, ÐOPA @ ÐOPB . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 seg OP ^ chord AB, seg OP @ seg OP . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 common side, Fig. 6.4, hypotenuse OA @ hypotenuse OB . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 Radii of the same circle, \D OPA @ D OPB . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 Set on the side of the hypotenuse, sec PA @ sec PB . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 c.s.c.t., Activity II: Each student in the group does this activity. Draw a circle in your notebook. Draw a rope from the circle. Connect the center of the rope and the center of the circle. Measure the angles the segment makes with the string. Discuss angular dimensions with your friends. Which property do the observations suggest?, , 77, , Fig. 6.5

Page 88:

Theorem: The segment connecting the center of a circle and the center of its chord is perpendicular to the chord. Given: seg AB is a chord of a circle centered at O and P is the center of the chord AB of the circle. That means seg AP @ seg PB., to prove: seg OP ^ chord AB, prove, : draw seg OA and seg OB., O, In D AOP and D BOP, seg OA @ seg OB . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 Radii of the same circle, B, A, P, sec OP @ sec OP. 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 common sides, Fig. 6.6, Mon AP @ Mon BP. 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 given, \D AOP @ D BOP . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 Test SSS, , \ÐOPA @ ÐOPB . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 c.a.c.t. 🇧🇷 🇧🇷 .(I), , , ÐOPA + ÐOPB = 180° . 🇧🇷 🇧🇷 Angles in the linear pair, \ÐOPB + ÐOPB = 180° . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 of (I), , \ 2 ÐOPB = 180°, , , \ ÐOPB = 90°, , \sec OP ^ Rope AB, Now, , Worked examples, Ex. (1) The radius of a circle is 5 cm. The length of a rope is 8 cm. Find the distance, the chord from the center., Let's draw a figure from the given information., Solution: , , O is the center of the circle., O, , the length of the chord is 8 cm., sec, OM ^ Rope PQ., P, Q, M, , Fig. 6.7, We know that a perpendicular drawn from the center of a circle to its chord cuts the chord in half, \ PM = MQ = 4 cm, circle radius is 5 cm, means OQ = 5 cm.... given, At right angle D OMQ with the Pythagorean theorem, OM2 + MQ2 = OQ2, \OM2 + 42 = 52, \OM2 = 52 - 42 = 25 - 16 = 9 = 32, \OM = 3 , So the distance of the chord from the center of the circle is 3 cm., , 78

Page 89:

Example (2) The radius of a circle is 20 cm. The distance of a chord from the center of the circle is 12 cm., Find the length of the chord., Solution: Let the center of the circle be O. Radius = OD = 20 cm., Distance of chord CD from O is 12 cm. seg OP ^ sec CD, , \ OP = 12 cm, Now CP = PD ...... traced perpendicularly from the, , center bisects the chord, At right angles D OPD, using the Pythagorean theorem, OP2 + PD2 = OD2 , (12)2 + PD2 = 202, , PD2 = 202 - 122, , PD2 = (20+12) (20-12), , O, C, , P, , D, , Fig. 6.8, , = 32 ´ 8 = 256, , , , \ PD = 16, , \ CP = 16, , CD = CP + PD = 16 + 16 = 32, \ the length of the rope is 32 cm., , Set of exercises 6.1, 1st, 2nd . , 3rd, 4th, 5th, , 6th, , The distance of chord AB from the center of the circle is 8 cm. The length of rope AB is 12 cm. Find the diameter of the circle. The diameter of a circle is 26 cm and the length of a chord of the circle is 24 cm. Find the distance of the string from the center. The radius of a circle is 34 cm and the distance from the rope to the center is 30 cm. Find the length of the string. The radius of a circle centered at O is 41 units. The length of a chord PQ is 80 units, find the distance of the chord from the center of the circle. In Figure 6.9, the center of two circles is O. Chord, O, AB of the larger circle intersects the smaller circle, at points P and Q. Show that AP = BQ, A P, Q B, Prove that if a diameter of a circle bisects two chords of the circle, these two chords, Fig. 6.9, are parallel to each other., , Exercise I, (1), (2), (3), (4), , Draw circles with suitable radii., Draw two equal chords in each circle., Draw perpendicular to each rope from the center., Measure the distance of each rope from the center., , 79

Page 90:

Let's learn., Relationship between congruent chords of a circle and their distances from the center, Exercise II : Measure the lengths of perpendiculars in chords in the following figures., N, , L, , A, , P, , O, , M, B , , M, , figure (i), , T, , figure (ii), , figure (iii), , found OL = OM in fig (i), PN = PT in fig (ii) and MA = MB in fig (iii) ?, Write the property you observed in this activity., , Let's learn., Properties of congruent chords, Theorem : Congruent chords of a circle are equidistant from the center of the circle., : In a circle with center O, Given, A, , AB chord @ CD chord, OP ^ AB, OQ ^ CD, Prove : OP = OQ, Construction : Join sec OA and sec OD., , P, , B, , O, , C, Q, , Fig. 6.10, D, 1, 1, : AP = AB, DQ = CD . 🇧🇷 🇧🇷 Drawn perpendicularly from the center, 2, 2, , Proof, , , of a circle for its chord bisects the chord., , AB = CD . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 given, \AP = DQ, \sec AP @ sec DQ . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 (I) . 🇧🇷 🇧🇷 Segments of equal length, , At right angle D APO and right angle D DQO, sec AP @ sec DQ . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 of (I), hypotenuse OA @ hypotenuse OD . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 Radii of the same circle, \D APO @ D DQO . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 Set on the side of the hypotenuse, , sec OP @ sec OQ . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 c.s.c.t., \OP = OQ . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 Length of congruent segments., Congruent chords in a circle are equidistant from the center of the circle., , 80

Page 91:

Theorem : The chords of a circle equidistant from the center of a circle are congruent., Given, : On a circle centered at O, P, B, A, seg OP ^ chord AB, sec OQ ^ chord CD, C, O, e OP = OQ, To prove: Rope AB @ Rope CD, Q, Construction: Draw sec OA and sec OD., Fig. 6.11, D, Proof, : (Complete the proof by filling in the blanks.), At right angles DOPA and right D OQD, hypotenuse OA @ hypotenuse OD . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 ., Mon OP @ Mon OQ . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 data, , \D OPA @ D OQD . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 ., , , \sec AP @sec QD . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 c.s.c.t., , \ AP = QD . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 (I), , But AP =, , 1, 1, AB and DQ = CD . 🇧🇷 🇧🇷 🇧🇷 ., 2, 2, , and AP = QD . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 from (I), \AB = CD, , \seg AB @ sec CD, , Note that both theorems are mutually contradictory. , Chords equidistant from the center of a circle are congruent., Activity: The two theorems above can also be proved for two congruent circles., 1. Congruent chords in congruent circles are equidistant from their respective centers., 2. Congruent chords Equidistant circles of their respective centers are congruent., Write 'Given', 'To be proved' and the proofs of these theorems., Worked example, Eg, In Figure 6.12, O is the center of the circle and , AB = CD. If OP = 4 cm, find the length of OQ., Solution: O is the center of the circle,, chord AB @ chord CD.....given, OP ^ AB, OQ ^ CD, , A, , B, , P, O, , C, , Q, , D, , Fig. 6.12, , 81

Page 92:

OP = 4 cm, means that the distance from AB to the center O is 4 cm., The congruent chords of a circle are equidistant from the center of the circle., , \ OQ = 4 cm, , Exercise 6.2, 1., , O o The radius of the circle is 10 cm. There are two tendons, each 16 cm long. How far are these chords from the center of the circle?, , 2nd, , In a circle with a radius of 13 cm, two equal chords are 5 cm apart from the center., Find the lengths of the chords. , , 3rd, , seg PM and seg PN are congruent chords of a circle centered on C. Show that the ray is PC, the bisector of ÐNPM., , Remember., In previous rules we verified the property that the bisectors of a triangle are simultaneously. We denote the point of correspondence with the letter I., Let's learn., Circle of a triangle, , A, , In fig. 6.13, the bisectors of all angles of a DABC meet at point I. Perpendiculars on three sides are identified from the point of correspondence taken from., , R, , P, , IP ^ AB,, , I, , B , , Q , , C, , IQ ^ BC,, , IR ^ AC, , We know that each bisector is equidistant from the sides of the angle it is removed. , , Fig. 6.13, Point I is on the bisector of ÐB ., , \ IP = IQ., Point I is on the bisector of ÐC \ IQ = IR, \ IP = IQ= IR, , That is, point I is equidistant from all the sides of DABC., \ If we draw a circle with center I and radius IP it touches the sides AB, AC,,, BC of DABC., This circle is called the incircle of the triangle ABC., , 82

Page 93:

Let's learn., Construct the inner circle of a triangle, Ex. Construct D PQR such that PQ = 6 cm, ÐQ = 35°,, , R, , QR = 5.5 cm. Draw a circle of D PQR., , 5.5, , I, , Draw an approximate figure and show all dimensions on it., (1) Construct D PQR from the given dimensions., , 65°, , P, , (2 ) Draw bisectors of any two angles of the triangle., , cm, Q, , M 6 cm, , Rough Fig. 6.14, , (3) Denote the intersection of the dividing lines, , R, , with I., , 5.5, , ( 4) Draw a Perpendicular IM of the point I, , cm, , I, , beside PQ. , (5) Draw a circle with center I and radius IM., , P, , M, , 65°, , 6 cm, , Q, , Fig. 6.15, , Remember !, The circle that includes all sides touched A triangle is called the triangle's inner circle, and the center of the circle is called the triangle's incenter. This correspondence point is indicated by the letter C., , Let's learn., , P, , In fig. 6.16, the perpendicular bisectors of , sides of D PQR intersect at point C. Therefore, C is the meeting point of , perpendicular bisectors, , , C, Q, , R, Fig. 6.16, , 83

Page 94:

Circumference, point C lies on the perpendicular bisectors of the sides of the triangle PQR. Team up with PC, QC and RC. We know that each point on the perpendicular bisector is equidistant from the endpoints of the segment. Point C lies on the perpendicular bisector of sec PQ. \ PC = QK . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 I, Point C lies on the perpendicular bisector of sec QR., , \ QC = RC . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 🇧🇷 II, , \PC=QC=RC . 🇧🇷 🇧🇷 🇧🇷 🇧🇷 From I and II, \ the circle with center C and radius PC passes through all corners of D PQR., This circle is called the circumcircle of the triangle, and the center of the circle is called the circumcircle of the triangle., , let's learn. , To draw the perimeter of a triangle, for example, construct D DEF such that DE = 4.2 cm, ÐD = 60°, ÐE = 70° and draw a perimeter from it. Draw an approximate figure. Write down the dimensions given., Approximate figure, , F, F, D, , 60°, , C, , 70°, , E, , Fig. 6.17, C, 70°, , 60°, D, , Construction steps : , (1) Draw D DEF of given measurements., (2) Draw bisectors of any two sides of the triangle., (3) Call the intersection of the bisectors C., (4) Join sec CF., (5) Circle with center point C and radius CF., , 4.2 cm, , E, , Fig. 6.18, , 84

Page 95:

Activity: Draw different triangles with different measurements and draw circles and circles on them. Complete the observation table and discuss., type of triangle, , equilateral triangle, , isosceles triangle, , scalene triangle, , position of center, , inner triangle, , inner triangle, , inner triangle, , position of perimeter, , inner triangle , , inside, outside, in triangle, , type of triangle, , acute triangle, , right triangle, , obtuse triangle, , location of center, location of circumcircle, , center of, hypotenuse, , remember !, , ·, , Incircle of a triangle touches all sides of, the triangle from the inside., , ·, , The center of an acute triangle is inside the triangle., , ·, , To construct the inner circle of a triangle, we need bisectors of any two angles of the triangle., , ·, , circumcircle of a right triangle, is the center of its hypotenuse., , ·, , circumcircle of an obtuse angle , triangle is in the äu Exterior of the triangle., , ·, , Circumcircle of a triangle passes through all the vertices of a triangle., , ·, , For to construct a circumcircle of a triangle, we need to draw perpendicular bisectors of any two sides of the triangle. , , •The center of each triangle is inside the triangle., , Task: Draw any equilateral triangle. Draw its incircle and circumcircle. What did you notice in this activity? (1) When drawing the inner circle and perimeter, do the bisection lines and the perpendicular coincide? (2) Do in and radius match? If so, what might be the reason for this? (3) Measure the incircle and circumcircle radii and note your ratio., , 85

Page 96:

Remember !, , The bisectors and perpendicular bisectors of an equilateral triangle coincide., The center and perimeter of an equilateral triangle coincide., The ratio of the surrounding radius to the inner radius of an equilateral triangle is 2:1, Exercise Theorem 6.3, 1 .Construct D ABC such that ÐB = 100°, BC = 6.4 cm, ÐC = 50° and construct your, incircle., 2. Construct D PQR such that ÐP = 70°, ÐR = 50°, QR = 7.3 cm. and build your perimeter. 3. Construct D XYZ so that XY = 6.7 cm, YZ = 5.8 cm, XZ = 6.9 cm. Construct its incircle., 4. In D LMN, LM = 7.2 cm, ÐM = 105°, MN = 6.4 cm, then draw D LMN and construct its circumcircle., 5. Construct D DEF such that DE = EF = 6 cm , ÐF = 45° and construct its circumference., , Problem Set 6, 1., , Choose the alternative correct answer and fill in the blanks., (i) The radius of a circle is 10 cm and the distance is a rope from the center is 6 cm. Therefore, the length of the string is ........., (A) 16 cm, (B) 8 cm, (C) 12 cm, (D) 32 cm, (ii) The point of coincidence of all the bisectors of a triangle are called ......, (A) centroid, (B) circumcenter, (C) center, (D) orthocenter, (iii) the circle that passes through all the vertices of a triangle is called ....., , (A) circumscribed circle (B) inner circle, (C) congruent circle (D) concentric circle, (iv) The length of a chord of a circle is 24 cm. If the distance of the chord from the center is 5 cm, then the radius of this circle is ...., , (A) 12 cm, (B) 13 cm, (C) 14 cm (D) 15 cm, ( v ) , , The largest chord in the circle of radius 2.9 cm is ....., (A) 3.5 cm, (B) 7 cm, (C) 10 cm, (D) 5.8 cm, , ( vi) The radius of a circle with center O is 4 cm. If l(OP) = 4.2 cm, let's say where will be the point P., , (A) in the center (B) inside the circle (C) outside the circle (D) in the circle, (vii) The lengths of chords lines that are on opposite sides of the center of a circle are 6 cm and 8 cm long. If the radius of the circle is 5 cm, then the distance between these chords is ....., (A) 2 cm, (B) 1 cm, (C) 8 cm, (D) 7 cm, , 86

Page 97:

2. Construct the incircle and circumcircle of an equilateral DSP with a side of 7.5 cm. Measure the radii of both circles and find the ratio of the radius of the circumcircle to the radius of the inner circle., 3., , Construct D NTS with NT = 5.7 cm, TS = 7.5 cm and, e radius of it . ÐNTS = 110° and draw a circle, Q, , 4th, , R, , In Figure 6.19, C is the center of the circle., seg QT is a diameter, CT = 13, CP = 5, find the chord length RS ., , S, , P, C, , Fig. 6.19, , T, 5., , , , In Fig. 6.20, P is the center of the circle., Chord AB and chord CD intersect, in diameter at point E. If ÐAEP @ ÐDEP, then prove that AB = CD, , B, , E, , C, , ° °, , D, , P, , A, Fig. 6.20, , C, 6 ., , In Figure 6.21, CD is a diameter of the circle centered at O. Diameter, CD is at point E perpendicular to chord AB. Show that D ABC is an isosceles triangle., , O, A, , E, D, , B, Fig. 6.21, , ICT tools or links, Draw various circles using Geogebra software. Check and experiment with chord properties. Draw the circumscribed circle and inner circle of multiple triangles., Experience how the inner circle and circumscribed circle change when a triangle is resized with the Move option., , qqq, 87

Page 98:

7, , Coordinate geometry, Let's learn., F, , • Axis, Origin, Quadrant, • Coordinates of a point in a plane., • Draw a point., , • Line parallel to the X axis ., • Line parallel to the Y axis., • Equation of a line., , Chintu and his friends were playing cricket on the ground in front of a tall building when a visitor arrived., Visitor: Hey Chintu, Dattabhau lives here, doesn't he?, Chintu: Yes, on the second floor. Do you see the window? This is his apartment. Visitor: But there are five windows on the second floor. It could be any one of them!, Chintu ः Your window is the third from the left, on the second floor., Chintu's description of the location of Dattabhaus' apartment is actually based on the most basic concept of coordinate geometry., That wasn't enough, it was enough to give the floor number to locate the house. Also, the serial number should be given from left or right. That is, two numbers had to be given in a specific order. Two ordinal numbers, i.e. the second from the bottom and the third from the left, had to be used., Let's learn., axis, origin, quadrant, We could indicate the location of Dattabhaus Haus using two ordinal numbers. Likewise, the location of a point can be fully described using its distances from two mutually perpendicular lines. To locate a point on a plane, a horizontal line of numbers is drawn on the plane. This number line is called the X axis., , 88

Page 99:

Rene Descartes (1596-1650), Rene Descartes, a 17th-century French mathematician, proposed the coordinate system to accurately describe the position of a point in a plane. It is called the Cartesian coordinate system. The word Cartesian is apparently derived from his name., He revolutionized the field of mathematics by establishing the relationship between algebra and geometry., The Cartesian coordinate system is the basis of analytic geometry. La Geometric was Descartes' first book on mathematics. In it, he used algebra to study geometry and proposed that a point in a plane can be represented by an ordered pair of real numbers. This ordered pair is the "cartesian coordinates" of a point. Coordinate geometry has been used in many fields such as physics, engineering, nautical science, seismology and art. It plays an important role in the development of technology in Geogebra. We clearly see the interrelationship between algebra and geometry in the Geogebra software; the name itself is a combination of the words "geometry" and "algebra". and perpendicular to the X axis is the Y axis. In general, the number O is represented by the same point on both number lines. This point is called the origin and is denoted by the letter O. On the X axis, positive numbers are shown to the right of O and negative numbers are shown to the left. Positive numbers are shown on the Y-axis above O and negative numbers below. The X and Y axes divide the plane into four parts, each called a quadrant. As shown in the figure, the quadrants are numbered counterclockwise., Points on the axes are not included in the quadrants., , 89, , Y-Axis, , 3, Second Quadrant, II, , First Quadrant , I , , , 2, 1, , -3, , - 2, , -1, , Third Quadrant, III, , 0, -1, -2, -3, , O, , 1, , 2, , 3 X-Axis, , Fourth quadrant, IV, , Fig.7.1

Page 100:

The coordinates of a point in a plane, Y, 4, Q (-3, 2), , X -4, ¢, , -3 -2 -1, , P (2, 3), , 2 R, 1, , S, , N, , 3, , The point P is represented in the plane defined by the X and Y axes. Its position can be determined by its distance from the two axes. To find these distances, we plot sec PM ^ X axis and sec PN ^ y axis. The coordinate of the M point on the x axis is 2 and the coordinate of the N point on the y axis is 3 . Therefore, the x coordinate of point P is 2 and the y coordinate of point P is 3 , 0 1 2, , 3, , 4 X, , -2, -3, -4, Y¢, , Fig.7.2, coordinate x first. According to this convention, the coordinate order of point P is defined as 2, 3. The position of point P is briefly described by the pair (2, 3). The order of the numbers in the pair (2, 3) is important. Such a pair of numbers is called an ordered pair. To describe the position of the Q point, we plot sec QS ^ X axis and sec QR ^ Y axis., The coordinate of the Q point on the X axis is -3 and the coordinate on the Y axis is 2. , Therefore, the coordinates of the Q point are (-3, 2)., Y, , Ex. Enter the coordinates of points E, F, G, T in the adjacent figure. Solution :, , · The coordinates of point E are (2, 1), · The coordinates of point F are ( -3, 3), · The coordinates of point G are (-4, -2)., · The coordinates from point T are (3, -1), , 4, F , , 3, 2 , E, , 1, -4, G , , -3, , -2, , -1, , -1, , X, 0 1, , -2, -3, -4, , Fig.7.3, , 90, , 2, , 3, T, , 4

Page 101:

Let's learn., Coordinates of points on the axes, Y, , (0,-3), , The x coordinate of point M is its distance from the y axis. The distance from the M point to the X axis is zero. Therefore, the y coordinate of M is 0. Therefore, the coordinates of the point M on the X axis are (3,0)., The y coordinate of the point N on the Y axis is 4 units from the X axis axis removed because N has a distance of 4. Its x-coordinate is 0 because its distance from the y-axis is zero. Therefore, the coordinates of the point, , (0,-4), , N on the Y axis are (0, 4)., , (0,4), , P, , N, , (0,3) , , Quadrant II, (-, +), , Quadrant I, (+,+), , (0.2), (0.1), , O, , (-4.0) (-3.0) (-2 ,0) (-1.0) (0.0) (1.0), , M, , (2.0) (3.0) (4.0), , X, , (0, -1) , , Quadrant III, (-,-), , (0,-2), , Quadrant IV, (+,-), , Fig.7.4, Now the 'O' origin is both on the X axis and on the Y axis. Therefore, its distance from the X-axis and Y-axis is zero. Therefore, the coordinates of O are (0, 0). Each point in a plane is assigned one and only one pair of coordinates (ordered pair). Remember, , The y coordinate of each point on the X axis is zero., The x coordinate of each point on the Y axis is zero., The coordinates of the origin are (0, 0)., Which quadrant or which axis are the points given below?, A(5,7), B(-6,4), C(4,-7), D(-8,-9), P(-3,0 ), Q( 0 ,8), Solution : The x coordinate of A (5, 7) is positive and its y coordinate is positive.., \ Point A is in the first quadrant., The x coordinate of B ( -6, 4) is negative and the y-coordinate is positive., \ Point B is in the second quadrant., ., The x-coordinate of C (4, -7) is positive and the y-coordinate is negative ., \ Point C is in the fourth quadrant. , The x-coordinate of D (-8, -7) is negative and the y-coordinate is negative., \ Point D is in the third quadrant., Eg., , 91

Page 102:

The y-coordinate of P(-3.0) is zero \ the point P is on the x-axis., The x-coordinate of Q(0.8) is zero \ the point Q is on the y-axis. , Activity, , As in Fig. 7.5, ask the girls to sit in rows to form the x-axis and the y-axis., , • Ask some boys to sit in the positions, •, , •, , marked by the colored dots in the four quadrants ., Now call the students in turn, using the first letter of each student's name. When his initial is called, the student stands up and gives his own coordinates. For example, Rajendra, (2, 2) and Kirti (-1, 0), while enjoying this field activity, students learn how to, , R, K, , Fig. 7.5, , indicate the location of a point in a plane ., , let's learn., To plot the points of given coordinates, suppose we need the points, , Y, , P(4,3) and Q(-2,2 ), 4, , steps to plot the points, ( i), , Draw the X axis and Y axis in the plane., Show the origin., , Q (-2,2), , (ii) To find the point P (4.3), draw a line parallel to the Y-axis through the point on the X-axis that represents the number 4. Through the point on the Y-axis that represents the number 3, draw a line parallel to the X-axis ., , P (4 ,3), , 3, , ·, , 2, 1, , -2, , -1, , -1, , 0, , -2, -3, , Fig. 7.6, , 92, , 1, , 2, , 3, , 4, , X

Page 103:

(iii), , The intersection of these two lines parallel to the Y and X axes is the point P (4,3). What quadrant is this point in? , , (iv), , In the same way, draw the point Q (-2, 2) . Is this point in the second quadrant ?, Using the same method, plot the points, , Ex, , R(-3, -4), S(3, -1), , In which quadrant or on which axis are the ones below points ?, (i) (5, 3), , (ii) (-2, 4), , (iii) (2, -5), , (iv) (0, 4), , (v ) ( - 3, 0), , (vi) (-2, 2.5), , (vii) (5, 3.5), , (viii) (-3.5, 1.5), , (x) ( 2, -4 ), , (ix) (0, -4), solution :, , quadrant / axis, , (i), , coordinates, (5,3), , quadrant I, , (ii) , , ( -2,4), , (iii), , coordinates, , quadrant/axis, , (vi), , (-2, -2,5), , quadrant III, , quadrant II, , (vii), , ( 5,3,5), , Quadrant I, , (2,-5), , Quadrant IV, , (viii), , (-3.5,1.5), , Quadrant II, , (iv), , (0.4 ) , , Y axis, , (ix), , (0, -4), , Y axis, , (v), , (-3,0), , X axis, , (x), , (2 ,- 4 ), , Quadrant IV, , Exercise Theorem 7.1 , 1., , Indicate in which quadrant or on which axis are the following points., , A(-3, 2),, E( 37, 35),, M( 12, 0), 2nd, , B(-5, -2), K(3.5, 1.5), D(2, 10),, · F(15,-18), G(3, -7),, H(0,-5),, N(0,9),, P(0,2,5),, Q (-7, -3), , In which quadrant are the following points?, (i), , whose two coordinates are positive., , (ii) whose two coordinates are negative., (iii) whose x-coordinates are positive and y-coordinates are negative. , (iv) whose x-coordinate is negative and the y-coordinate is positive., 3., , Draw the coordinate system in a plane and draw the following points., L(-2, 4), , , M(5, 6),, , N(-3, -4),, , P(2, -3),, , 93, , Q(6, -5),, , S(7, 0 ), T(0, -5)

Page 104:

Let's learn., Lines parallel to the X axis, , On a graph paper, draw the following points, A (5, 4), B (2, 4), C (-2, 4), D (-4 , 4) , E (0, 4), F (3, 4), , · Note the coordinates of the, , Y, , given points., , · Did you notice that the y coordinates of all the points are the same ?, , D, , 5, , C, , 4, , A, , ·, , 3, , · All points are collinear., · Which axis is this line parallel to?, · The y-coordinate of each point on line DA is 4. It's constant. So the line DA, described by the equation y = 4. If the y-coordinate of any point is 4, it lies on the line DA., , F, , B, , E, , 2, 1, -4 , , -3 , , -2, , 0, , -1, -1, , 1, , 2, , 3, , 4, , 5, , X, , -2, -3, -4, , The equation of parallel straight lines to the X axis at a distance of 4 units, , Fig. 7.7, , of the X-axis is y = 4., , Let's discuss., , · can we draw a straight line parallel to the X-axis at a distance of 6 units from and below the x-axis ?, , 1, , · All points (-3, -6), (10, -6), ( 2 , -6 ) will be on this line ?, · What would be the equation of this line ?, , remember !, if b > 0, and we draw a line y = b passing through the point (0, b), it will pass through the X axis and be parallel to it . If b < 0, then the line y = b is below and parallel to the x-axis., The equation of a line parallel to the x-axis is of the form y = b., , 94

Page 105:

Let's learn., Lines parallel to the Y axis, , On a piece of graph paper, write the following points., P(-4, 3), Q(-4, 0),, , R(-4, 1 ) , , , S(-4, -2), T(-4, 2),, , U(-4, -3), , · Note the coordinates of the points., · You noticed that the x coordinate of , ·, ·, ·, , are all points equal ?, are all points collinear ?, which axis is this line parallel to ?, the x-coordinate of each point on the PS line is - 4. It is constant. Therefore, the PS line can be described by the equation x = - 4. Any point whose x-coordinate is -4 is on the PS line. The equation of the line parallel to the Y axis at a distance of 4 units and to the, , Y, P, , 3, , T, , 2, , R, Q, , 1, -4, , -3, , -2 , , -1, -1, , S, , -2, , U, , -3, -4, , to the left of the Y axis is x = - 4., , 0, , 1, , 2 , , 3 , , X, , Fig. 7.8, , Let's discuss ., , Can we draw a line parallel to the Y-axis 2 units from it and to the right of it?, 1, If all points (2, 10), (2,8), ( 2, - 2 ) are on this line?, What would be the equation of this line?, Remember!, If we draw the line x = the one parallel to the Y Axis, o passes through the point (a, 0) and , if a > 0, then the line is to the right of the y-axis. If a < 0, then the line is to the left of the y-axis., The equation of a line parallel to the y-axis is of the form x = a., , 95

Page 106:

Remember!, (1) The y-coordinate of any point on the x-axis is zero. On the other hand, any point whose y-coordinate is zero is on the x-axis. Therefore, the x-axis equation is y = 0. (2) The x-coordinate of every point on the y-axis is zero. On the other hand, any point whose x-coordinate is zero is on the y-axis. So the Y-axis equation is x = 0., , Let's learn., Graph a linear equation, , Y, , Ex, , 3, 2, , x=2, , Graph the equations , x = 2 and y = -3. Solution: (i) Draw the x-axis and y-axis on graph paper. (ii) Since x = 2, draw a line on , to the right of the Y-axis at a distance of 2 units from it and parallel to it. (iii) Given y = -3, draw a line below the X-axis at a distance of 3 units from it and parallel to it., (iv) These lines, parallel to the two axes, are the graphs of the given equations. (v) Write the coordinates of the point P, the intersection of these two lines., (vi) Check that the coordinates of the point P, (2, -3)., , 1, -3, , -2, , 0, , -1, , 2, , 1 are , , 3, , -1, -2, , y = -3, , P, , -3, , Fig. 7.9, , The graph of a linear equation in general Form., Activity : On graph paper, draw the, points (0,1) (1,3) (2,5). Are they collinear? If so, draw the line through them. · Which quadrants does this line pass through? · Write the coordinates of the point where it intersects the Y axis in the third quadrant that lies on this line. Write the coordinates of the point., , 96, , Y, 5, , (2, 5), , 4, 3, , (1, 3), , 2, 1 (0, 1), -3 -2 - 1 , , -1, , 0 1, , 2, , 3, , X, , -2, -3, -4, -5, , Fig. 7.10, , X

Page 107:

Ex. 2x - y + 1 = 0 is a linear equation in two variables in general form. Let's draw the graph of this equation. Solution: 2x - y + 1 = 0 means y = 2x + 1. Let's take some x values and find the corresponding y values. Example: If x = 0 , then we plug that value of x into the equation and get y = 1. Likewise, let's find the values of y if 0, 1, 2, 1 , -2 are some values of x are, and 2, , write these values in the following table as ordered pairs., x, , 0, , 1, , 2, , 1, 2, , -2, , y, , 1, , 3 , , 5, , 2, , -3 , , (x, y), , (0.1), , (1.3), , (2.5), , (1, 2), , (-2 , -3), , 2, , Now, let's plot these points. Let's check if these points are collinear. Let's draw this, line. The line is the graph of the equation 2x - y + 1 = 0., , ICT tools or links, , , Use Geogebra software to plot the x and y axes. Draw several points., , Find and examine the coordinates of the points in the "algebraic view". Read the equations of the lines parallel to the axes. Use the move option to vary the positions of the lines., What are the equations for the x and y axes?, , Exercise Set 7.2, 1., On graph paper, plot the points A (3, 0), B (3, 3), C(0.3). Connect A, B and B, C. What is the figure formed?, 2. Write the equation of the line parallel to the Y-axis at a distance of 7 units to the left of it., 3. Write the equation of the line parallel to the x-axis at a 5 units away from it and below the x-axis. 4. The point Q( -3, -2) lies on a line parallel to the Y axis. Write the equation of the line and draw its graph., , 97

Page 108:

5. The X axis and the line x = -4 are parallel lines. How far apart are they?, 6. Which of the equations below have graphs parallel to the x-axis and which have graphs parallel to the y-axis?, (i) x = 3, , (ii) y - 2 = 0 , , (iii) x + 6 = 0, , (iv) y = -5, , 7. Draw the points A(2, 3), B(6, -1) and C(0, 5) on the graph paper ). If these points are collinear, draw the line that contains them. Write the coordinates of the points where the line intersects the x-axis and the y-axis. 8. Draw the graphs of the following equations in the same coordinate system. Write the coordinates of their points of intersection., x + 4 = 0,, , y - 1 =0,, , 2x + 3 = 0,, , 3y - 15 =0, , 9. Draw the graphs of the equations given below ., (i) x + y = 2, , (ii) 3x - y = 0, , (iii) 2x + y = 1, Problem set 7, , 1., , Choose the alternative correct answer to the following questions ., , (i) What coordinate shape does a point have on the X axis?, , (A) (b, b) (B) (o, b) (C) (a, o) (D) (a , a ), (ii) Every point on the line y = x has the form ....., , (A) (a, a) (B) (o, a) ( C) (a, o) ( D) ( a , - a), (iii) What is the equation of the x axis?, (B) y = 0, (C) x + y = 0 (D) x = y, , (A) x = 0, (iv ) In which quadrant is the point (-4, -3)?, , (A) First (B) Second (C) Third (D) Fourth, (v ) What is the nature of the line connecting the points (-5.5) , (6.5), (-3.5), (0.5) contains?, , , (A) passes through the origin, , (B) parallel to the Y axis, , , , (C) parallel to the X axis, , (D) None of these, , (vi) Which of the points P(-1,1), Q(3,-4), R(1,-1), S (-2,-3), T (-4,4) are in the fourth, quadrant?, , , (A) P and T, , (B) Q and R, , (C) S only, , 98, , (D) P and R

Page 109:

2. Some points are shown in Figure 7.11., , Y, 3, , Use them to answer the following questions:, (i), , 2, , Q, , Write the coordinates of the points, , P, , 1 , , Q and R., -4, , (ii) Write the coordinates of the points, , -3 -2 -1, , T and M., , (iv) What are the points whose x and, , 2 , , M, 3, , 4, , X, R, , -2, , S, , (iii) Which point is in the third quadrant ?, , 0 1, -1 T, -3, , Fig.7.11, , y -coordinates are equal ?, , 3. Without plotting the points on a graph, indicate which quadrant or axis the next point is on., (i) (5, -3) , , ( ii) (-7, -12) , , ( iii) (-23, 4), , (iv) (-9, 5), , (v) (0, -3), , (vi) (-6, 0), , Y, , 4 . Draw the following points in one and the same coordinate system., A(1, 3), B(-3, -1), C(1, -4), D( -2, 3), E(0, -8), F(1, 0), , 3, , L, , 2, , P, , 1, , 5. On the adjacent graph, the LM line is parallel, , to the Y axis. (Fig. 7.12), (i) How far is the LM line from the Y axis?, (ii) Write the coordinates of points P, Q and R., (iii) What is the difference? between the, , -4, , -3 -2 -1, , -1, , 0 1, , 2 3, , R, , X, , Q, , -2, -3, , M, , -4, , Fig.7.12, , x coordinates of points L and M?, , 6. How many lines are there parallel to the x axis and are 5 units apart?, 7*. If "a" is a real number, what is the distance between the y-axis and the line x = a ?, qqq, 99

Page 110:

8, , Trigonometry, , Let's learn., , • Introduction to trigonometry • Relationships between trigonometric ratios, • Trigonometric ratios, • Trigonometric ratios of specific angles, Introduction to trigonometry, How far is this ship from the shore?, , What will it be , the height of this tree?, How do you measure?, , We can measure distances using a rope or walking on the ground, but how do you measure the distance between a ship and a lighthouse? How do you measure the height of a tall tree? Note the images above. The questions in the pictures are related to mathematics., Trigonometry, a branch of mathematics, is useful for finding answers to such questions., Trigonometry is used in various branches of engineering, astronomy, navigation, etc., The word trigonometry derives from of three Greek words ab “Tri” means three, “Gona” means sides and “Metron” means measures., , Let's remember., We study triangle. The topic of trigonometry starts with the right triangle, the Pythagorean theorem and similar triangles, so let's remember these topics. The side opposite ÐA is BC and the side opposite ÐC is AB. Using the Pythagorean theorem, we can write the following statement for this, triangle., (AB)2 + (BC)2 = (AC)2 , , B, , 100, , Fig. 8.1, , C

Page 111:

A, , If DABC ~ DPQR, then their corresponding sides have the same proportions., AB, PQ, , So =, , P, , BC AC, =, QR PR, , C Q, , B, , R, , Fig 8.2 , Let's see how to find the height of a tall tree using properties of similar triangles., Activity: This experiment can be carried out on a clear and sunny day., Look at the figure on the side., Tree height is QR, Pole height is BC. Stick a stick into the ground as shown in the picture. Measure your height and length, your shadow. Also measure the length of the tree's shadow. The sun's rays are parallel. D PQR and D ABC are therefore equiangular, that is, similar triangles. Sides of similar triangles are proportional., QR BC, ., =, PR AC, , So we get an equation,, tree height = QR =, , B, stick, , So we get, , Q, , R, , P , , BC, × PR, AC, , A, , C, , Fig.8.3, , We know the values of PR, BC and AC. Substituting these values into this equation gives us the length of the QR, i.e. the height of the tree. do in the morning at 8'O clock. Can you say why?, Activity: You can do this activity and find out the height of a tall tree in your area. If there is no tree on the site, determine the height of a pole., Lamp, Pole, , Pole, , Fig. 8.4, , 101

Page 112:

Let's learn., Terms related to right triangle, Im right angle D ABC, ÐB = 90°, Ð A and Ð C are acute angles., C, , A, , Opposite, , Hypotenuse, , of ÐA, , Adajcent Side, of ÐC, , B, ¯, Adjacent side of Ð A, , A, , Fig. 8.5, Ex, , ¯, , ¯, , Hypotenuse, , C, , ¯, B, Opposite side of ÐC, , Fig. 8.6, , In Fig. 8.7, DPQR is a right triangle. WriteP, , Facing page Ð P =....., Facing page Ð R =....., Page next to Ð P = ....., Page next to Ð R = ..... , , Q, , Fig. 8.7, , R, , Trigonometric ratios, Right triangles are shown in the adjacent Fig.8.8. ÐB is their common angle. So all right triangles are similar. D PQB ~ D ACB, , \, , PB PQ BQ, = =, AB AC BC, , \, , PQ PB, =, AC AB, QB PB, =, BC AB , , \, \, , E, A , P, , PQ AC, =, ... alternating, PB AB, QB BC, =, ... alternating, PB AB, , 102, , B, , Q, , C, , Fig. 8.8, , F

Page 113:

The triangular figures in 8.9 and 8.10 are separate triangles from figure 8.8., A, , hypotenuse, , hypotenuse, , ¯, , opposite side, of ÐB, , B, , B, Q, ¯, adjacent side of Ð B, , C , ¯, Neighboring side of Ð B, , In D ACB,, , In D PQB,, , Opposite side of ÐB, AC, =, Hypotenuse, AB, , PQ, Opposite side of ÐB, = , Hypotenuse, PB, , The ratios, , \, , of ÐB, , Fig.8.10, , Fig.8.9, (i), , opposite side, , ¯, , P, , PQ, AC and , are the same. , PB, AB, , Opposite of ÐB, PQ, AC, =, =, , Hypotenuse, PB, AB, , This ratio is called the 'sine' ratio of Ð B and is abbreviated as 'sin B'., ( ii ) In D PQB and D ACB,, Adjacent side of ÐB, BC, BQ Adjacent side of ÐB, =, e, =, Hypotenuse, Hypotenuse, AB, PB, , \, , BQ, BC, Adjacent side of ÐB, = , =, Hypotenuse, PB, AB, , This ratio is called the 'cosine' ratio of Ð B and is abbreviated as 'cos B', PQ, BQ, , (iii) =, , AC , =, BC, , Opposite side of ÐB, adjacent side of ÐB, , This ratio is called the tangent ratio of ÐB, and is abbreviated as tan B., Ex. :, A, , q, B, , Fig.8.11, , C, , Sometimes we write the measures of the acute angles of a right triangle using the Greek letters q (theta), a (alpha), b (beta) ), etc. In the accompanying figure D ABC, measure of the acute angle C, the letter q is denoted . Thus, we can write the ratios sin C, cos C, tan C as sin q, cos q and tan q, respectively., , 103

Page 114:

AB, ,, AC, , sin C = sin q =, , cos C = cos q =, , BC, ,, AC, , tan C = tan q =, , AB, BC, , Remember!, , sin ratio = , , opposite side, hypotenuse, , ·, , sin q =, , opposite side of Ðq, hypotenuse, , · , , cos ratio =, , adjacent side, hypotenuse, , · , , cos q =, , adjacent side of Ðq , hypotenuse, , , , tan ratio =, , opposite side, adjacent side, , , , tan q =, , opposite side of Ðq, opposite side of Ðq, , ·, , Exercise Theorem 8.1, 1st, , Fig. 8.12, 2., , In Fig.8.12, ÐR is the right angle of D PQR. Write the following ratios., (i) sin P (ii) cos Q (iii) tan P (iv) tan Q, , R, , P, , b, , Y, , Q, , At right angles D XYZ, ÐXYZ = 90° and a,b,c are the lengths of the sides shown in the figure. Write the following ratios,, (i) sin X (ii) tan Z (iii) cos X (iv) tan X, , X, c, , a, Z, , Fig. 8.13, , 3., , Right angle D LMN, ÐLMN = 90°, ÐL = 50° and ÐN = 40°, write the following ratios., (i) sin 50°, (ii) cos 50°, (iii) tan 40°, (iv) cos 40° , , L, 50°, , M, , 40°, , Fig. 8.14, , N, , P, , 4., , q, a, , R, , Q, , Fig. 8.15, , S , , In Figure 8.15,.ÐPQR = 90°,, ÐPQS = 90°, ÐPRQ = a and ÐQPS = q, Write the following trigonometric ratios., (i) sin a, cos a, tan a, (ii ) sin q, cos q , tan q, , 104

Page 115:

Let's learn., Relationship between trigonometric ratios, P, , In Fig.8.16 , D PMN is a right triangle., Ð M = 90°, ÐP and Ð N are complementary, , (90- q)°, , angles ., , \ If Ð N = q then Ð P = 90 - q, , q, , N, , M, , Fig.8.16, , sin q =, , PM, ........(1 ) , PN , , sin(90 - q) =, , NM, ........(4), PN, , cos q =, , NM, .......(2), PN , , cos (90 - q) =, , PM, ........(5), PN, , tan q =, , PM, ........(3), NM, , tan ( 90 - q) =, , NM, ........(6), PM, , \ sin q = cos (90 - q) ........ of (1) and (5 ), cos q = sin(90 - q) ........ from (2) and (4), also note that tan q, , , similar,, , ´ tan (90 - q) =, , PM, NM , , ´, , NM, ........ of (3) and (6), PM, , \ tan q ´ tan (90 - q) = 1, sin q, =, cos q , , PM , PN, NM, PN, , =, , PM, PN, , ´, , PN, NM, , =, , PM, = tan q, NM, , Remember !, cos(90 - q ) = sin q ,, sin q, = tan q,, cos q, , sin(90 - q) = cos q, tan q, , 105, , ´ tan(90 - q) = 1

Page 116:

*For more information, 1, 1, sin q = cosec q, cos q = sec q,, , 1, = cot q, tan q, , This means that cosec q, sec q and cot q are inverse proportions of sin q are , , cos q and tan q., sec q = cosec (90 - q) cosec q = sec (90 - q), , tan q = cot (90 - q), , cot q = tan (90 - q) the angle is half the hypotenuse and the side opposite the 60° angle is, , 3, the hypotenuse., 2, , A, In Fig. 8.17, D ABC is a rectangle, , 60°, , triangle. ÐC = 30°, ÐA = 60°, ÐB = 90° ., C, , \ AB =, , B, , 30°, , 1, AC and BC = 3 AC, 2, 2, , Fig. 8.17, , Let's learn., Trigonometric ratios of angles of 30° and 60°, P, , Right angle, 60°, , a, Q, , D PQR when ÐR = 30°,, , ÐP = 60°, ÐQ = 90° and PQ = a, , 2a, , 1, PR, 2, 1, a = PR, 2, , then PQ =, 30°, , 3a, , R, , Fig. 8.18, , \ PR = 2a, , \ If PQ = a, then PR = 2a and QR =, 106, , QR =, , 3, PR, 2, , QR =, , 3, 2, , ´ 2a, , QR = 3a, 3a

Page 117:

(II) Trigonometric ratios of the 60° angle, , (I) Trigonometric ratios of the 30° angle, PQ, PR, , a 1, =, 2a 2, , sin 30° = =, , sin 60° =, , QR , =, PR, , QR, PR, , 3a, =, 2a, , 3, 2, , cos 60° =, , PQ, =, PR, , PQ, QR, , a, =, 3a, , 1 , 3, , tan 60° =, , QR, =, PQ, , cos 30° = =, tan 30° = =, , 3a, 2a, , =, , 3, 2, , a, 1, =, 2a , 2, 3a, , a, , = 3, , At right angles D PQR, ÐQ = 90°. Therefore ÐP and ÐR are mutually complementary angles. Again, check the relationship between the sine and cosine ratios of complementary angles. , cos q = sin (90 - q), cos 30° = sin (90°- 30°) = sin 60°, cos 30°= sin 60°, , Remember !, , sin 30° =, sin 60 ° = , , 1, 2, , cos 30° =, , 3, 2, , cos 60° =, , 1, 3, , 3, 2, , tan 30° =, , 1, 2, , tan 60° = 3 , , (III) Trigonometric ratios of the angle of 45°, A, 45°, , 2a, , a, , B, , 45°, , a, Fig.8.19, , C, , In right angles D ABC, ÐB = 90°, ÐA =45°,, ÐC = 45° \ This is an isosceles triangle. Assume that AB = a, so BC = a. Using the Pythagorean theorem, let's find the length of AC, AC2 = AB2 + BC2, = a2 + a2, AC2 = 2a2, \AC = 2 a, , 107

Page 118:

In Fig. 8.19 ÐC = 45°, sin 45° =, cos 45° =, , AB, =, AC, BC, =, AC, , a, 2a, , a, 2a, , =, , =, , 1 , , 2 , , tan 45° =, , AB, =, BC, , a, a, , =1, , 1, 2, , Remember !, sin 45° =, , 1, ,, 2, , cos 45 ° =, , 1, ,, 2, , tan 45° = 1, , (IV) Angular ratios of angles 0° and 90°, A, A, A, B, , B, , C, , C, , B, , C, , Fig.8.20, AC, . Hold, AB, , At right angles D ACB, ÐC = 90° and ÐB = 30°. We know that sin 30°=, , the length of side AB is constant, as the measure of ÐB continues to decrease, the length of AC, which is the inverse of ÐB, also decreases. Therefore, if the length of ÐB decreases, then the value of sin q also decreases., , \ if the length of ÐB becomes 0°, then the length of AC becomes 0., , \ sin 0° =, , AC, 0 , AB = AB = 0 \ sin 0° = 0, A, , A, , A, , B, , 30°, , C, , B, , 70°, , Fig.8.21, , 108, , C, , B , , 85°C

Page 119:

Now look at Fig. 8.21. In this right triangle, as the length of ÐB increases, so does the length of AC. When the measure of ÐB becomes 90°, the length of AC becomes equal to AB., , \ sin 90° = AC , , \ sin 90° = 1, AB, We know the relations between trigonometric ratios of complementary angles q = cos (90 - q), , , , and cos q = sin (90 - q), , , \ cos 0° = sin (90 - 0)° = sin 90° = 1 and cos 90° = sin ( 90 - 90 )° = sin 0°= 0, , remember !, sin 0° = 0, , , sin 90° = 1,, , cos 0° = 1,, , cos 90° = 0, , we know that,, sin q, sin 0, 0, tan q =, , \, tan 0 = cos 0 = = 0, cos q, 1, sin 90°, 1, but tan 90° = cos 90° =, 0 , but we can don't divide 1 by 0. Note that q is an acute angle. As the value increases and the value of 90° is reached, tan q also increases infinitely. Therefore, we cannot find the definitive value of tan 90. , 90°, , sin, , 0, , 1, 2, , 1, 2, , 3, 2, , 1, , cos, , 1, , 3, 2 , , 1, 2, , 1, 2, , 0, , tan, , 0, , 1, 3, , 1, , 109, , 3, , undefined

Page 120:

Worked examples: Ex. (1) Find the value of 2tan 45° + cos 30° - sin 60°, 2tan 45° + cos 30° - sin 60°, , Solution :, , = 2 ´ 1, , +, , 3, 2, , -, , 3, 2, , =2+0, = 2, Ex. (2) Find the value of, , cos 56°, sin 34°, , 56° + 34° = 90° means that 56 and 34 are the measures of the complementary angles., , Solution :, , sin q = cos (90 - q), , \ sin 34° = cos(90-34)° = cos 56°, \ cos 56° = cos 56° = 1, sin 34°, , cos 56°, , Ex. 3 At right angle DA ACB, if ÐC = 90°, AC = 3, BC = 4., , A, , Find the ratios sin A , sin B, cos A, tan B, Solution : In rectangle D ACB, with theorem of Pythagoras,, AB = AC +BC, = 32 + 42 = 52, \ AB = 5, 2, , 2, , 2, , 3, , BC, 4, AC, 3, = cos A =, =, sin A =, AB, 5, AB, 5, , and sin B =, , 3, AC, = , 5, AB, , tan B =, , then find cos q and tan q., P, , 5, 13 , , At right angles D PQR, ÐR= q, sin q =, , Q, , Fig. 8.22, , AC, 3, =, BC , 4, , Ex. 4 In the right triangle D PQR, ÐQ = 90°, ÐR= q and when sin q =, solution :, , 4, , C, , q, Fig. 8.2 3, , \, , R, , 110, , 5, 13, , 5, PQ, =, PR, 13, , B

Page 122:

To remember! "Square of" sin q means (sin q)2. It is written as sin2 q., We prove the equation sin2 q + cos2 q = 1 using the Pythagorean theorem, where q is an acute angle of a right triangle., Check that the equation is true even if q = 0° or q = 90°. Since the equation sin2 q + cos2 q = 1 holds for any value of q. It is therefore a basic trigonometric identity. Theorem 8.2, 1., , In the following table, a ratio is given in each column. Find the two remaining ratios in the column and complete the table., 11, 61, , sin q, cos q, , 1, 3, , 35, 37, , tan q, 2nd, , 3, 5, , 1 , 2 , , 21, 20, , 1, , 8, 15, , Find the values of 4, tan 2 60° + 3 sin 2 60°, 5, tan 60, (iv), sin 60 + cos 60, , (i) 5 sin 30° + 3 sin 45° , , (ii), , (iii) 2 sin 30° + cos 0° + 3 sin 90° , (v) cos 2 45° + sin 2 30° , , 3., , If sin q =, , 4, then find cos q, 5, , 4., , If cos q =, , 15, then find sin q, 17, , (vi) cos 60° × cos 30 ° + sin 60 °× sin 30°, , 112, , 1, 2 2

