Xam Idea Mathematics Standard Class 10 Term 1 MCQ

2. The length of each side of a rhombus whose diagonals are of lengths 10 cm and 24 cm is (a) 25 cm (b) 13 cm (c) 26 cm (d) 34 cm 3. If DABC ~ DEDF and DABC is not similar to DDEF, then which of the following is not true? [NCERT Exemplar] (a) BC. EF = AC. FD (b) AB. EF = AC. DE (c) BC. DE = AB. EF (d) BC. DE = AB. FD 4. If DABC ~ DDEF, ar (DABC) = 9 , BC = 21 cm, then EF is equal to ar (DDEF) 25 (a) 9 cm (b) 6 cm (c) 35 cm (d) 25 cm 5. D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD = 3 cm, BD = 5 cm, BC = 12.8 cm and DE ||BC. Then length of DE (in cm) is (a) 4.8 cm (b) 7.6 cm (c) 19.2 cm (d) 2.5 cm 6. If in two triangles ABC and PQR, AB = BC = CA , then QR PR PQ (a) DPQR ~ DCAB (b) DPQR ~ DABC (c) DCBA ~ DPQR (d) DBCA ~ DPQR 7. If DPRQ ~ DXYZ, then (a) PR = RQ (b) PQ = PR (c) PQ = QR (d) QR = PR XZ YZ XY XZ XZ YZ XZ XY 8. A square and a rhombus are always (a) similar (b) congruent (c) similar but not congruent (d) neither similar nor congruent 9. In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE. Then, the two triangles are [NCERT Exemplar] (a) congruent but not similar (b) similar but not congruent (c) neither congruent nor similar (d) congruent as well as similar 10. It is given that DABC ~ DDFE, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm. Then, which of the following is true? [NCERT Exemplar] (a) DE = 12 cm, ∠F = 50° (b) DE = 12 cm, ∠F = 100° (c) EF = 12 cm, ∠D = 100° (d) EF = 12 cm, ∠D = 30° 11. Two circles are always (b) neither similar nor congruent (d) none of these (a) congruent (c) similar but may not be congruent 12. In figure, two line segments AC and BD intersect each other at the point P such that PA= 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to [NCERT Exemplar] A 6 cm D (b) 30° 5 cm 30° P 50° B 3 cm 2.5 cm C (a) 50° (c) 60° (d) 100° 98 Mathematics–X: Term–1

13. If DABC ~ DQ RP, aarr ((DDPAQBCR)) = 9 , AB = 18 cm and BC = 15 cm, then PR is equal to 4 [NCERT Exemplar] (a) 10 cm (b) 12 cm (c) 20 cm (d) 8 cm 3 14. In DLMN and DPQR, ∠L = ∠P, ∠N = ∠R and MN = 2QR. Then the two triangles are (a) Congruent but not similar (b) Similar but not congruent (c) neither congruent nor similar (d) Congruent as well as similar 15. In DABC and DRPQ, AB = 4.5 cm, BC = 5 cm, CA = 6 2 cm, PR = 12 2 cm, PQ = 10 cm, QR = 9 cm. If ∠A = 75° and ∠B = 55°, then ∠P is equal to (a) 75° (b) 55° (c) 50° (d) 130° 16. If DABC ~ DEDF and DABC is not similar to DDEF, then which of the following is not true? (a) BC . EF = AC . FD (b) AB . EF = AC . DE (c) BC . DE = AB . EF (d) BC . DE = AB . FD 17. If DPQR ~ DXYZ and PQ = 5 , then ar (DXYZ) is equal to XY 2 ar (DPQR) (a) 4 (b) 2 (c) 25 (d) 5 25 5 4 2 18. It is given that ar(DABC) = 81 square units and ar(DDEF) = 64 square units. If DABC ~ DDEF, then (a) AB = 81 (b) AB2 = 9 DE 64 DE2 8 (c) AB = 9 (d) AB = 81 units, DE = 64 units DE 8 19. DABC and DBDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is (a) 2 : 1 (b) 1 : 2 (c) 1 : 4 (d) 4 : 1 20. In triangle ABC, if AB = 6 3 cm, AC = 12 cm and BC = 6 cm, then ∠B is (a) 120° (b) 60° (c) 90° (d) 45° 21. In DABC, DE  BC in given figure, then BD is equal to A 4 cm 6 cm DE 3 cm BC (a) 2 cm (b) 8 cm (c) 3 cm (d) 9 cm 22. If Manish goes 3 km towards East and then 4 km towards North. His distance from starting point is (a) 3 km (b) 4 km (c) 5 km (d) 2 km 23. In a triangle ABC if AB = 13 cm, BC = 12 cm and AC = 5 cm, then the triangle is right angled at (a) A (b) B (c) C (d) can’t say Triangles 99

