Oasis School Mathematics-8 147 Example 3 In the adjoining figure, BA||DC, find the value of x. Solution: Here, BA || DC, ∠BOD = 130° Through O, draw OP || BA || DC. Now, ∠BOP = ∠ABE = 3x [Corresponding angles being BA || OP] ∠DOP = ∠CDF = 2x [Corresponding angles being DC || OP] Now, ∠BOD = ∠BOP + ∠DOP [Whole parts axiom] or, 130° = 3x + 2x or, 130° = 5x ∴ x = 130° 5 = 26° Hence, the value of x is 26°. Example 4 State, with reasons, whether AB is parallel to CD or not. Solution: Here, mark the angles a, b and c as shown in the adjoining diagram. Here, a = 120°, c = 60°. Now, b + c = 180° [Being linear pair] or b + 60° = 180° ∴ b = 180° – 60° = 120°…....(i) Again, a = 120°.......(ii) From (i) and (ii), we get a = b Hence, AB || CD [∵ a = b, being corresponding angles equal] Exercise 14.3 1. Copy the following figures in your copy and mark the angle corresponding to the marked one. (a) (b) (c) (d) E A P C F 2x 1300 3x D O B A C B D a=1200 c = 600 b
148 Oasis School Mathematics-8 2. In the given figure, write the corresponding angles of the following angles. (i) ∠PRB (ii) ∠BRS (iii) ∠PRA (iv) ∠ARS 3. Identify whether corresponding angles formed in the given figure are equal or not. Justify your answer. (a) (b) (c) 4. Identify whether the given pair of lines are parallel or not. Justify your answer. 5. Find the values of unknown angles in each of the following. S A B P Q C D R M P Q N 600 600 130º A F C G B D E 120º (a) (b) (a) (b) (e) (c) (f) (g) 800 x y b a 500 x 420 2x (d) y0 320 x0 220 x 500 250 x0 600 y 4x0 (h) x 62º 58º Answer 1. Consult your teacher. 2. Consult your teacher. 3. Consult your teacher. 4. Consult your teacher. 5. (a) x = 250 , (b) x = 800 , y = 800 (c) x = 280 (d) x = 320 , y = 320 (e) a = 500 , b = 500 (f) x = 150 , y = 600 (g) x = 420 (h) x = 1200
Oasis School Mathematics-8 149 14.4 Co–interior angles: Let's find the nature of black marked angles in all of the figures below: • Each pair are non–adjacent angles. • Both angles are interior angles. • Both lie on the same side of the transversal. These pair of angles are co-interior angles. Again, If a transversal cuts two parallel lines, sum of co-interior angles so formed is 1800 . In the given figure, d and e are co-interior angles. c and f are co-interior angles. Since AB||CD d + e = 180° c + f = 180° Conversely, if the sum of co-interior angles is 1800 , then two lines are parallel. b and h, a and g are two pairs of co-exterior angles Experimental Verification If a transversal intersects two parallel lines, then the sum of two co-interior angles so formed is 180°. Draw two pairs of parallel lines AB and CD. Also draw a transversal EF which meets AB and CD at G and H respectively. Now, ∠AGH and ∠GHC are a pair of co-interior angles. Similarly, ∠BGH and ∠GHD are a pair of co-interior angles. B D F A C E G H Fig (ii) E B D F Fig (i) C A G H To verify: ∠AGH + ∠GHC = 180° ∠BGH + ∠GHD = 180° b f a e d g c h B D F C A E
150 Oasis School Mathematics-8 Observations: Measure the angles and tabulate the values as below: Figure ∠AGH ∠GHC ∠BGH ∠GHD ∠AGH+∠GHC ∠BGH+∠GHD Remarks i. ............0 ............0 ............0 ............0 ............0 ............0 ∠AGH+ ∠GHC=1800 ∠BGH+ ∠GHD=1800 ii. ............0 ............0 ............0 ............0 ............0 ............0 ∠AGH+ ∠GHC=1800 ∠BGH+ ∠GHD=1800 Conclusion: If a transversal intersects two parallel lines then the sum of two co-interior angles so formed is 1800 . Conversely: If a transversal cuts two other straight lines such that the sum of two interior angles on the same side of the transversal is two right angles, then the two lines are parallel. Consider the following figure. In the given figure, ∠ABD + ∠BDC = 180° [Sum of co-interior angles is 180°, then AB||CD.] Worked Out Examples Example 1 From the given figure, find two pairs of co-interior angles. Solution: In the given figure, two pairs of co-interior angles are a and d, b and c. Example 2 In the given figure, identify whether AB||CD or not. Solution: Here, ∠BGH = 120°, ∠GHD = 60°. ∠BGH + ∠GHD = 120° + 60° = 180° Since the sum of co-interior angles is 180°, AB||CD. 85° B A D E C 95° G 60º 120º B D H F C A E e d f c b g a h
Oasis School Mathematics-8 151 Example 3 In the given figure, AB || DE. Find the value of ∠BCD. Here, AB || DE, ∠ABC = 130°, ∠CDE = 110° Draw a line XY through C parallel to AB and DE. Now, ∠ABC + ∠BCX = 180° or 130° + ∠BCX = 180°(sum of co-interior angles being AB||XY) ∴ ∠BCX = 180° – 130° = 50° Again, ∠CDE + ∠DCY = 180° (Sum of co-interior angles being DE//CY) or 110° + ∠DCY = 180° ∴ ∠DCY = 180° – 110° = 70° Lastly, ∠BCD + ∠BCX + ∠DCY = 180° (Sum of angles on same side of a straight line) or ∠BCD + 50° + 70° = 180° or ∠BCD + 120° = 180° ∴ ∠BCD = 180° – 120° = 60° ∴ ∠BCD = 60° Example 4 In the given figure, AB||CD, if ∠A = 120º, find the values of x, y and z. Solution: Here, ∠A = 120° ∠A + ∠D = 180° [Co-interior angles] or, 120° + x = 180° x = 180° – 120° = 60° Again, x + y = 180° [Co-interior angles] or, 60° + y = 180° or, y = 1800 – 600 = 1200 Again, y + z = 1800 [Co-interior angles] or, 1200 + z = 1800 or, z = 1800 – 1200 z = 600 ∴ x = 600 , y = 1200 , z = 60° 1300 1100 B A X C Y D E 1300 1100 B A X C Y D E A D B C 1200 x y z
152 Oasis School Mathematics-8 Exercise 14.4 1. Copy the following figures in your exercise book and shade the angle co-interior to the marked one. (a) (b) (c) (d) 2. In the given figure, write the co-interior angle of (i) ∠AGH (ii) ∠GHD. 3. State whether AB is parallel to CD or not, give reasons. 4. In each of the following figures, find the values of the unknown angles. E G A C D F H B 300 1500 A C Q B D P A B D C P Q 500 1500 (a) (b) R S x 1100 500 z y 1200 a b a b 1200 c (a) (b) (c) (d) Answer 1. Consult your teacher. 2. Consult your teacher. 3. (a) parallel (b) not parallel 4. (a) a = 600 , b = 1200 (b) x = 700 (c) y = 500 , z = 1300 (d) a = 600 , b = 120, c = 600