Page 123:

Exercise 8, 1., , Choose the correct answer alternative for the following multiple choice questions., (i) Which of the following statements is correct?, , (A) sin q = cos (90- q) , , (B ) cos q = tan (90-q), , (C) sin q = tan (90-q) , , (D) tan q = tan (90-q), , (ii) Which of the following is the value of sin 90° ?, (A), , 3, 2, , (B) 0 , , (C), , 1, 2, , (D) 1, , (iii) 2 tan 45° + cos 45° - sin 45° = ?, (A) 0 , (B) 1, (C) 2, cos 28°, (iv), = ?, sin 62°, (A) 2, (B) -1, (C) 0 , 2º, , (D) 3, , (D) 1, , T, , At right angle D TSU, TS = 5, ÐS = 90°,, SU = 12 then find sin T, cos T, tan T , Da similarly, find sin U, cos U, tan U, , S, 3, , Right angle D YXZ, ÐX = 90°, XZ = 8 cm,, YZ = 17 cm, find sin Y, cos Y, tan Y , , sin Z, cos Z, tan Z., , X, 8, Z, , 4., , Fig. 8.26, , At right angles D LMN when ÐN = q, ÐM = 90°,, cos q = 24 , 25 , find sin q and tan q, , L, , Fill in the blanks., , M, , U, Y, , 17, Fig. 8.27, , Find (sin2 q) and (cos2 q) analogously. , , 5., , (i) sin20° = cos, , °, , (ii) tan an30° ´ tan, , ° =1, , (iii) cos40° = sin, , Fig. 8.28, , N, , °, qqq, 113

Page 124:

9, , Surface area and volume, , Let's study., , • Surface area of a cone, • Volume of a cone, , • Surface area of a sphere, • Volume of a sphere, , Remember., We learned how to calculate the surface area and find the volume of a parallelepiped, a cube and a cylinder, in the previous patterns., • The length, width and height of a parallelepiped are l , b and h, respectively., , cuboid , , (i) area of perpendicular faces of a cuboid = 2(l + b) ´ h, , h, b, , l, , Fig.9.1, , cube, , Here we consider only 4 faces. , (ii) Total area of a parallelepiped = 2(lb + bh + lh ), here we consider all 6 faces., (iii) volume of a parallelepiped = l ´ b ´ h, • If l is the edge of a cube ,, , l , , (i) total area of a cube = 6l 2, (ii) Area of the vertical faces of a cube = 4l 2, (iii) Volume of a cube = l 3, , Fig.9.2, Cylinder, , • The radius of the cylinder is r and the height is h., , h, r, , (i) curved surface of a cylinder = 2prh, (ii) gesa mtsurface of a cylinder = 2pr(r + h), (iii) Volume of a cylinder = pr2h, , Fig.9.3, , 114

Page 125:

Practical set 9.1, 1. The length, width and height of a cuboid medicine box are 20 cm, 12 cm and 10 cm respectively. Find the surface area of the vertical faces and the total surface area of this box. 2. The total surface area of a parallelepiped-shaped box is 500 square units. Its width and height are 6 units and 5 units, respectively. How long is this box?, 3. The side of a cube is 4.5 cm. Find the surface area of all vertical faces and the total surface area of the cube. 4. The total surface area of a cube is 5400 cm². Find the surface area of all vertical faces of the cube. 5. The volume of a parallelepiped is 34.50 cubic meters. The width and height of the parallelepiped are 1.5 m and 1.15 m, respectively. Determine its length. 6. What is the volume of a cube with an edge of 7.5 cm? 7. The radius of the base of a cylinder is 20 cm and its height is 13 cm. (p = 3.14), 8. The curved surface of a cylinder measures 1980 cm2 and the radius of its base measures 15 cm. Find the height of the cylinder. (p =, , 22, )., 7, , Let's learn., , Terms related to a cone and its relation, A, , h, O, , l, r B, , Fig.9.4, , a Cone is shown in adjacent Fig.9.4. The center of the circle representing the base of the cone is O and A is the vertex (apex) of the cone. Seg OB is a radius and seg OA is perpendicular to the radius at O means that AO is the perpendicular height of the cone. The oblique height of the cone is the length of AB, denoted by (l)., D AOB is a right triangle., \ By the Pythagorean theorem, AB 2 = AO 2 + OB 2, , \l 2 = h 2 + r 2, , That is, (slant height)2 = (perpendicular height)2 + (base radius)2, Surface of a cone, A cone has two surfaces: (i) circular base and (ii) curved surface., From these two we can find the area of the base of a cone because we know the formula for the area of a circle., How do you find the curved area of a cone? How do you derive a formula for this?, , 115

Page 126:

To find a formula for the curved surface of a cone, let's look at the mesh of the curved surface, which is a sector of a circle. If a cone is cut along edge AB, we obtain its mesh as shown in Fig.9.5., , A, B, , D, , Compare Figures 9.4 and 9.5, C, Did you notice the following?, Fig.9.5 , (i) The radius AB of the sector is equal to the oblique height of the cones., (ii) The arc BCD of the sector is equal to the perimeter of the base of the cone., (iii) Curved surface of the cone = area of sector A-BCD ., It means to find the curvature surface of a cone we must find the area of its network, which is the area of the sector., Try to understand how it is done through the activity below., Activity : Observe the following figures., , l, , l, , l, , icons, Fig. 9.6, , l, , net with curved surface, Fig. 9.7, , perimeter of the base of the circle = 2pr, , pieces of net Fig. 9.8, , B, , A, , As shown in Fig. 9.8, make parts of the mesh as small as possible. Connect them as shown in Fig.9.9., , l, , By connecting the small mesh pieces of the cone, we get approximately a rectangle ABCD., The total length of AB and CD is 2pr., , D, , \ length of the side AB of rectangle ABCD is pr, , pr, Fig. 9.9, , and length of side CD is also pr, length of side BC of the rectangle = oblique height of the cone = l., , curved surface of the cone is equal to the area of the rectangle., \ curved surface of the cone = area of the rectangle = AB ´ BC = pr ´ l = prl, 116, , C

Page 127:

We can now derive the formula for the total surface area of a cone. total surface area of the cone = curved surface area + base area, = prl + pr2, = pr(l + r), notice something? When a cone is not closed (like a joker cap or birthday party cap) it has only one surface, which is the curve. Then we get the surface area of the cone by the formula prl. Activity: Prepare a cylinder from a sheet of cardboard, leaving one side open. Prepare an open cardboard cone that has the same radius as the base and height of the cylinder. Pour fine sand into the cone until it fills the cone. Drain cone into cylinder. Repeat the process until the cylinder is filled with sand. Note how many cones of sand are needed to fill the cylinder., , r, h, , r, l, , h, , Fig. 9.10, To fill the cylinder you will need three cones of sand learn., volume of a cone , If the radii of the base and the heights of the cone and the cylinder are the same, then 3 ´ volume of the cone = volume of the cylinder, \ 3 ´ cone volume = pr2h, , \ cone volume = , , 1, 3, , ´ pr2h, , Remember !, (i) area of the base of a cone = pr2, , (ii) curved surface of a cone = prl, , (iii) total surface area of a cone = pr (l + r) (iv) volume of a cone =, , 117, , 1, ´ pr2h, 3

Page 128:

Worked examples: Ex. (1) The radius of the base (r) and the vertical height (h) of the cone are given. Find its oblique height (l), (i) r = 6 cm, h = 8 cm, (ii) r = 9 cm, h = 12 cm, solution:, (i) r = 6 cm, h = 8 cm, , l2 = r2 + h2, \l2 = (6)2 + (8)2, \l2 = 36 + 64, \l2 = 100, \l = 10 cm, , (ii) r = 9 cm, h = 12 cm, l2 = r2 + h2, \l2 = (9)2 + (12)2, \l2 = 81 + 144, \l2 = 225, \l = 15 cm, , Ex. (2) Find (i) the height of the slope, (ii) the curved surface, and (iii) the total surface area of a cone when its base radius is 12 cm and its height is 16 cm. (p = 3.14), solution :, (ii) curved surface = prl, (i)r = 12 cm, h = 16 cm, l2 = r2 + h2, = 3.14 ´ 12 ´ 20, 2, 2 , 2, = 753.6 cm2, \l = (12) + (16), , \l2 = 144 + 256, \l2 = 400, \l = 20 cm, , (iii) total surface area of the cone, , = pr (l + r), , = 3.14 ´ 12(20+12), , = 3.14 ´ 12 ´ 32, , = 1205.76 cm2, , Ex. (3) The total surface area of a cone is 704 cm² and the radius of its base is 7 cm, find the oblique height of the cone. (p =, solution :, , 22, ), 7, , total cone area = pr (l + r), 22, 7, , \, , 704 =, , ´ 7 (l + 7), , \ , , 704, =l+7, 22, , \, , 32 = l + 7, , \ 32 - 7 = l, \, , l = 25 cm, , 118

Page 129:

Ex. (4) The base of a cone measures 1386 cm² and its height is 28 cm., 22, , ), Find its surface. (p =, 7, solution :, area of base of cone = pr 2, 22, 7, , \, , 1386 =, , \, , 1386 × 7, = r2, 22, , \l 2 = ( 21) 2 + (28)2, \l 2 = 441 + 784, \l 2 = 1225, \l = 35 cm, , ´ r2, , cone shell = prl, , \ 63 ´ 7 = r 2, \, , 441 = r 2 , , \, , r = 21 cm, , 22, ´ 21 ´ 35.7, , , , =, , , , = 22 ´ 21 ´ 5, , , , = 2310 cm², , Exercise set 9 ,2, 1., , The vertical height of a cone is 12 cm and its oblique height is 13 cm. Calculate the radius of the base of the cone., , 2., , Calculate the volume of a cone when its area of total surface area is 7128 cm² and the radius of the base is 28 cm ( p =, , 22, ), 7, , 3, , The curved surface of a cone is 251.2 cm2 and the radius of the base is 8 cm (p = 3.14 ), , 4, , What is the cost of manufacturing a sheet metal closed cone with a base radius of 6 m and a slant height of 8 m when the fabrication rate is Rs.10 per meter square?, , 5,,, volume of a cone is 6280 cubic centimeters and the radius of the base of the cone is 30 cm Find its vertical height. (p = 3.14), , 6. , , The surface area of a cone is 188.4 cm² and its oblique height is 10 cm. Find its vertical height (p = 3.14)., , 7., , The volume of a cone is 1212 cm3 and its height is 24 cm. Find the surface of the cone., (p = 22 ), 7, , 8., , The curved surface of a cone is 2200 cm² and its oblique height is 50 cm. Find the total surface area of the cone. (p =, , 9, , 22, ), 7, , 25 people sit in a cone-shaped tent. Each person needs an area of 4 square meters. the tent floor. If the height of the tent is 18m, find the volume of the tent, , 119