24. If DABC ~ DPQR and ar (DABC) = 9 , also AB =4 cm, then PQ is equal to ar (DPQR) 4 (a) 8 units (b) 8 units (c) 3 units (d) 8 units 3 2 2 25. If in DABC and DDEF, AB = DE , then both will be similar when BC FD (a) ∠B = ∠E (b) ∠A = ∠D (c) ∠B = ∠D (d) ∠A = ∠F 26. In DABC, AB = 6 7 cm, BC = 24 cm and CA = 18 cm. Then angle A is (a) an acute angle (b) an obtuse angle (c) a right angle (d) can’t say 27. It is given that DABC ~ DPQR, with BC = 1 . Then, ar (PRQ) is equal to [NCERT Exemplar] QR 3 ar (BCA) (a) 9 (b) 3 (c) 1 (d) 1 3 9 28. If S is a point on side PQ of a DPQR such that PS = QS = RS, then [NCERT Exemplar] (a) PR . QR = RS2 (b) QS2 + RS2 = QR2 (c) PR2 + QR2 = PQ2 (d) PS2 + RS2 = PR2 29. If in two triangles DEF and PQR, ∠D = ∠Q and ∠R = ∠E, then which of the following is not true? (a) EF = DF (b) DE = EF (c) DE = DF (d) EF = DE PR PQ PQ RP QR PQ RP QR 30. Given that DABC ~ DDEF. If DE = 2AB and BC = 3 cm then, EF equals (a) 12 cm (b) 2 cm (c) 1.5 cm (d) 6 cm 31. The straight line distance between A and B is (See figure) B 1 12 12 A 2 (a) 5 3 units (b) 5 units (c) 3 5 units (d) 5 2 units 32. In a DABC, +A = 25°, +B = 35° and AB = 16 units. In DPQR, +P = 35°, +Q = 120° and PR = 4 units. Which of the following is true? (a) ar (DABC) = 2ar (DPQR) (b) ar (DABC) = 4ar (DPQR) (c) ar (DABC) = 8ar (DPQR) (d) ar (DABC) = 16ar (DPQR) 33. The altitude of an equilateral triangle, having the length of its side as 12 cm, is (a) 6 2 cm (b) 6 cm (c) 8.5 cm (d) 6 3 cm 34. All the equilateral triangles are (a) similar (b) congruent (c) both (a) and (b) (d) none 35. A DPQR is similar to another triangle ABC such that ar (DPQR) = 4ar (DABC) . The ratio of their perimeter is given as (a) 2 : 1 (b) 1 : 2 (c) 4 : 1 (d) none of these 100 Mathematics–X: Term–1