Oasis School Mathematics-8 153 Miscellaneous Exercise 1. Find the unknown angles from the given figures. 2. Find the values of the unknown angles in each of the following figures. 3. Evaluate each of the unknown angles from the following figures. 320 600 a0 (a) x 1600 1200 (c) 162 220 0 x (e) 1000 1100 x0 (f) 250 x0 1050 (d) 2x0 1500 3x0 (b) 4. Study the given figure and answer the given questions. (a) Write the name of a pair of parallel line. (b) If a line EF though E is drawn parallel to AB and CD then write the corresponding angle equal to ∠BAE and ∠DCE. (c) If ∠BAE = 110° and ∠ECD = 115°, find the value of ∠AEC. 600 3x y x 800 (a+50)0 (2a+30)0 (a) (e) (b) (d) (f) (c) 700 x a 1200 p q 1220 320 x x0 700 (a) (b) (c) x z 1200 y 1200 500 x y z B D A E C
154 Oasis School Mathematics-8 5. In the given figure, (a) Write the vertically opposite angle, alternate angle corresponding angle and co-interior angle of ∠ARS. (b) If ∠ARS = 70° and ∠RSD = 70°, identify whether AB is parallel to CD or not. Justify your answer. (c) Calculate the value of ∠DSQ with reason. 6. In the given figure, AB‖CD (a) What is the relation of ∠AEC and ∠AEB? Calculate the value of y, using above relation. (b) Write the corresponding angle of ∠ECD. Using this relation find the value of x. (c) What is the relation among x, y and z? Calculate the value of z, using this relation. (d) Calculate the value of w. x 150° E C D A B z 35° w y Answer 1. (a) x = 200 (b) x = 800 , y =800 (c) a = 200 (d) a = 1200 , p = 600 , q = 1200 (e) x = 900 (f) x = 1100 2. (a) x = 1100 (b) x = 600 , y = 1200 z = 1200 (c) x = 500 , y = 600 , z = 700 3. (a) a = 280 , (b) x = 300 , (c) x = 1000 (d) x = 1400 , (e) x = 1300 , (f) x = 400 , (g) x = 300 4. (a) Consult your teacher. (b) Consult your teacher. (c) 225° 5. (a) Consult your teacher. (b) Consult your teacher. (c) 30° 6. (a) Consult your teacher. (b) Consult your teacher. (c) 115° (d) 11° Project Work Take rectangular sheet of paper. By paper folding cutting method, verify that: If a transversal cuts two parallel lines, • alternate angles so formed are equal. • corresponding angles so formed are equal. • sum of two co-interior angles is 1800 . P B D Q C A R S
Oasis School Mathematics-8 155 Unit 15 Triangle, Quadrilateral and Polygon Flash Back A triangle is a closed plane figure bounded by three line segments. It is denoted by a Greek letter ∆(Delta). A triangle has three sides and three angles. The sides opposite to vertex A, B, C are denoted by a, b, c respectively. 15.1 Triangles Type of triangle according to their sides a. Scalene triangle : In the given figure, AB = 2cm, BC = 4cm and AC = 3cm. In the figure alongside, i.e. AB ≠ BC ≠ AC. ∆ABC is a scalene triangle If all the three sides of a triangle are different in length, then it is a scalene triangle. b. Isosceles triangle: In the figure alongside, AB = AC = 3cm ∆ABC is an isosceles triangle. If any two sides of a triangle are equal, then it is an isosceles triangle. c. Equilateral triangle: In the figure, AB = BC = AC. ∆ABC is an equilateral triangle. If all the sides of a triangle are equal, then it is an equilateral A B C 3cm 2cm 4cm A B C 3cm 4cm 3cm A B 2cm 2cm 2cm C c b a A B C You know! • Construction of angles with the help of protractor and compass. • Types of triangles on the basis of its sides and on the basis of its angle. • Identification of different types of quadrilaterals (parallelogram,rhombus, square, rectangle, trapezium, etc.)
156 Oasis School Mathematics-8 triangle. Each angle of an equilateral triangle is always 60°. According to angles a. Right angled triangle: In ∆ABC, ∠B = 90°. ∆ABC is a right angled triangle. If one angle of a triangle is 900 , then it is a right angled triangle. b. Obtuse angled triangle: In ∆ABC, ∠B is 1200 which is obtuse. So ∆ABC is an obtuse angled triangle. If one angle of a triangle is more than 900 and less than 1800 , then it is an obtuse angled triangle. c. Acute angled triangle: In ∆ABC, ∠A = 550 ∠B = 600 and ∠C = 650,all three angles are less than 900 . i.e. all are acute. ∆ABC is an acute angled triangle. Hence, If all the angles of a triangle are less than 90°, then it is called an acute angled triangle. Note: Type of triangle Number of acute angles Number of right angles Number of obtuse angles Right angled triangle 2 1 0 Acute angled triangle 3 0 0 Obtuse angled triangle 2 0 1 Review Exercise 15.1 1. Answer the following questions. (a) How many acute angles does an acute angled triangle have? (b) How many obtuse angles does an obtuse angled triangle have? (c) How many right angles does a right angled triangle have? (d) In ∆ABC, if AB = 5cm, BC = 6cm, AC = 4.5cm, what type of triangle is ∆ABC on the basis of its sides? (e) In ∆PQR, PQ = 6.5cm, QR = 6.5 cm, PR = 6cm, which type of triangle is this? (f) In ∆XYZ, XY = YZ = XZ = 6.8cm, which type of triangle is ∆XYZ? A B C A B 120° C A B 55° 60° 65° C
Oasis School Mathematics-8 157 2. Identify the type of given quadrilateral. (a) (b) (c) (d) (e) (f) 15.2 Experimental verification of sides and angles of triangles Sum of all interior angles of a triangle is 180°. Given: Draw two triangles ABC of different size as shown in the figure. A B C Fig. (i) Fig. (ii) B A C To verify: ∠A + ∠B + ∠C = 180° Observations: Measure ∠A, ∠B and ∠C with the help of a protractor and tabulate the values as follows . Figure ∠A ∠B ∠C Remarks (i) ........................... ° ........................... ° ........................... ° ∠A + ∠B + ∠C = 1800 (ii) ........................... ° ........................... ° ........................... ° ∠A + ∠B + ∠C = 1800 Conclusion: The sum of all interior angles of a triangle is 180°. D C A B D A C B D A C B Answer Consult your teacher. Experiment : 1 A D B C D A C B A B D C
158 Oasis School Mathematics-8 Experiment : 2 If two sides of a triangle are equal, then the angles opposite to them are also equal or the base angles of an isosceles triangle are equal. Draw two isosceles triangles ABC in which AB = AC with different measurements as shown in the figures below. To verify : ∠B = ∠C. Observations: Measure ∠B and ∠C with the help of a protractor and tabulate them as follows: Figure ∠B ∠C Remarks (i) ........................... ° ........................... ° ∠B = ∠C (ii) ........................... ° ........................... ° ∠B = ∠C Conclusion: The base angle of an isosceles triangle are equal. Experiment : 3 Each of the base angles of an isosceles right angled triangle is 45°. Draw two right angled triangle ABC in which AB = BC and ∠B = 900 with different shapes and sizes as shown in figures below. To verify : ∠A = ∠C = 45° C Fig. (ii) A B A B Fig. (i) C A B C Fig. (i) C Fig. (ii) A B