Page 130:

10. Dried fodder for livestock is piled up in a cone in a field. The height of the cone is 2.1 m and the diameter of the base is 7.2 m. Find the volume of the lining. If it is to be covered with polythin in the rainy season, how much minimum polythin sheet is needed?, (p =, , 22, e, 7, , 17,37 = 4,17.), let's learn., surface of a sphere , surface area of a sphere = 4pr2, , \ surface area of a hollow hemisphere = 2pr2, , Fig. 9.11, , total surface area of a solid hemisphere, = surface area of hemisphere + area of circle, = 2pr2 + pr2 = 3pr2 , , Take a sweet lime (Mozambique), cut it into two equal parts. , , Take one of the parts. Place your circular face on a piece of paper. Draw its circular border. Copy three more of these circles. Cut each sweet lemon half into two equal parts again., , Now you have 4 sweet lemon quarters. Separate the peel from a quarter part. Cut it into the smallest pieces possible. Try to cover one of the drawn circles with the small pieces. Note that the circle is almost covered. The activity suggests that the curved surface of a sphere = 4 pr2

Page 131:

Worked examples: (1) Find the surface of a sphere with a radius of 7 cm. (p =, solution :, , (2) Find the radius of a sphere with a surface area of 1256 cm² (p = 3.14), solution : surface area of sphere = 4pr2, \ 1256 = 4 ´ 3.14 ´ r2, , 22, ), 7, , surface area of sphere = 4pr2, 22, 7, 22, 7, , , , =4´, , , , =4´, , , , = 88 ´ 7 , , ´ (7)2 , , \ r2 =, , ´7´7, , , , =, , \ 100 = r2, \, 10 = r, \radius of the sphere is 10 cm., , = 616, , surface of the sphere = 616 q. cm., Activity :, , x, , Make a cone and a hemisphere out of cardboard so that the radii of the cone and, , , , hemisphere are the same and the height of the cone is equal to the radius of the hemisphere., , , , Fill in the cones with fine sand. Pour sand into the hemisphere. How, , , , does it take many cones to completely fill the hemisphere ?, , r, r, , h, , Fig. 9.12, , \volume of sphere, , Two cones filled with sand are needed to fill the hemisphere ., \ 2 ´ volume of cone = volume of hemisphere., , = 2 ´ volume of hemisphere., , \ volume of hemisphere = 2 ´ volume of the cone, , =, , 1, ´ pr2h, 3, 1, , = 2 ´ ´ pr2 ´ r, 3, 2 , = pr3, , 3, , , , =2´, , 121, , 4 3 , pr, 3, 4, 3, , \ volume of sphere = pr3

Page 132:

Remember !, 2, , • volume of hemisphere = 3 pr3, • total surface area of hemisphere = 2 pr2 + pr2 = 3 pr2, Worked examples:, Ex. (1) Find the volume of a sphere with , , 22 , radius 21 cm. (p =, ), 7, 4, solution: sphere volume = pr3, 3, 22, 4, , = ´, ´ (21)3, 7, 3, 22, 4, = ´, ´ 21 ´ 21 ´ 21, 7, 3, , = 88 ´ 441, , Ex. (2) Find the radius of a sphere whose volume is 113040 cubic centimeters. (p = 3.14), solution : volume of the sphere =, 113040 =, , 4, 3, , 4 3, pr, 3, , ´ 3,14 ´ r3, , 113040 ´ 3, = r3, 4 ´ 3 ,14, 28260 ´ 3 , = r3, 3.14, , \ 9000 ´ 3 = r3, \ r3 = 27000, \ r = 30 cm, \ radius of sphere is 30 cm., , \ volume of sphere = 38808 cubic centimeters, , Ex (3) Find the volume of a sphere whose surface area is 314 cm². (Assume p = 3.14), Solution: Surface area of sphere = 4pr2, 314 = 4 ´ 3.14, , \, \, \, , Volume of sphere =, , ´ r2, , 4 3, pr, 3 , , 314, 4 ´ 3.14 = r2, , =, , 4, 3, , ´ 3.14 ´ 53, , 31400, = r2, 4 ´ 314, 100, = r2, 4, , =, , 4, 3, , ´ 3.14 ´ 125 , , 25, , r, , = 523.33 cubic centimeters., , = r2, = 5 cm, , 122

Page 133:

Exercise Theorem 9.3, , 3., , Find the surfaces and volumes of spheres with the following radii., (i) 4 cm. (ii) 9 cm. (iii) 3.5 cm., (p = 3.14), if the radius of a solid hemisphere is 5 cm, then find its curved surface and its total surface. (p = 3.14) If the surface area of a sphere is 2826 cm2, find its volume. (p = 3.14), , 4., , Find the surface area of a sphere when its volume is 38808 cubic centimeters. (p =, , 5º, , volume of a hemisphere is 18000 p cubic centimeters. Find its diameter., , 1º, 2º, , 22, ), 7, , Problem set 9, 1º, , if the diameter of a roll rammer is 0.9 m long and 1.4 m long, how much area of a field is pressed at 500 revolutions? 🇧🇷 The external length, width and height of the tank are 60.4 cm, 40.4 cm and 40.2 cm respectively. How much maximum volume of water will it contain ?, , 3., , If the ratio of the radius of the base to the height of a cone is 5:12 and its volume is 314 cubic meters. Determine its perpendicular and oblique heights (p = 3.14)., , 4., , Determine the radius of a sphere when its volume is 904.32 cubic centimeters. (p = 3.14), , 5., , The total surface area of a cube is 864 cm². Find its volume., , 6., , Find the volume of a sphere when its surface area is 154 cm²., , 7., , The total surface area of a cone is 616 cm². If the sloped height of the cone is three times the radius of its base, determine its sloped height., , 8., , The inside diameter of a well is 4.20 meters and its depth is 10 meters. Find the inner surface of the well. Find the cost of plastering inside at the rate of Rs. 52 per square meter, , 9., , The length of a road roller is 2.1 m and its diameter is 1.4 m. It took 500 revolutions of the road roller to level a floor. How much ground area was leveled by the road roller? Find the leveling cost at the rate of Rs. 7 per m², , qqq, 123

Page 134:

Answers, 1., , Basics of geometry, exercise set 1.1, , 1., , (i) 3, , (ii) 3, , (iii) 7, , (iv) 1, , (v) 3, , (vi) 5, , (vii) 2, , (viii) 7, , 2nd, , (i) 6, , 3rd, , (i) P-R-Q, , (ii) Not collinear, , (v) X-Y-Z, , ( vi ) Not collinear, , 4., , (ii) 8, , 18 and 2, , (iii) 10, , 5. 25 and 9, , (iv) 1, , (v) 3, , (vi ) 12 , , (iii) A-C-B, , 6. (i) 4.5, , (ii) 6.2, , (iv) Not collinear, (iii) 2 7, , 7th triangle, , Exercise set 1.2, 1st, , (i ) No, , (ii) No, , (iii) Yes, , 2nd 4, , 3rd 5, , 4th BP < AP < AB, , 5th, , (i) RS Radius or RT Radius (ii) PQ Radius (iii) ) QR Seg ( iv) QR Ray and RQ Ray etc., (v) RQ Ray and RT Ray etc.. (vi) SR Ray , ST Ray etc. Point C, Point D and Point P (ii) Point L and Point U, Point P and Point R, (iii), , d (U,V) = 10, d (P,C) = 6, d (V, B) = 3 , d (U,L) = 2, Exercise Theorem 1.3, , 1º, , (i), , , , If a quadrilateral is a parallelogram, then the opposite angles of this quadrilateral are , congruent., , (ii ) If quadrilateral a is a rectangle, then the diagonals are congruent., (iii) If a triangle is an isosceles triangle, then the segment joining the vertex of a triangle and the midpoint is , 2º, , (i), , , , perpendicular to the base of the base., If alternate angles formed by two lines and their transversals are congruent, then the lines are parallel., , (ii) If two parallel lines are intersected by a transversal, the interior angles are as formal, , , complementing a congruent quadrilateral then this quadrilateral is a rectangle., Problem set 1, 1st, 2nd, 3rd, 4th, , (i) A (ii) C (iii) C (iv) C (v) B, ( i) False, (ii) False, (iii) True, (iv) False, (i) 3 (ii) 8 (iii) 9 (iv) 2 (v) 6 (vi) 22 (vii) 165, -15 and 1.5. (i) 10.5 (ii) 9.1, 6.-6 and 8, , 124

Page 135:

2. Parallel lines, exercise set 2.1, 1st, 2nd, 3rd, 5th, , (i) 95° (ii) 95° (iii) 85° (iv) 85°, Ða = 70°, Ðb = 70 °, Ðc = 115°, Ðd = 65°, Ða = 135°, Ðb = 135°, Ðc = 135°, (i) 75° (ii) 75° (iii) 105° (iv) 75°, exercise set 2.2 , , 1st, , 4th ÐABC = 130°, , nº, , series of exercises 2, 1st, 5th, , (i) C (ii) C (iii) A (iv) B (v) C, 6. f = 100 ° g = 80°, x = 1260°, , 4th, , x = 130°, , y = 50°, , 3rd triangles, exercise series 3.1, 1st, 5th, 7th, , 110°, 2nd 45°, 3rd 80°, 60°, 40°, 6th ÐDRE = 70°, ÐARE = 110°, 60°, 80°, 40°, ÐAOB = 125°, 9° 30°, 70°, 80°, , 4th 30th, 60th, 90th, , Exercise set 3.2, 1st, 2nd, 3rd, 4th, , (i) SSC test, (ii) SAS test, (iii) ASA test, (iv) sideways test hypotenuse., (i) ASA test, ÐBAC @ ÐQPR, AB side @ PQ side, AC side @ PR side, (ii) SAS test, ÐTPQ @ ÐTSR, ÐTQP @ ÐTRS, PQ side @ SR side, hypotenuse side test , ÐACB @ ÐQRP, ÐABC @ ÐQPR, AC Side @ QR Side, SSS Test, ÐMLN @ ÐMPN, ÐLMN @ ÐMNP, ÐLNM @ ÐPMN, Exercise Set 3.3 , , 1st, 2nd, , x = 50°, y = 60°, mÐABD = 110°, mÐACD = 110 ° ., 7.5 units n, , 3. 6.5 units, , 4. l(PG) = 5 cm , l(PT) = 7.5 cm, , Exercise set 3.4, 1., , 2 cm, , 2. 28°, , 3. ÐQPR , ÐPQR, , 4th largest side NA, smallest side FN, , Exercise set 3.5, 1st, 2nd, , XY, YZ, XZ , ÐX @ ÐL, ÐY @ ÐM,, = =, LM MN LN, l(QR ) = 12 cm, l (PR) = 10 cm, , 125, , ÐZ @ ÐN