36. If the three sides of a triangle are a, 3 a, 2 a then the measure of the angle opposite to the longest side is (a) 60° (b) 90° (c) 45° (d) 30° 37. QA and PB are perpendicular on AB, if AO = 10 cm, BO = 6 cm and PB = 9 cm, then measure of AQ (see figure) is P AO B (a) 15 cm Q (c) 10 cm (d) none of these (b) 25 cm 38. The areas of two similar triangles are 144 cm2 and 81 cm2. If one median of the first triangle is 16 cm, length of corresponding median of the second triangle is (a) 9 cm (b) 27 cm (c) 12 cm (d) 16 cm 39. O is a point on side PQ of a DPQR such that PO = QO = RO, then which of the following is true? (a) RO2 = PR× QR (b) PR2 + QR2 = PQ2 (c) QR2 = OQ2 + QR2 (d) OP2 + OR2 = PR2 40. If in two triangles ABC and DEF, AB = BC = CA then DF FE ED (a) DABC ~ DDEF (b) DABC ~ DEDF (c) DABC ~ DEFD (d) DABC ~ DDFE 41. It is given that DABC ~ DDEF and BC = 1 . Then ar (DDEF) is equal to EF 5 ar (DABC) (a) 5 (b) 25 (c) 1 (d) 1 25 5 42. If DABC ~ DDEF then which of the following is not true? (a) BC.DF = AC.EF (b) AB.DF = AC.DE (c) BC.DE = AB.EF (d) BC.DE = AB.FD 43. In the given figure AD = AE and +ADE = 70° and +ACB = 50° then +BAC is equal to BD EC A DE (a) 70° B C (d) 60° (b) 50° (c) 80° Triangles 101

44. In the given figure AD = 3 cm, AE = 5 cm, BD = 4 cm. CE = 4 cm, CF = 2 cm, BF = 2.5 cm then which of the following is true? A 3 cm 5 cm D E 4 cm 4 cm B 2.5 cm F 2 cm C (a) DE ; ; BC (b) DF ; ; AC (c) EF ; ; AB (d) none of them 45. Two triangles BAC and BDC, right angled at A and D respectively, are drawn on the same base BC and on the same side of BC. If AC and DB intersect at P, then AP × PC is equal to (a) AB × CD (b) AB × BC (c) BC × CD (d) DP × PB 46. XY is drawn parallel to the base BC of a DABC cutting AB at X and AC at Y. If AB = 4 BX and YC = 2 cm, then AY is equal to (a) 2 cm (b) 6 cm (c) 8 cm (d) 4 cm 47. If D, E and F are mid points of sides BC, CA and AB repspectively of DABC, then the ratio of the areas of triangles DEF and ABC is (a) 2 : 3 (b) 1 : 4 (c) 1 : 2 (d) 4 : 5 BX XC 48. ABCD is a parallelogram with diagonal AC. If a line XZ is drawn such that XZ ; ; AB then is equal to AZ D Y BX C (a) AY (b) DZ (c) AZ (d) AC AC AZ ZD AY 49. Sum of squares of the sides of rhombus is equal to (a) Sum of diagonals (b) Difference of diagonals (c) Sum of squares of diagonals (d) none of them 102 Mathematics–X: Term–1

50. In the given figure, if AD = BC, then AB2 + CD2 equals C D BA (a) AD2 + BC2 (b) AD2 + CD2 (c) BD2 + AC2 (d) none of them 51. Sides of triangles are given below. Which of these is a right triangle? (a) 7 cm, 5 cm, 24 cm (b) 34 cm, 30 cm, 16 cm (c) 4 cm, 3 cm, 7 cm (d) 8 cm, 12 cm, 14 cm 52. The ratio of areas of two similar triangles is equal to (a) ratio of their corresponding sides (b) ratio of their corresponding altitudes (c) ratio of the square of their corresponding sides (d) ratio of the squares of their perimeter 53. The areas of two similar triangles are 144 cm2 and 64 cm2. If one median of the first triangle is 12 cm, length of corresponding median of the second triangle is (a) 9 cm (b) 8 cm (c) 7 cm (d) 16 cm ar (DABD) 54. In DABC, if AD is the bisector of +A then ar (DACD) is equal to (a) AB (b) AB (c) AC (d) AB BC AD BC AC 55. The ratio between the area of an equilateral triangle described on one diagonal of square and area of an equilateral triangle described on one side of the square is (a) 1 : 2 (b) 4 : 1 (c) 2 : 1 (d) 1 : 1 56. In an equilateral triangle ABC if AD = BC, which of the following is true? (a) 2AB2 = 3AD2 (b) 4AB2 = 3AD2 (c) 3AB2 = 4AD2 (d) 3AB2 = 2AD2 57. If DABC is an equilateral triangle such that AD = BC, then which of the following is true? (a) 3 DC2 (b) 2 DC2 (c) 3CD2 (d) 4DC2 2 58. DABC is a right triangle right angled at A and AD = BC . Then, BD/DC is equal to (a) (AB/AC)2 (b) AB/AC (c) d AB 2 (d) AB AD AD n 59. Vatsal claims that congruent figures are similar as well. Vinayak claims that similar figures are congruent as well. Who is/are correct? [CBSE Question Bank] (a) Only Vatsal (b) Only Vinayak (c) Both Vatsal and Vinayak (d) Neither Vatsal nor Vinayak Triangles 103