Oasis School Mathematics-8 159 Observations: Measure ∠A and ∠C with the help of a protractor and tabulate them as follows: Figure ∠A ∠C Remarks i. ........................... ° ........................... ° ∠A = ∠C = 450 ii. ........................... ° ........................... ° ∠A = ∠C = 450 Conclusion : Each of the base angles of an isosceles right triangle is 45°. Experiment : 4 Each angle of an equilateral triangle is 600 . Draw two equilateral triangles ABC of different measurements. To verify : ∠A = ∠B = ∠C = 600 Observation: Measure ∠A, ∠B and ∠C of each figure and tabulate their values as follows: Figure ∠A ∠B ∠C Remarks i. ................0 ................0 ................0 ∠A = ∠B = ∠C = 600 ii. ................0 ................0 ................0 ∠A = ∠B = ∠C = 600 Conclusion: Hence, each angle of an equilateral triangle is 600 . Altitude and Median of a Triangle Altitude: In the given ∆ABC, AD ⊥ BC, BE ⊥ AC and CF ⊥ AB AD, BE and CF are altitudes of ∆ABC. Perpendicular drawn from any vertex to its opposite side is called the altitude of a triangle. • A triangle has three altitudes. • All three altitudes intersect at the same point. A B C E F D C Fig. (ii) A B A B Fig. (i) C
160 Oasis School Mathematics-8 Median: In the given triangle ABC, D, E and F are the mid points of BC, AC and AB respectively. AD, BE and CF are its medians. The straight line joining a vertex to the mid–point of opposite side is the median of a triangle. Experimental Verification on Medians and Altitudes Experiment : 1 The line joining the vertex to the mid–point of the base of an isosceles triangle is perpendicular to the base. Draw two isosceles triangles ABC, where AB = AC. Find the mid–point of BC (D) and join AD. A B C D Fig. (i) A B D C Fig. (ii) To verify : AD ⊥ BC i.e. ∠ADB = ∠ADC = 900. Observations : Measure ∠ADB and ∠ADC from each of the figures and tabulate the values as below: Figure ∠ADB ∠ADC Remarks i. ..............0 ..............0 AD ⊥ BC ii. ..............0 ..............0 AD ⊥ BC Conclusion: The line joining the vertex to the mid–point of the base of an isosceles triangle is perpendicular to the base i.e. median of an isosceles triangle is perpendicular to the base. Its converse is also necessary to verify experimentally. • A triangle has three medians. • The three medians always intersect each other at the same point. A F E B D C
Oasis School Mathematics-8 161 Experiment : 2 Perpendicular drawn from the vertex of an isosceles triangle bisects the base. A B C D Fig. (i) A B D C Fig. (ii) Draw two isosceles triangle ABC, where AB = AC. Draw AD ⊥ DC. To Verify : BD = DC. Observations : Measure the length of BD and DC in each of the above figure and tabulate in the given table. Figure BD DC Remarks i. ............ cm ........... cm BD = DC ii. ......... cm ........... cm BD = DC Conclusion: Hence, the perpendicular drawn from the vertex of an isosceles triangles bisects the base. • Sum of three angles of a triangle is 180°. • Base angles of an isosceles triangles are equal. • Each of the base angle of an isosceles right angle triangle is 45°. • Each angle of an equilateral triangle is 60°. • Median of an isosceles triangle is perpendicular to the base • Perpendicular drawn from the vertex of an isosceles triangle bisects the base. Remember ! Worked Out Examples Example 1 In the given figure, find the value of each angle. Solution: Here, ∠A = 4x, ∠B = 5x and ∠C = 6x A B C 5x 6x 4x
162 Oasis School Mathematics-8 ∠A + ∠B + ∠C = 1800 [Sum of three angles of a triangle is 1800 ] or, 4x + 5x + 6x = 1800 or, 15x = 1800 or, x = 180° 15 = 120 ∴ ∠A = 4x = 4 × 120 = 480 ∠B = 5x = 5 × 120 = 600 ∠C = 6x = 6 ×120 = 720 Example 2 In the given figure, ABC is a triangle in which AB = AC, ∠A = 70°, find ∠B. Solution: Here, Given : In ∆ABC, AB = AC, ∠A = 70° Let, ∠B = ∠C = x [Base angles of an isosceles triangle] Now, ∠A +∠B +∠C = 180° [Sum of all interior angles of a triangle] or, 70° + x + x = 180° or, 70° + 2x = 180° or, 2x = 180 – 700 or, x = 110° 2 = 55° ∴ ∠B = ∠C = 550 . Example 3 In the given figure, ∆ABD is an equilateral triangle and ∆BCD is an isosceles triangle in which BC = CD. Find ∠CBA. Solution: Here, ∆ABD is an equilateral triangle ∠ABD = 60° [Each angle of an equilateral triangle is 60°] Again, ∠CDB = ∠CBD [Being, BC = CD] Now, ∠BCD + ∠CDB + ∠CBD = 180° [Sum of all interior angles of a triangle] or 80° + ∠CBD + ∠CBD = 180° [∠CDE = ∠CBE] or, 2∠CBD = 180° – 80° ∠CBD = 100° 2 = 50° ∴ ∠CBA = ∠CBD + ∠ABD [Whole part axiom] = 50° + 60° = 110° A B C 70° B A D C 800
Oasis School Mathematics-8 163 Example 4 From the given figure, find the value of each angle. Solution: ∠C = ∠B = 2x [In ∆ABC, base angle of an isosceles triangle] Now, ∠A + ∠B + ∠C = 180° [Sum of all interior angles of a triangle] or x + 2x + 2x = 180° or 5x = 180° or x = 180° 5 = 36° Hence, vertical angle = ∠A = x° = 36° and Each base angle = ∠B = ∠C = 2x° = 2 × 36° = 72°. Example 5 In the given figure, AB = AC. If ∠BAC = 400 , find the value of x. Solution: Here, AB = AC and ∠BAC = 400 . Here, ∠ABC = ∠ACB [ Base angles of an isosceles triangle] Now, ∠ABC + ∠ACB + ∠BAC = 1800 [Sum of all interior angles of a triangle] or ∠ACB + ∠ACB + 40° = 1800 [ ∵ ∠ABC = ∠ACB] or 2∠ACB = 1800 – 400 or ∠ACB = 140° 2 or ∠ACB = 700 Again, ∠ACB+∠ACD = 1800 [ ∵ Being linear pair] Again, 700 + ∠ACD = 1800 or, ∠ACD = 1800 – 700 ∴ x = 1100 A B C x 2x A B D C 400 x