Page 136:

Exercise set 3, 1st, , (i) D, , (ii) B, , (iii) B, , 5th squares, Exercise set 5.1, 1st, 2nd, 3rd, 4th, 6th, , mÐXWZ = 135°, mÐYZW = 45°, l (WY) = 10 cm, x = 40°, ÐC = 132°, ÐD = 48°, 25 cm, 50 cm, 25 cm, 50 cm, 60°, 120°, 60°, 120 °, , ÐA = 70° , ÐB = 110° , ÐC = 70° , ÐR = 110°, exercise set 5.3, , 1st, 2nd, 3rd, 4th, 5th, , BO = 4 cm , ÐACB = 35°, QR = 7.5 cm, ÐPQR = 105°, ÐSRQ = 75°, ÐIMJ = 90°, ÐJIK = 45°, ÐLJK = 45°, Side = 14.5 cm, Circumference = 58 cm, (i) Incorrect (ii) ) Incorrect (iii) Correct ( iv) Correct (v) Correct (vi) Incorrect, Exercise 5.4, , 1st, , ÐJ = 127°, ÐL = 72°, , 2nd ÐB = 108° , ÐD = 72°, set of exercises 5.5, , 1st, , XY = 4.5 cm,, YZ = 2.5 cm,, XZ = 5.5 cm, exercise set 5, , 1st, (i) D, (ii) C, 4th 24cm , 32cm, 24cm, 32cm, , (iii) D, 2nd 25cm , 3rd 6.5 2cm, 5th PQ = 26cm, 6th ÐMPS = 65°, , 6th circle , Exercise set 6,1, 1st, , 20 cm, , 2, 5 cm, , 3, 32 pieces, , 4, 9 pieces, , Joint exercise 6.2, 1st, , 12 cm, , 2, 24 cm , problem set 6, , 1., , (i) A, (ii) C, , (iii) A, (iv) B, , (v) D , , (vi) C, , 126, , (vi i) Q or J, 2. 2:1, , 4. 24 units

Page 137:

7. Coordinate Geometry, Exercise Set 7.1, 1st, 2nd, , Point A: Quadrant II, Point B: Quadrant III, Point K: Quadrant I, Point D: Quadrant I, Point E: Quadrant I, Point F: Quadrant IV , Point G: IV Quadrant, H Point: Y Axis, Point M: X Axis, Point N: Y Axis, Point P: Y Axis, Point Q: Quadrant III, (i) Quadrant I, (ii) Quadrant III, (iii) Quadrant IV (iv) Quadrant II, Exercise Set 7.2, , 1st, 6th, 7th, 8th, , Square, 2nd x = -7, (i) Y axis, (ii) X axis, , for axis X(5.0) , to the Y axis (0.5), (-4.1), (-1.5, 1), (-1.5, 5) , (-4, 5), , 3. y = -5, (iii) Y axis,, , 4. x = -3, (iv) X axis,, , 5. 4, , Problem 7, 1 ., 2nd, 3rd, 5th , , (i )C, (ii)A, (iii)B, (iv)C, (v)C, (vi)B, (i)Q(-2,2), R(4,-1) (ii) T (0,-1), M(3,0) (iii) Point S (iv) Point O, (i) Quadrant IV (ii) Quadrant III, (iii) Quadrant II (iv) Quadrant II, (v) Axis Y (vi) X axis, 7. |a|, (i) 3 (ii) P(3,2), Q(3,- 1), R (3,0) (iii) 0 6. . y = 5, y = -5, , 8th trigonometry, theorem of exercise 8.1, 1st, , (i), , QR, PQ, , (ii), , QR, PQ, , (iii), , QR, PR , , (iv), , 2., , (i), , a, , c, , (ii), , b, , a, , (iii), , b, , c, , (iv) a, b , , 3rd, , (i), , MN, LN, , (ii), , LM, LN, , (iii), , LM, MN, , (iv), , 4th, , (i), , PQ RQ PQ , ,, ,, PR PR RQ, , (ii), , PR, QR, , MN, LN, , QS PQ QS, ,, ,, PS PS PQ, Exercise set 8.2, , 1st, , 2 21 8 1 , 60 1, 3 20 15 4 2 2, ,, ,, ,, ; cos q :, ,, ,, ,, ,, , ,, 3, 61, 3 29 17 3, 2 2 29 17 5, , sin q :, , 12 1, ,, ,, 37, 2, , tan q :, , 12 11 1, ,, ,, ,, 3, 35 60, , 2,, , 3, 4, , 127

Page 138:

2., , (i), , 93, 11, (ii), 20, 2, , (iii) 5, , (iv), , 2 3, 3 +1, , (v), , 3, 3, (vi), 2, 4, , 3rd, , 3, 5, , 4th, , 8, 17, , Problem 8, 1st, 2nd, 3rd, 4th, 5th, , (i) A (ii) D (iii) C (iv) D, 12, 5, 12, 5, 5, 12, sin T =, , cos T =, , tan T =, , sin U =, , cos U =, , tan U =, 5, 12 , 13, 13, 13, 13, 8, 8, 15, 15, 8, 15, sin Y =, , cos Y =, , tan Y =, , sin Z =, , cos Z =, , tan Z =, 15, 17, 17, 17, 17, 8, 7, 49, 576, 7, sin q =, , tan q =, , sin2 q =, , cos2 q =, 25, 625 , 625, 24, (i) 70, , (ii) 60, , (iii) 50, , 9. Surface area and volume, exercise set 9.1, 1., , 640 cm², 1120 cm², , 2. 20 units, , 3. 81 cm², 121.50 cm² , , 4., , 3600 cm², , 5, 20 m, , 6, 421.88 cm³, , 7, , 1632.80 cm², 4144.80 cm², , 8, 21 cm, , set of exercises 9.2, 1.5 cm, , 2, 36960 cubic centimeters, , 5. 15 cm, , 6., , 8 cm, , 3. 10 cm, 6 cm, , 4. ` 2640, , 7. 550 cm² 8. 2816 cm², 9856 cm³ , , 9, 600 cubic meters 10. 28.51 cubic meters, 47.18 m², exercise set 9.3, 1. (i) 200.96 cm², 267.95 cm³, , (ii) 1017, 36 cm², 3052.08 cm³. , , (iii) 153.86 m², 179.50 cm³, 2.157 cm², 235.5 cm², 3.14130 cm³. 4. 5544 cm², 5. 60 cm, , task set 9, 1. 1980 cm², 4. 6 cm, 2. 96801.6 cm³, 5. 1728 cm³, , 3. 12 m, 13 m, 6. 179 .67 cubic centimeters, , 7.21 cm, , 8. 132 square meters, ` 6864 9. 4620 square meters, ` 32340, , qqq, 128

Page 139:

Handy Notebook for Standard IX, , Handy Notebook with Journal - Math, •, , English, Medium, , •, , •, , Price, ` 55.00, , •, •, •, , Based on government-approved curriculum and textbook, Inclusion of practices based on all chapters according to the assessment scheme., With many different activities, pictures, pictures/diagrams, etc., Inclusion of Purpose/Multiple, Choice Questions, Inclusion of Useful Questions, For Oral Examination, More Practice Questions, and Separate Writing Space, Answers, , Handy Notebooks are available from regional Textbook Bureau depots. (1) Maharashtra State Textbook Stores and Distribution Center, Senapati Bapat Marg, Pune 411004 25659465, (2) Maharashtra State Textbook Stores and Distribution Center, P-41, Industrial Estate, Mumbai - Bengaluru, Highway, Opposite Sakal Office, Kolhapur 416122 2468576 (3) Maharashtra State Textbook Stores and, Distribution Center, 10, Udyognagar, S.V. Road, Goregaon (West), Mutter bai 400062 28771842, (4) Maharashtra State Textbook Stores and Distribution Center, CIDCO, Lot No. 14, W Sector 12, Wavanja Street, Nova Panvel Street, Dist. Rajgad, Panvel 410206 274626465 (5) Maharashtra State Textbook Stores and Distribution Center near Lekhanagar, Lot no. 24, 'MAGH' Sector, CIDCO, New Mumbai-Agra Road, Nashik, 422009 2391511 (6) Maharashtra State Textbook Stores and Distribution Center, M.I.D.C. Sheds No. 2 and 3, Near Railway Station, Aurangabad 431001 2332171 (7) Maharashtra State Textbook Stores and Distribution, Centre, Opposite Rabindranath Tagore Science College, Maharaj Baug Road, Nagpur 440001, 2547716/2523078 (8) Maharashtra State Textbook Stores and Center of Distribution, Lot No. F-91, M.I.D.C., Latur 413531 220930 (9) Maharashtra State Textbook Stores and Distribution Center, Shakuntal Colony, Behind V.M.V. College, Amravati 444604 2530965, , E-Learning Materials (Audio-Visual) for Standards One through Twelve are available from the Textbook Bureau, Balbharati..., • Register your needs using the Q.R. Code given below. • Register your e-learning material needs using the Google Play Store and downloading the ebalbharati app. www.ebalbharati.in, www.balbharati.in, ebalbharati

Page 140:

MATHEMATICS, Part -, , PATTERN NINE, , 9, , 2, , 61.00