60. Consider the statements below. (ii) All squares are similar. (i) All circles are similar. (iv) All equilateral triangles are congruent. (iii) All right triangles are congruent. Which statement is/are correct? [CBSE Question Bank] (a) (i) and (ii) (b) (ii) and (iv) (c) (i) and (iii) (d) (iii) and (iv) 61. Two similar figures are shown. R 10 m 76° 5m C Sx y 4m D 58° 2m 1 m 130° 76° P 98° 2m 6m Q A 3m 98° B What are the values of x and y? [CBSE Question Bank] (a) x = 58°, y = 130° (b) x = 98°, y = 76° (c) x = 82°, y = 84° (d) x = 130°, y = 84° 62. Two quadrilaterals are such that their diagonals bisect each other. What additional information is required to conclude that the quadrilaterals are similar? [CBSE Question Bank] (a) Opposite angles are equal (b) Opposite sides are equal (c) Diagonals bisect at right angle and adjacent angles are equal (d) Diagonals are equal and opposite sides are equal 63. Consider the figure below. C A 2.4 cm 70° 1.6 cm O 70° D 0.8 cm 4.8 cm B Which of the following statement is correct about the triangles in the figure? (a) DA OB ~ D DOC be cause DAOO BO [CBSE Question Bank] CO = (b) DAOB ~ DDOC because +AOB= +DOC (c) DAOB ~ DDOC because AO = BO and +BAO = +CDO DO CO (d) DAOB ~ DDOC because AO = BO and +AOB= +DOC DO CO 104 Mathematics–X: Term–1

64. Consider the figure below. A (7x + 1)° B 60° O C (3x + 29)° D Which of the following statement help proving that triangle ABO is similar to triangle DOC? (i) ∠B = 70° and (ii) ∠C = 70° [CBSE Question Bank] (a) Statement (i) alone is sufficient, but statement (ii) alone is not sufficient. (b) Statement (ii) alone is sufficient, but statement (i) alone is not sufficient. (c) Each statement alone is sufficient. (d) Both statement together is sufficient, but neither statement alone is sufficient. 65. In the figure below, PQ || CB. C 3.3 cm P 2.4 cm A 1.3 cm Q B To the nearest tenth, what is the length of QB? [CBSE Question Bank] (d) 2.2 cm (a) 1.4 cm (b) 1.7 cm (c) 1.8 cm 66. In the given figure, QR || AB, RP || BD, CQ = x + 2, QA = x, CP = 5x + 4, PD = 3x. Ax x+2 Q BR C 5x + 4 5x P D The value of x is _____________. [CBSE Question Bank] (d) 9 (a) 1 (b) 6 (c) 3 Triangles 105

67. Observe the two triangles shown below. P F 4.8 y E y 6 3.2 4 Q RD Which statement is correct? [CBSE Question Bank] (a) Triangles are similar by SSA (b) Triangles are similar by SAS (c) Triangles are not similar as sides are not in proportion (d) No valid conclusion about similarity of triangles can be made as angle measures are not 68. Consider the triangles below. [CBSE Question Bank] P 40° A 10 cm 5 cm 5 cm B CQ 10 cm R Which statement is correct? (a) For triangles to be similar, the measure of ∠A = 40°. (b) For triangles to be similar, the measure of ∠A = 100°. (c) Triangles are similar as all isosceles triangles are similar. (d) Triangles are similar as corresponding sides of the triangles are in the ratio 1 : 2. 69. Rohit is 6 feet tall. At an instant, his shadow is 5 feet long. At the same instant, the shadow of a pole is 30 feet long. How tall is the pole? [CBSE Question Bank] (a) 12 feet (b) 24 feet (c) 30 feet (d) 36 feet 70. Ankit is 5 feet tall. He places a mirror on the ground and moves until he can see the top of a building. At the instant when Ankit is 2 feet from the mirror, the building is 48 feet from the mirror. How tall is the building? [CBSE Question Bank] (a) 96 feet (b) 120 feet (c) 180 feet (d) 240 feet 71. The area of two similar triangles are a and k2a. What is the ratio of the corresponding side lengths of the triangles? [CBSE Question Bank] (a) 1 : k (b) 1 : k2 (c) 1 : a (d) 1 : a2 72. In the figure below, PQ || BC. A P Q B C 106 Mathematics–X: Term–1