164 Oasis School Mathematics-8 Exercise 15.2 1. Answer the following questions: (a) What is the sum of three angle of a triangle? (b) What is the value of each angle of an equilateral triangle? (c) In the given figure, which two angles are equal? 2. Find the value of x and y in each of the given figures. (a) (b) (c) (d) A x 550 650 B C P 600 2x 3x Q R X Y Z 6x 7x 2x D E F 2x 3x 3. Find the values of unknown angles from the given figures. x 560 (a) A B C 500 x X Y Z (b) 60 y 0 x D E F (c) R x 1500 y S P Q (d) (f) x S R P w Q x z y (e) A D C B 4. Find the value of x from the given triangles. 5. Find the values of x, y, z from the figures below. A B C 600 600 600 4 cm x cm (a) A (b) x P Q R S (c) 600 x Y Z W X B C 300 x0 500 400 B A C D (a) P R S T Q 300 650 550 y z x (b) (c) 850 350 250 x
Oasis School Mathematics-8 165 6. Evaluate x, y, z from the following figures. (b) 550 650 x y z M N O Q P (a) 350 x 600 y z D G H E F (c) 500 x P S Q R y 7. Find the value of x, y from the given figures. 8. (a) In ∆ABC, AB = AC and AE||CB; prove that ∠DAE = ∠EAB. (b) In the given figure, AB//CD. GI is the bisector of ∠BGH and HI is the bisector of ∠GHD. Prove that ∠GIH = 900 . 9. Study the given figure and answer the questions given below. (a) What is the relation among the angles ∠ BAC, ∠BCA and ∠EBD? (b) Using this relation, find the value of x. (c) Find the value of y. (a) x 500 M B C A (b) Q R S P 300 x y D E A B C E G F A C B I D H y 48° E D B C A 15° x 25° Answer 1. Consult your teacher 2. (a) x = 600 (b) x = 240 (c) x = 120 (d) x = 180 3. (a) x = 620 (b) x = 650 (c) x = 600 , y = 600 (d) x = 300 , y = 300 (e) w = 600 , x = 600 , y = 1200 , z = 600 (f) x = 1350 4. (a) x = 4cm (b) x =1200 (c) x = 1200 5. (a) x=1200 (b) x = 600 , y = 300 , z = 1200 (c) 35° 6. (a) x = 600 , y = 850 , z = 850 (b) x = 600 , y = 550 , z = 650 (c) x = 650 , y = 650 7. (a) x = 400, (b) x =600, y = 600 8. Consult your teacher 9. (b) x = 40°, (c) y = 98°
166 Oasis School Mathematics-8 15.3Types of Quadrilateral a. Parallelogram In the given figure, whereAB = DC,AB||DC,AD = BC,AD||BC, then ABCD is a parallelogram. Hence, a parallelogram is a quadrilateral whose opposite sides are equal and parallel. b. Rectangle In the given figure, ABCD is a rectangle where, AB = DC, AB||DC, AD= BC, AD||BC, ∠A = ∠B = ∠C =∠D = 90°. A rectangle is a quadrilateral whose opposite sides are equal and parallel and each angle is right angle. c. Square Here, ABCD is a square. Where AB = BC = CD = AD and ∠A = ∠B = ∠C = ∠D = 90°. A square is a rectangle whose all four sides are equal. d. Rhombus Here, ABCD is a rhombus. Where, AB = BC = CD = AD and AB||DC, AD ||BC. A rhombus is a parallelogram whose all sides are equal. e. Trapezium In the given figure, ABCD is a trapezium, where AB||DC. A trapezium is a quadrilateral in which a pair of opposite sides are parallel. D C A B Opposite angles of a parallelogram are equal. In a parallelogram ABCD, ∠A = ∠C and ∠B =∠D. D A C B Every rectangle is a parallelogram. D A C B Every square is a rectangle. D A C B Every square is a rhombus. Every rhombus is a parallelogram. A D B C
Oasis School Mathematics-8 167 f. Kite In the given figure, ABCD is a kite, where AB = AD and BC = DC. A kite is a quadrilateral in which two pairs of adjacent sides are equal. ExperimentalVerification on Parallelogram, Rectangle andSquare Experiment : 1 The opposite sides of a parallelogram are equal. Draw two parallelograms ABCD where AB || CD and AD || BC with different sizes with the help of set square and ruler. A Fig. (i) D B C A Fig. (ii) D C B To verify : AB = CD and AD = BC. Observation: Measure AB, CD, AD, BC and tabulate their values as follows: Figure AB CD AD BC Results i. ..........cm ..........cm ..........cm ..........cm AB = CD and AD = BC ii. ..........cm ..........cm ..........cm ..........cm AB = CD and AD = BC Conclusion: Hence, the opposite sides of a parallelogram are equal. Experiment : 2 The opposite angles of a parallelogram are equal. Draw two parallelograms ABCD, where AB || CD and AD || BC with different sizes with the help of set square and ruler. A Fig. (i) D B C A Fig. (ii) D C B To verify : ∠A = ∠C and ∠D = ∠B. Observation: Measure ∠A, ∠B, ∠C, ∠D and tabulate their values as follows: A B D C
168 Oasis School Mathematics-8 Figure ∠A ∠B ∠C ∠D Results i. ..........cm ..........cm ..........cm ..........cm ∠A = ∠C and ∠D = ∠B ii. ..........cm ..........cm ..........cm ..........cm ∠A = ∠C and ∠D = ∠B Conclusion : Hence, the opposite angles of a parallelogram are equal. Experiment : 3 The diagonals of a parallelogram bisect each other. Draw two parallelograms ABCD of different sizes by using set square and ruler. In each figure, draw diagonals AC and BD which intersect each other at O. A D B C Fig. (i) O A D B C Fig. (ii) O To verify : AO = OC and BO = OD Observation : Measure AO, OC, BO, OD and tabulate them as follows: Figure AO OC BO OD Results i. ....cm ...cm ...cm ...cm AO = OC and BO = OD ii. ....cm ...cm ...cm ...cm AO = OC and BO = OD Conclusion : Hence, the diagonals of a parallelogram bisect each other. Experiment : 4 Diagonals of a rectangle are equal in length. Given: • Draw three rectangles ABCD of different size. • Draw the diagonals AC and BD in each rectangle. To verify : AC = BD Observations: Measure the lengths of AC and BD from each rectangle and tabulate their values as below: Fig. (ii) A D B C Fig. (i) A B D C
Oasis School Mathematics-8 169 Figure AC BD Remarks i. ....cm ...cm AC = BD ii. ....cm ...cm AC = BD Conclusion: Hence, the diagonals of a rectangle are equal. Experiment : 5 Diagonals of a square bisect each other at right angles. Given: • Draw three squares ABCD of different size. • Join the diagonals AC and BD of each square which meet at point O. A O D B C Fig. (i) A D B C O Fig. (ii) To verify : AO = OC, DO = OB, AC ⊥ BD. Observation : Measure AO, OB, DO, OC and ∠AOB, ∠BOC, then tabulate their values as below: Figure AO OB DO OC ∠AOB ∠BOC Remarks i. ....cm ...cm ...cm ...cm ......° ......° AO = OC, DO = OB, AC ⊥ BD ii. ....cm ...cm ...cm ...cm ......° ......° AO = OC, DO = OB, AC ⊥ BD Conclusion : Hence, the diagonals of a square bisect each other at right angle. • Opposite sides of a parallelogram are equal. • Opposite angles of a parallelograms are equal. • Diagonals of a parallelogram bisect each other. • Every rectangle is a parallelogram. • Every rhombus is a parallelogram. • In a trapezium one pair of opposite sides are parallel. Remember !
170 Oasis School Mathematics-8 I. Activity: To find the properties of rectangle by paper folding. Material required: • A4 size paper • coloured chart paper • glue • scissors. Procedure: • Take a rectangular sheet of paper and fold it as shown below a, b, c and d. • Paste the paper (e) on the chart paper and give the name ABCD as in 'f'. (a) (b) (c) (d) (e) Fill up the given table after measuring it. ∠A AC ∠B BD AB AO BC BO CD CO AD DO ∠C ∠D 1. Are opposite sides equal? 2. Are opposite sides parallel? 3. Are all angles equal? 4. How many angles are right angles? 5. Are the diagonals equal? 6. Do the diagonals bisect each other? Conclusion: ____________________________________ II. Activity: To find the properties of rhombus by paper folding. Material required: • A4 size paper • coloured chart paper • glue • scissors. Procedure: • Take a rectangular sheet of A4 size paper. • Fold it as shown below. • Fold it again to get four layers. • Mark two points P and Q as shown in the figure. A D B C O (f)
Oasis School Mathematics-8 171 • Fold along PQ and make a crease. • Open it and you find a rhombus. • Cut the rhombus and paste it on the chart paper. • Give the name as shown in figure 'g'. (a) (b) (c) (d) (e) (f) Fill up the given table after measuring it. ∠AOB ∠AOD ∠BOC ∠COD AB AO BC BO CD CO AD DO Answer the following questions. 1. Are all the four sides equal? 2. Are opposite angles equal? 3. Do the diagonals bisect each other? 4. Do the diagonals bisect each other at right angle? 5. Are all angles equal? 6. Are the diagonals equal in length? Conclusion: _______________________________________ Worked Out Examples Example 1 In the given figure, ABCD is a parallelogram. If ∠A = 1100 , find ∠B, ∠C and ∠D. Solution: ABCD is a parallelogram ∠A + ∠B = 180° [Sum of co-interior angles is 1800 ] or 110° + ∠B = 180° [∵∠A = 110°] A D B C O (g) A B D C 1100
172 Oasis School Mathematics-8 ∴ ∠B = 180° – 110° = 70° Again, ∠C = ∠A = 110° [Opposite angles of a parallelogram are equal] and ∠D = ∠B = 70° Hence, ∠B = ∠D = 70°, ∠C = 110° Example 2 In the given figure, ABCD is a square. If AO = OD = 5cm and AB = 5 2cm, find (i) OB (ii) OC (iii) ∠AOB (iv) AD (v) CD Solution: Here, OD = OB [Diagonals of square bisect each other] ∴ OB = 5 cm Again, AO = OC [Diagonals of square bisect each other] ∴ OC = 5 cm Again, ∠AOB = 90° [Diagonals of a square bisect each other atright angle] Hence, AB = CD = AD [All sides of square are equal.] ∴ AD = CD = 5 2cm Exercise 15.3 1. Fill in the blanks. (a) The sum of all the interior angles of a quadrilateral is ……… . (b) In a parallelogram, ……… sides are equal and ……… angles are equal. (c) Value of each angle of a rectangle is ……… . (d) If opposite sides are equal and each angle is 90°, then it is a ……… . (e) In a square, ……… angles are equal and ……… sides are equal. (f) In a rhombus, ……… angles are equal and ……… sides are equal. (g) In a kite, ……… pair of adjacent sides are equal. (h) In a trapezium ……… pair of opposite sides are parallel. 2. Tick (√) or cross (×) in the box. Features Parallelogram Rectangle Rhombus Square Opposite sides are equal. √ √ √ √ Opposite angles are equal. A D O B C 5cm 5cm 5 2cm
Oasis School Mathematics-8 173 All sides are equal. All angles are equal. Each angle is 90°. Diagonals bisect to each other 3. Find the value of x, y in the given figures. (a) (b) (c) (d) 4. Find the size of x, y, z, etc. in the following figures. 5. Find the values of x and y in the given figures. (a) A B D C O 3x+1 2x+5 (b) P Q S R O 7cm 2x-1 (c) 12cm P S Q R O y x 6. Evaluate a, b, x, y, z in the given figures. A B D C 300 250 x y b a (b) (c) x S P R Q 300 x (a) 7. In a quadrilateral ABCD, if ∠A = ∠C and ∠B = ∠D then prove that ABCD is a parallelogram. 6 cm y 5 cm y 3.2 cm 2.6 cm 2x+1 3x-2 y+1 x x 2y-1 4.6 cm x (2x+20)0 600 (b) 1200 2y 3x z x (c) z y (a) 600 2x 1500 (e) 3x-5 4x+10 (f) 800 x y z (d) A D B C
174 Oasis School Mathematics-8 8. Study the given figure and answer the questions given below. (a) Identify the given figure with reason. (b) Is AC = BD? why? (c) What type of triangle is ∆AOB? Given reason. (d) If AO = 5 2cm, what is the length of AB? 9. Study the given figure and answer the questions given below. (a) Identify the given picture with reason. (b) What type of triangle is ∆ABD? If AB = 6cm, and AF 8cm. What is the length of AC? (c) Is AC = BD? Give reason. (d) Is AO = OC and BO = OD? Give reason. (e) Find the conclusions from above result. 10. Study the given figure and answer the questions given below. (a) Identify the given figure,; with reason. (b) Which of the given relation is not true? (i) AB = DC (ii) AD = BC (iii) A C = BD (iv) AO = CO and BO = OD (c) If ∠ADB = 25° and ∠BCD = 110°, find the value of ∠BDC. (d) What additional feature should it have to be a rectangle? 11. Study the given figure and answer the questions given below. (a) Identify the given figure with reason. (b) Which one of the following relation is not true? (i) AO = OC, BO = OD (ii) AC⊥BD (iii) AC = BD (c) What type of triangle is ΔABC? If ∠BAC = 62°, what is the value of ∠BCA and ∠ABC? (d) What additional feature should it have to be a square? A B C O D A D C O B A B C O D A B C O D
Oasis School Mathematics-8 175 Answer 1. Consult your teacher. 2. Consult your teacher. 3. (a) x = 5 cm, y = 6 cm (b) x = 2.6 cm, y = 3.2 cm (c) x = 3 cm, y = 2 cm (d) x = 4.6 cm 4. (a) x= 600 , y = 1200 , z = 1200 (b) x = 200 (c) x = 400 , y = 300 , z = 600 (d) x =1000 , y = 800 , z = 800 (e) x = 150 (f) x = 250 5. (a) x = 4 cm (b) x = 4 cm (c) x = 6 cm, y = 6 cm 6. (a) x = 300 (b) x = 450 (c) x = 250 , y = 300 , a = 1250 , b = 1250 8. (a) consult your teacher (b) Consult your teacher (c) Consult your teacher (d) 10cm 9. (a) Consult your teacher (b) 10cm (c) Consult your teacher (d), (e) Consult your teacher. 10. (a), (b) consult your teacher (c) 45° (d) Consult your teacher. 11. (a), (b) Consult your teacher (c) 62° and 56° (d) Consult your teacher. Project Work • Make a rectangle, a square, a rhombus and a parallelogram by paper folding method. • By the actual measurement of their sides and angles, draw out their properties. When two adjacent sides and angle between them are given: Example : Construct a parallelogram ABCD in whichAB = 4 cm, BC = 6 cm and ∠ABC = 60°. 15.4 Construction of Parallelogram Type: I Rough Sketch A D B C 60º 4 cm 6 cm 6 cm 6 cm 4 cm 4 cm C D X A B ∴ ABCD is the required parallelogram.
176 Oasis School Mathematics-8 15.5 Construction of Rhombus • Draw a line segment BC = 6 cm. • Make∠XBC = 60° at the point B. • From B cut BA = 4 cm, on BX. • Opposite sides of parallelogram are equal so cut off CD = 4 cm and AD = 6 cm from C and A to get D. • Join AD and CD. Steps: • Draw a line segment QR = 4.9 cm. • At Q, draw an angle ∠XQR = 30°. • Cut off QS = 6.1 cm from Q on QX: Opposite sides of a parallelogram are equal so cut SP = 4.9 cm from S and QP = RS from Q. • Join PQ, PS and SR. Steps: Type: II When a side, a diagonal and angle between them are given : Example : Construct a parallelogram PQRS in which QR = 4.9 cm, QS = 6.1 cm, ∠SQR = 30°. X S R Q P 6.1 cm 4.9 cm 4.9 cm 30º ∴ PQRS is the required parallelogram. Arhombus is also parallelogram. Its all four sides are equal in length. Process of construction of rhombus is same as the construction of parallelogram. 4.9cm 4.9cm Rough Sketch P S Q R 30º 6.1 cm Think! Opposite sides of parallelogram are equal.