The ratio of the perimeter of triangle ABC to the perimeter of triangle APQ is 3:1. Given that the numerical value of the area of triangle APQ is a whole number, which of the following could be the area of the triangle ABC? [CBSE Question Bank] (a) 28 (b) 60 (c) 99 (d) 120 73. The ratio of the areas of two similar triangles, ABC and PQR shown below is 25 : 144. What is the ratio of their medians AM and PN? [CBSE Question Bank] AP B M CQ N R (a) 5 : 12 (b) 5 : 16 (c) 12 : 5 (d) 25 : 144 74. The ratio of the areas of two similar right triangles is 9 : 16. The length of one of the sides of the smaller triangle is 15 cm. How much longer is the length of the corresponding side of the larger triangle from smaller triangle? [CBSE Question Bank] (a) 2 cm (b) 3 cm (c) 4 cm (d) 5 cm 75. Observe the right triangle ABC, right angled at B as shown below. A x P 5 x+5 BC What is the length of PC? [CBSE Question Bank] (d) 7.5 cm (a) 2.5 cm (b) 4.5 cm (c) 6 cm 76. Observe the right triangle ABC, right angled at A as shown below. If BP ⊥ AC, then which of the following is NOT correct? B [CBSE Question Bank] AP C (a) ΔAPB ~ ΔABC (b) ΔAPB ~ ΔBPC (c) BC2 = CP . AC (d) AC2 = AB . CB 77. Consider the figure below. B AP C Triangles 107

Mr Shah follows the below step to prove AB2 + BC2 = AC2. (i) ΔAPB ~ ΔABC (ii) AP = AB (iii) AB2 = AP . AC. [CBSE Question Bank] AB AC Which of these could be his next step? (a) Prove ΔABC ~ ΔPAB (b) Prove ΔAPB ~ ΔCPB (c) Prove ΔBPC ~ ΔABC (d) Prove ΔAPB ~ ΔBPC 78. From point X, Alok walks 112 m east to reach at point Y. From point Y, Alok walks 15 m toward north to reach point Z. What is the straight-line distance between position when he started and his position now? N [CBSE Question Bank] Z 15 m WE X 112 m Y (a) 113 m (b) 117 m S (d) 127 m (c) 123 m 79. Which set of lengths forms a right triangle? [CBSE Question Bank] (a) 5 cm, 12 cm, 16 cm (c) 3 cm, 3 cm, 4 cm (b) 7 cm, 24 cm, 25 cm (d) 6 cm, 7 cm, 9 cm 80. Consider the following three clamis about a triangle ABC with side lengths m, n and r. (i) ABC is a right triangle provided n2 − m2 = r2. (ii) Triangle with side lengths m + 2, n + 2 and r + 2 is a right-angle triangle. (iii) Triangle with side lengths 2m, 2n and 2r is a right-angle triangle. Which of these is correct? [CBSE Question Bank] (a) Statement (i) would be correct if n > m, n > r and statement 2 would be correct if ABC is a right triangle. (b) Statement (i) would be correct if r > m, r > n and statement 2 would be correct if ABC is a right triangle. (c) Statement (i) would be correct if n > m, n > r and statement 3 would be correct if ABC is a right triangle. (d) Statement (i) would be correct if r > m, r > n and statement 3 would be correct if ABC is a right triangle. 81. In the adjoining figure, D, E and F are the mid-points of the side BC, AC and AB respectively of ∆ABC then ar(∆DEF) : ar(∆ABC) is A FE B DC (a) 1 (b) 1 (c) 1 (d) None of these 2 4 9 108 Mathematics–X: Term–1