Oasis School Mathematics-8 177 • Draw a line segment BC = 6.5cm. • AT B, draw an angle of 450. • From B, cut by an arc of 6.5cm to get A. • From Aand C cut by an arc of 6.5cm to get D. • Join AD and CD. Steps: Type: I When a side and an angle is given Example : Construct a rhombus ABCD where AB = 6.5 cm and ∠B = 45°. Rough Sketch 6.5 cm 6.5 cm 450 6.5 cm 6.5 cm B C A D 6.5 cm 6.5 cm 6.5 cm 6.5 cm 45° B C A D ∴ ABCD is the required rhombus. Exercise 15.4 1. Construct a parallelogram PQRS from the following data. (a) PQ = 5 cm, QR = 6 cm, ∠PQR = 60°. (b) PQ = 6.2 cm, PS = 6.8 cm, ∠SPQ = 75°. (c) QR = 4 cm, QS = 6 cm, ∠SQR = 60°. (d) QR = 5 cm, RP = 7 cm, ∠PRQ = 45°. (e) PQ = 7cm, QS = 6.5cm, ∠PQS = 600 . 2. Construct a rhombus ABCD from the given information. (a) AB = 6cm, ∠A = 60° (b) BC = 7.3cm, ∠C = 75° (c) CD = 6.4cm, ∠C = 30° (d) AD = 5.6cm, ∠D = 120° Answer Consult your teacher. Think! All sides of rhombus are equal. ∴ Each side is equal to 6.5cm.
178 Oasis School Mathematics-8 Rough Sketch B A D C 5.7cm 5.7cm Rough Sketch A D C B 6 cm 3 cm 6 cm C B A D Type: I When two adjacent sides are given: Example : Construct a rectangle ABCD in which AB = 6 cm and AD = 3 cm. ∴ ABCD is the required rectangle. ∴ ABCD is a required square. • Draw a line segment AB of length 6 cm. • Draw angle of 90° at vertex A. • Cut AD = 3 cm from A. • Opposite sides of a rectangle are equal so cut DC = 6 cm and BC = 3 cm to get vertex C. • Join DC and BC. Steps: Construction of Square Type: I When one side is given: Example: Construct a square ABCD where AB = 5.7 cm. 6 cm D B C A Think! Each angle of rectangle is 900. 15.5 Construction of Rectangle
Oasis School Mathematics-8 179 • Draw a rough sketch. • Draw a diagonal BD = 5.2 cm. • Diagonals bisect the angles of a square, construct ∠DBX = ∠BDY = 45°. • Mark the point of intersection asA. • Take AB (or AD) as radius and cut BC = DC = AB. • Join BC and CD. Steps: • Draw a line segment BC of length 5.7cm • At B, draw an angle 900. • From B, cut by an arc of 5.7 cm to get A. • From Aand C cut by the arc of 5.7 cm to get point D. • Join AD and DC. Steps: Type: II When one diagonal is given: Example: Draw a square ABCD in which one diagonal BD = 5.2 cm Rough Sketch 5.2 cm B A D C 5.2 cm C B D A Y X ∴ ABCD is the required square. Think! All sides of square are equal. Each angle of square is 900.
180 Oasis School Mathematics-8 Exercise 15.5 1. Construct a rectangle ABCD in which, (a) AB = 6 cm, BC = 5 cm (b) AB = 5.2 cm, BC = 3.4 cm. (c) CD = 6.8cm, AD = 5.8cm. (d) AD = 6.4cm, CD = 7cm. 2. Construct a square ABCD in which, (a) AB = 5.1 cm, (b) BC = 3.9 cm. (c) AC = 6 cm. (d) BD = 8 cm. 15.6 Polygons Introduction: A polygon is a figure bounded by three or more straight lines. For example, (a) It is not a polygon because two line segments AC and BD intersect at point P which is not an end point. (b) It is a polygon. A B P D C B A G F E D Types of polygons C Regular polygon: If all the sides of a polygon are equal, then it is called a regular polygon and angles of regular polygon are also equal. In the given figure, AB = BC = CD = DE = EF = AF Then, ∠A = ∠B =∠C = ∠D =∠E = ∠F Convex polygon: If every angle of a polygon is less than 180°, then it is called a convex polygon. Concave polygon: If at least one angle of a polygon is greater than 180°, then it is called a concave polygon. In the given figure, ∠AED is greater than 180°. E F A B F D E D C A B D A B C E Answer Consult your teacher.
Oasis School Mathematics-8 181 Interior angle of a polygon Angles formed within the polygon by its sides are called its interior angles. ∠ABC, ∠BAC and ∠ACB are the interior angles of polygon (i). Again, ∠BAE, ∠AED, ∠EDC etc. are the interior angles of the polygon (ii). Unless otherwise stated, a polygon means a convex polygon. Sum of interior angles of a polygon Draw all diagonals to form triangles through a single vertex of given polygon. 4 sides, 2 triangles 6 sides, 4 triangles It is observed that the number of triangles formed is 2 less than the number of its sides in polygon. So, if number of sides of polygon is n, then the number of triangles formed will be (n – 2) But the sum of angles of a triangle = 180° ∴ Sum of angles of (n – 2) triangles = (n – 2) × 180° Hence, sum of interior angles of a polygon having n sides = (n – 2) × 180° If the n-gon is regular, it has n-equal sides and n-equal angles. ∴ Each interior angle of polygon = (n – 2) n × 180° Exterior angles of a polygon Exterior angle: If one of the sides of a regular polygon is extended outside, the angle so formed is called the exterior angle. A B C C B A E D Interior angle Interior angle D C E F A B Concave polygon Exterior angle Exterior angle
182 Oasis School Mathematics-8 Sum of the exterior angle of a polygon Let θ be the interior angle and α be the exterior angle. Now, at each vertex of a polygon. Exterior angle + interior angle=180° [both the angles together form a straight angle] i.e. α + θ = 180° or α + (n – 2) × 180° n = 180° [∵ Interior angle (θ) = (n – 2) n × 180°] or α = 180° – (n – 2) × 180° n or α = 180°n – 180°n + 360° n ∴ α = 360° n . Hence, the exterior angle of regular polygon = 360° n . If the sides of a polygon are produced in order, the sum of exterior angles so formed is always 360°. The table below shows some polygons, number of sides and their names. Number of sides Name Figure Number of sides Name Figure 5 Pentagon 11 Undecagon 6 Hexagon 12 Dodecagon B A E D F C θ α • Sum of the interior angles of a polygon = (n – 2) 180° • Sum of the exterior angles of a polygon = 360° • An interior angle of a regular polygon = (n – 2) × 180° n • An exterior angle of a regular polygon = 360° n Remember !
Oasis School Mathematics-8 183 7 Heptagon 13 13 – gon 8 Octagon 14 14 – gon 9 Nonagon 15 Quin –decagon 10 Decagon Worked Out Examples Example 1 Using the formula, find the size of each interior angle and exterior angle in a regular hexagon. Solution: Here, no. of sides of hexagon (n) = 6 We know that, each interior angle of a regular hexagon = (n – 2) n × 180° = (6 – 2) 6 × 180° (where n = 6) = 4 6 × 180° = 120° and each exterior angle of hexagon = 360° n = 360° 6 = 60° Hence, each interior angle and exterior angle of regular hexagon is 120° and 60° respectively. Example 2 Find the number of sides of a polygon, the sum of whose interior angles is 1260°. Solution: Sum of interior angles of polygon = 1260°
184 Oasis School Mathematics-8 Number of sides of polygon (n) = ? We know that, Sum of interior angle of a polygon = (n – 2) × 180° or, 1260° = (n – 2) × 180° or, 1260° 180° = n – 2 or 7 = n – 2 ∴ n =7 + 2 = 9. Hence, no. of sides of polygon (n) = 9. Example 3 In the adjoining figure, find the value of x. Solution: Here, number of sides of given polygon (n) = 6 We know that, Sum of interior angles of polygon = (n – 2) × 180° or 95° + 125° + 160° + x + 145° + 110° = (6 – 2) × 180° or x + 635 = 4 × 180° or x = 720° – 635° ∴ x = 85° Exercise 15.6 1. Using the formula, find the size of each interior angle and exterior angle in the following regular polygons. (a) Pentagon (b) Octagon (c) Decagon (d) Square (e) Do-decagon (f) Heptagon 2. Find the value of unknown angle in the given figure. (a) (b) (c) b a b a b a 3. (a) Find the number of sides of a polygon, the sum of which interior angles are; (i) 1800° (ii) 720° (b) If an exterior angle of a regular polygon is 60°, find the number of sides of the polygon. xº 125º 160º 95º 110º 145º
Oasis School Mathematics-8 185 (c) If an interior angle of a regular polygon is 108°, find the number of sides of the polygon. (d) If an interior angle of a regular polygon is 150°, find the number of sides of the polygon. 4. Fill in the following table and establish the formula for the sum of interior angles in any polygon. Polygon Number of sides (n) Number of triangles Sum of interior angles Polygon Number of sides (n) Number of triangles Sum of interior angles Triangle 3 1 (3–2)×180° Quadrilateral Pentagon 4 2 (4–2)×180° Hexagon Heptagon Octagon 5. Study the given figure and answer the questions given below. (a) What type of polygon is given in the figure? (b) Calculate the value of x. (c) If x = b? Give reason (d) Find the value of a. (e) What does 'a' represent in the figure? 6. Study the following questions and answer the questions given below. (a) What is the formula to find the sum of interior angle is polygon? (b) What is the name of given polygon? Find the sum of the angle of given polygon. b x a G F E D B C A xº 150º 60º yº 50º 100º
186 Oasis School Mathematics-8 (c) Calculate the value of y. (d) Using the result of (b) and (c), find the value of x. 7. Study the following questions and answer them (a) What is the difference between the polygon and regular polygon? (b) What is the sum of the angles a, b, c, d and x? (c) Calculate the value of a, b, c, and d? (d) Using the result of (b) and (c) calculate the value of x. (e) Calculate the value of y. Answer 1. (a) 1080 , 720 (b) 1350 , 450 (c) 1440 , 360 (d) 900 , 900 (e) 1500 , 300 (f) 128.570 , 51.430 2. (a) a = 1200 , b = 600 (b) a = 1080 , b = 720 (c) a = 600 , b = 1200 3. (a) 12 (b) 6 (c) 5 (d) 12 4. Consult your teacher. 5. (a) Consult your teacher (b) 120° (c) Consult your teacher (d) 60° 6. (a) Consult your teacher (b) 720° (c) 260° (d) 70° 7. (a) Consult your teacher (b) 540° (c) a = 126°, b = 84°, c = 110°, d = 100° (d) 120° (e) 60° 54º 96º 70º 80º a x b c d y
Oasis School Mathematics-8 187 Unit 16 Congruency and Similarity 16.1 Introduction Let's compare the following pair of figures. Similar in shape and equal in size Similar in shape but different in size Different in both shape and size. Similar in shape and equal in size Similar in shape but different in size Different in both shape and size. Similar in shape and equal in size Similar in shape but different in size Different in both shape and size. The pair of figures in the first column are congruent figures. The pair of figures in the second column are similar figures. The pair of figures in the third column are neither congruent nor similar. Hence, two figures are said to be congruent figures if they are similar in shape and equal in size. Two figures are said to be similar figure if they are similar in shape even though different in size. You know! • Similar figure • Congruent figure
188 Oasis School Mathematics-8 Let's investigate ! • Are all circles similar? • Are all rectangles similar? • Are all triangles similar? Congruency and Similarity in different shapes Exercise 16.1 3cm A B P Q 3cm 3cm 4cm 60° 4cm 3cm 60° Congruent lines Congruent angles Similar lines Similar angles Similar pictures Similar cubes Cube Cube 1. Identify whether the given pair of figures are congruent or not. Justify your answer. (a) (b) (c) (d) 4cm A B 4cm P Q Q P R B C A D C A B S R P Q B C A Y Z X
Oasis School Mathematics-8 189 (e) (f) 2. Identify whether the following pair of figures are similar or not. (a) (b) (c) (d) (e) (f) (g) (h) 3. Draw the following figures and identify whether they are congruent or not. Justify your answer. (a) A circle of radius 3cm and a circle of radius 4cm. (b) A square having a side 4cm and a square having a side 5cm. (c) A rectangle ABCD where AB = 5cm, BC = 4cm and a rectangle PQRS where PQ = 5cm, QR = 4cm. 4. Draw the given figures in your copy and draw another congruent figure of each figure. A B U P Q R T S E C D A A B B C C A D B C D C A B 5cm 5cm 5cm 5cm 12cm 5cm 13cm B C A Q R P P Q S R A B D C
190 Oasis School Mathematics-8 16.2 Congruent Figures ∴ First pair are not congruent triangles and the second pair are congruent triangles. Two geometric figures having the same shape and size are called congruent figures. When two triangles have the same shape and size then they are called congruent triangles. Symbol for congruency is ≅. If two triangles are congruent it means that the sides and angles of the one are equal to the corresponding sides and angles of the other and their areas are equal. Conditions for Congruency of Triangles I. Two sides and the included angle (S.A.S. axiom) In ∆ABC and ∆PQR, AB = PQ = 5cm (S) ∠B = ∠Q = 600 (A) BC = QR = 4 cm (S) ∴∆ABC ≅ ∆PQR (By S.A.S axiom) Corresponding sides (Opposite to equal angles) AC = PR (Opposite to ∠B and ∠Q) Corresponding angles (Opposite to equal sides) ∠A = ∠P (Opposite to BC and QR) ∠C = ∠R (Opposite to AB and PQ) If any two sides of a triangle and angle between them are equal to any two sides and angle between them of another triangle, then the two triangles are congruent. Similar in shape, different in size Similar in shape, equal in size A B C Q R 600 4cm 4cm 600 5cm 5cm P 5. (a) Draw four pairs of congruent figures in your copy. (b) Draw four pairs of similar figures in your copy. Answer Consult your teacher. • Sides opposite to equal angles of congruent triangles are corresponding sides. • Angles opposite to equal sides of congruent triangles are corresponding angles. • Corresponding sides and corresponding angles of congruent triangles are equal.