82. If DABC ∼ DDEF such that AB = 1.2 cm and DE = 1.4 cm, the ratio of the areas of DABC and D DEF is [CBSE 2020 (30/3/1)] (a) 49 : 36 (b) 6 : 7 (c) 7 : 6 (d) 36 : 49 83. In figure, ABC is an isosceles triangle, right-angled at C. Therefore [CBSE 2020 (30/5/1)] B (a) AB2 = 2AC2 C A (d) AB2 = 4AC2 (b) BC2 = 2AB2 (c) AC2 = 2AB2 84. In figure, DE  BC. If AD = 3 and AE = 2.7 cm, then EC is equal to [CBSE 2020 (30/5/3)] DB 2 A 2.7 cm DE BC (a) 2.0 cm (b) 1.8 cm (c) 4.0 cm (d) 2.7 cm 85. It is given that DABC ~ DDFE, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm. Then, which of the following is true. (a) DE = 12 cm, ∠F = 50° (b) DE = 12 cm, ∠F = 100° (c) EF = 12 cm, ∠D = 100° (d) EF = 12 cm, ∠D = 30° 86. If in triangles ABC and DEF, AB = AC , then they will be similar when EF DE (a) ∠A = ∠D (b) ∠A = ∠E (c) ∠B = ∠E (d) ∠C = ∠F Answers 1. (c) 2. (b) 3. (c) 4. (c) 5. (a) 6. (a) 7. (c) 8. (d) 9. (b) 10. (b) 11. (c) 12. (d) 13. (a) 14. (b) 15. (c) 16. (c) 17. (a) 18. (c) 19. (d) 20. (c) 21. (a) 22. (c) 23. (c) 24. (a) 25. (c) 26. (c) 27. (a) 28. (c) 29. (b) 30. (d) 31. (c) 32. (d) 33. (d) 34. (a) 35. (a) 36. (b) 37. (a) 38. (c) 39. (b) 40. (d) 41. (b) 42. (d) 43. (d) 44. (c) 45. (d) 46. (b) 47. (b) 48. (c) 49. (c) 50. (c) 51. (b) 52. (c) 53. (b) 54. (d) 55. (c) 56. (c) 57. (c) 58. (b) 59. (a) 60. (a) 61. (a) 62. (c) 63. (d) 64. (c) 65. (c) 66. (a) 67. (b) 68. (b) 69. (d) 70. (b) 71. (a) 72. (c) 73. (a) 74. (d) 75. (d) 76. (d) 77. (c) 78. (a) 79. (b) 80. (c) 81. (b) 82. (d) 83. (a) 84. (b) 85. (b) 86. (b) Triangles 109

C AS E-BASED QUESTIONS 1. Read the following and answer any four questions from (i) to (v). Vijay’s House Tower Ajay’s House Vijay is trying to find the average height of a tower near his house. He is using the properties of similar triangles. The height of Vijay’s house if 20 m when Vijay’s house casts a shadow 10 m long on the ground. At the same time, the tower casts a shadow 50m long on the ground and the house of Ajay casts 20 m shadow on the ground. [CBSE Question Bank] (i) What is the height of the tower? (a) 20 m (b) 50 m (c) 100 m (d) 200 m (ii) What will be the length of the shadow of the tower when Vijay’s house casts a shadow of 12m? (a) 75 m (b) 50 m (c) 45 m (d) 60 m (iii) What is the height of Ajay’s house? (a) 30 m (b) 40 m (c) 50 m (d) 20 m (iv) When the tower casts a shadow of 40m, same time what will be the length of the shadow of Ajay’s house? (a) 16 m (b) 32 m (c) 20 m (d) 8 m (v) When the tower casts a shadow of 40m, same time what will be the length of the shadow of Vijay’s house? (a) 15 m (b) 32 m (c) 16 m (d) 8 m 2. Read the following and answer any four questions from (i) to (v). Rohan wants to measure the distance of a pond during the visit to his native. He mark points A and B on the opposite edges of a pond as shown in the figure below. To find the distance between the points, he makes a right-angled triangle using rope connecting B with another point C at a distance of 12 m, connecting C to point D at a distance of 40m from point C and then connecting D to the point A which is at a distance of 30 m from D such that ∠ADC = 90° . [CBSE Question Bank] 110 Mathematics–X: Term–1

A B 12 m C 30 m 40 m D (i) Which property of geometry will be used to find the distance AC? (a) Similarity of triangles (b) Thales Theorem (c) Pythagoras Theorem (d) Area of similar triangles (ii) What is the distance AC? (a) 50 m (b) 12 m (c) 100 m (d) 70 m (iii) Which is the following does not form a Pythagorian triplet? (a) 7,24,25 (b) 15,8,17 (c) 5,12,13 (d) 21,20,28 (iv) The length AB is (a) 12 m (b) 38 m (c) 50 m (d) 100 m (v) The length of the rope used (a) 120 m (b) 70 m (c) 82 m (d) 22 m 3. Read the following and answer any four questions from (i) to (v). SCALE FACTOR A scale drawing of an object is the same shape as the object but a different size. The scale of a drawing is a comparison of the length used on a drawing to the length it represents. The scale is written as a ratio. SIMILAR FIGURES The ratio of two corresponding sides in similar figures is called the scale factor. = Length in image Scale Factor Corresponding length in object If one shape can become another using Resizing then the shapes are similar. Translation or Slide Reflection or Flip Rotation or Turn Triangles 111

Hence, two shapes are Similar when one can become the other after a resize, flip, slide or turn. (i) A model of a boat is made on the scale of 1 : 4. The model is 120 cm long. The full size of the boat has a width of 60 cm. What is the width of the scale model? (b) 25 cm (c) 15 cm (d) 240 cm (a) 20 cm (ii) What will effect the similarity of any two polygons? (a) They are flipped horizontally (b) They are dilated by a scale factor (c) They are translated down (d) They are not the mirror image of one another (iii) If two similar triangles have a scale factor of a : b. Which statement regarding the two triangles is true? (a) The ratio of their perimeters is 3a : b (b) Their altitudes have a ratio a : b (c) Their medians have a ratio a : b (d) Their angle bisectors have a ratio a2 : b2 2 (iv) The shadow of a stick 5 m long is 2 m. At the same time the shadow of a tree 12.5 m high is Tree Stick Shadow Shadow (a) 3 m (b) 3.5 m (c) 4.5 m (d) 5 m (v) Below you see a student’s mathematical model of a farmhouse roof with measure-ments. The attic floor, ABCD in the model, is a square. The beams that support the roof are the edges of a rectangular prism, EFGHKLMN. E is the middle of AT, F is the middle of BT, G is the middle of CT, and H is the middle of DT. All the edges of the pyramid in the model have length of 12 m. T H G 12 m EF D M C N 12 m KL A 12 m B What is the length of EF, where EF is one of the horizontal edges of the block? (a) 24 m (b) 3 m (c) 6 m (d) 10 m 112 Mathematics–X: Term–1