Oasis School Mathematics-8 191 II. Two angles and included side (A.S.A. axiom) In ∆ABC and ∆PQR ∠B = ∠Q = 600 (A) BC = QR = 5cm (S) ∠C = ∠R = 500 (A) ∴∆ABC ≅ ∆PQR (By A.S.A. axiom) Corresponding sides (Opposite to equal angles) AB = PQ ( Opposite to ∠C and ∠R) AC = PR (Opposite to ∠B and ∠Q) Corresponding angles (Opposite to equal sides) ∠A = ∠P (opposite to BC and QR) If two angles and their adjacent side of one triangle are respectively equal to two angles and their adjacent side of another triangle, then the two triangles are congruent. III. Three sides (S.S.S. axiom) In ∆ABC and ∆PQR, AB = PQ = 5cm BC = QR = 4cm AC = PR = 6 cm ∴∆ABC ≅ ∆ PQR (By S.S.S. axiom) Corresponding angles (opposite to equal sides) ∠A = ∠P (Opposite to BC and QR) ∠B =∠Q (Opposite to AC and PR) ∠C = ∠R (Opposite to AB and PQ) If three sides of a triangle are respectively equal to three sides of another triangle, then the two triangles are congruent triangles. IV. Right angle, hypotenuse and any other side (R.H.S. axiom) In ∆ABC and ∆PQR, ∠B = ∠Q = 90° (R) AC = PR (H) AB = PQ (S) ∴∆ ABC ≅ ∆ PQR (By R.H.S. axiom) A B C Q R 600 600 500 500 5cm 5cm P A B C Q R 4cm 5cm 6cm 5cm 6cm 4cm P A C B R Q P
192 Oasis School Mathematics-8 Corresponding sides (Opposite to equal angles) BC = QR (Opposite to ∠A and ∠P) Corresponding angles (Opposite to equal sides) ∠A = ∠P (Opposite to BC and QR) ∠C = ∠R (Opposite to AB and PQ) If the hypotenuse and one of two remaining sides of a right angle triangle are respectively equal to hypotenuse and one ofremaining two sides, then two triangles are congruent. V. Two angles and a side (A.A.S. axiom) In ∆ABC and ∆PQR, ∠C = ∠R (A) ∠B = ∠Q (A) AB = PQ (S) ∴∆ABC ≅ ∆ PQR (By A.A.S. axiom) Corresponding sides (Opposite to equal angles) AC = PR (Opposite to ∠B and ∠Q) BC = QR (Opposite to ∠A and ∠P) Corresponding angles (Opposite to equal sides) ∠A = ∠P (Opposite to BC and QR) If two angles of a triangle and a side of a triangle are respectively equal to two angles and a side of another triangle, then the two triangles are congruent. Example 1 From the given condition, prove that ∆ABC ≅ ∆PQR. Solution: In ∆ABC and ∆PQR ∠BAC = ∠QPR (A) AB = PQ (S) ∠ABC = ∠PQR (A) ∴ ∆ABC ≅ ∆PQR, by A.S.A. axiom. A B C 4cm Q R P Worked Out Examples A B C Q R P
Oasis School Mathematics-8 193 Example 2 If the following pair of triangles are congruent, find the value of x. Solution: In ∆ABC and ∆PQR AB = PQ(S) ∠ABC = ∠PQR (A) BC = QR (S) ∴ ∆ABC ≅ ∆PQR, [by S.A.S. axiom.] AC = PR [Corresponding sides of congruent triangles] or, 2x – 1 = x + 3 or, 2x – x = 3 + 1 ∴ x = 4 Example 3 In the given figure, AB //CD. Make ∆ABO ≅ ∆CDO and find the value of x and y. Solution: In ∆ABC and ∆PQR ∠BAO = ∠ODC [Alternate angles] AB = CD [Given] ∠ABO = ∠OCD [Alternate angels] ∆ABO ≅ ∆CDO [By A.S.A. axiom] AO = OD [Corresponding sides of congruent triangles] or, 3x – 2 = x + 2 or, 3x – x = 2 + 2 or, 2x = 4 or, x = 4 2 ∴ x = 2 Again, BO = OC [Corresponding sides of congruent triangles] or, y + 3 = 2y + 1 or, y – 2y = 1 – 3 or, –y = –2 or, y = 2 ∴ x = 2, y = 2 A B C (2x-1)cm (x+3)cm Q R P A C D y+3 x+2 2y+1 3x-2 B O
194 Oasis School Mathematics-8 Example 4 In the given triangle, AB = AC, AD⊥BC, prove that- (i) ∆ADB ≅ ∆ADC, (ii) ∠ABD = ∠ACD. Given: In the given figure, AB = AC and AD⊥BC To prove : ∠ABD = ∠ACD Proof: Statements Reasons 1. In ∆ADB and ∆ADC (i) ∠ADB = ∠ADC (R) (ii) AB = AC (H) (iii) AD = AD (S) 1. (i) Being both right angles (ii) Given (iii) Common sides 2. ∆ADB ≅ ∆ADC 2. By R.H.S. axiom 3. ∠ABD = ∠ACD 3. Corresponding angles of congruent triangles Proved. Example 5 In the given figure, ABCD is a parallelogram, prove that (i) ABC ≅ ADC, (ii) AB = CD, AD = BC, (iii) ∠ABC = ∠ADC Solution: Given: ABCD is a parallelogram. To prove : (i) ∆ABC ≅ ∆ADC, (ii) AB = CD, AD = BC, (iii) ∠ABC = ∠ADC Proof: Statements Reasons 1. In ∆ABC and ∆ADC (i) ∠BAC = ∠ACD R) (ii) AC = AC (S) (iii) ∠ACB = ∠DAC (A) 1. (i) Opposite sides of a parallelogram (ii) Opposite sides of a parallelogram (iii) Common sides 2. ∆ABC ≅ ∆ADC 2. By R.S.A. axiom 3. AB = DC, AD = BC 3. Corresponding sides of congruent triangles. 4. ∠ABC = ∠ADC 3. Corresponding sides of congruent triangles. Proved. A B D C A D C B
Oasis School Mathematics-8 195 Exercise 16.2 1. From the given conditions, use necessary axiom and make ∆ABC ≅ ∆PQR. A B C Q P R A P B C Q R A B C P Q R (a) (d) (b) (e) (c) A B C P Q R A P B C Q R 2. In each of the following, make ∆ABC ≅ ∆PQR and also write corresponding sides and angles. (a) (c) (b) (d) (e) A B C 500 600 5cm P Q R 500 600 5cm 3. (a) In the given figure, prove that (i) ∆ABC ≅ ∆DEF (ii) Find the value of 'x' and 'y'. (b) In the given figure, Prove that: (i) ∆XYZ ≅ ∆PQR. (ii) find the value of 'a' and 'x'. (c) In the given figure, Prove that: (i) ∆ABC ≅ ∆ADC (ii) Find the value of 'x' and 'y'. B A C 5cm 4cm 6cm Q P R 4cm 5cm 6cm A B C 5cm 3cm P Q R 3cm 5cm P Q R 5cm 60º 80º C B A 5cm 60º 80º Q R P 4cm 3cm A B C 4cm 3cm F D E 600 5cm 700 (2x-1)cm (2y+3)cm A B C 70 60 0 0 5cm (x+3)cm (3y+1)cm X Y Z P Q R (3a+2)cm (x+10)0 (4a–2)cm (2x-10)0 A B C (2y+5)cm (4y–1)cm (3x+2)cm (2x+7)cm D
196 Oasis School Mathematics-8 (d) In the given figure, AB = CD, AB//CD, OA = (2x–1)cm, OB = (3y–x)cm, OC = xcm and OD = 2cm. Find the value of x and y. 4. (a) In the given figure, AB = AC and AD⊥BC prove that ∆ABD ≅ ∆ACD. (b) In the given figure, PS = QR and PQ = SR, prove that ∆PQR ≅ ∆PRS. (c) In the given figure, AC//DB and AO = BO. Prove that : (i) ∆ACO ≅ ∆BDO, (ii) CO = OD (d) In the given figure, AB//CD and AB = CD. Prove that : (i) ∆AOB ≅ ∆COD, (ii) AO = OD. (e) In the given figure, AB = AC and AD⊥BC. Prove that : BD = DC. (f) In the given figure, YW = WZ and XW⊥YZ. Prove : (i) ∆XYW ≅ ∆XWZ, (ii) XY = XZ (g) In the given figure, AB = DC and ∠ABC = ∠DCB. Prove that: (i) ∆ABC ≅ ∆DBC, (ii) AC = BD. B A D C O (2x-1)cm (3y-x)cm x cm 2 cm A B C P D Q S R A D C B O A B C D O A B C D X Y Z W A D B C Answer 1. Consult your teacher 2. Consult your teacher 3. (a) x = 4cm, y = 2cm, (b) x = 200 , a = 4cm, (c) x = 5cm, y = 3cm d) x = 3/2cm, y = 1 cm