4. Read the following and answer any four questions from (i) to (v). Two trees are standing parallel to each other. The bigger tree 8 m high, casts a shadow of 6 m. B D AC E (i) If AB and CD are the two trees and AE is the shadow of the longer tree, then (a) TAEB + TCED (b) TABE + TCED (c) TAEB + TDEC (d) TBEA + TDEC (ii) Since AB CD , so by basic proportionality theorem, we have (a) AE = BD (b) AC = DE (c) AE = AB (d) AE = BE CE DE AE BE CE CD CE DE (iii) If the ratio of the height of two trees is 3 : 1, then the shadow of the smaller tree is (a) 2 m (b) 6 m (c) 8 m (d) 8 m 3 (iv) The distance of point B from E is (a) 10 m (b) 8 m (c) 18 m (d) 10 m 3 ar (TABC) (v) If TABC + TPQR, ar (TPQR) = 4 , PQ = 10 cm, then AB is equal to 25 (a) 4 cm (b) 2 cm (c) 5 cm (d) 8 cm 5 5. Read the following and answer any four questions from (i) to (v). A ladder was placed against a wall such that it touches a point 4 m above the ground. The distance of the foot of the ladder from the bottom of the ground was 3 m. Keeping its foot at the same point, Akshay turns the ladder to the opposite side so that it reached the window of his house. E A 4m D B 3m C (i) The theorem which can be used for find the length of the ladder is (a) Thales Theorem (b) Converse of Thales Theorem (c) Pythagoras Theorem (d) Converse of Pythagoras Theorem (ii) The length of the ladder, in metre is (a) 4 m (b) 5 m (c) 9 m (d) 2 m (iii) If the window of the house is 3 m above the ground, then the distance of the point C from D is (a) 3 m (b) 4 m (c) 5 m (d) 3.5 m Triangles 113

(iv) In an isosceles right triangle PQR, right angled at P, then (a) QR2 = 2PQ2 (b) QP2 = 2PR2 (c) QP2 = 2QR2 (d) PR2 = 2QR2 (v) If OA2 = OB2 + AB2, then (a) TOBA is an equilateral triangle. (b) TOAB is an isosceles right triangle. (c) TOAB is a right triangle right angled at O. (d) TOAB is a right triangle right angled at B. 6. Read the following and answer any four questions from (i) to (v). Two buildings (say A and B) are located 12 m apart. The height of the two buildings are 32 m and 41 m. E DC AB (i) The distance between the top of the two buildings can be calculated using (a) Thales Theorem (b) Pythagoras Theorem (c) Converse of Thales Theorem (d) Converse of Pythagoras Theorem (ii) The length EF in the figure is (a) 32 m (b) 41 m (c) 41 m (d) 9 m 2 (iii) The distance DF is equal to (a) 15 m (b) 12 m (c) 9 m (d) 21 m (iv) In a triangle PQR, PQ = 7 cm, QR = 25 cm, RP = 24 cm, then the triangle is right angled at (a) P (b) Q (c) R (d) can’t say (v) ABC is an equilateral triangle of side ‘2a’ units. The length of each of its altitude is (a) a units (b) 2a units (c) 2 a units (d) 3 a units 7. Read the following and answer any four questions from (i) to (v). A farmer had a triangular piece of land. He put a fence, parallel to one of the sides of the field as shown in the figure. A DE BC 114 Mathematics–X: Term–1

(i) Which of the following statements is true? (a) AD = AE , using Thales Theorem (b) AD = AE , using Pythagoras Theorem DB EC DB EC (c) AD = AE , using Pythagoras Theorem (d) AD = AE , using Thales Theorem AB EC AB EC (ii) If the point D is 20 m away from A, where as AB and AC are 80 m and 100 m respectively, then (a) AE = 20 m (b) EC = 25 cm (c) AE = 25 cm (d) EC = 60 cm (iii) If AD = x + 1, DB = 3x – 1, AE = x + 3, EC = 3x + 4, then (a) x = 5 (b) x = 7 (c) x = 8 (d) x = 4 (iv) Which of the following is not true? (a) AD = AE (b) AD = AB (c) AB = AC (d) BD = AE AB AC AE AC BD EC AD EC (v) If P and Q are the mid points of sides YZ and XZ respectively, then (a) PQ ;; XY (b) PQ ;; YZ (c) PQ ;; ZX (d) None of these 8. Read the following and answer any four questions from (i) to (v). The ratio of two corresponding sides in similar figures is called scale factor. Scale factor = Length of image Actual length of object (i) A model of a car is made on the scale 1 : 8. The model is 40 cm long and 20 cm wide. The actual length of car is (a) 320 cm (b) 160 cm (c) 5 cm (d) 2.5 cm (ii) If two similar triangles have a scale factor of 2 : 5, then which of the following statements is true? (a) The ratio of their medians is 2 : 5. (b) The ratio of their altitudes is 5 : 2. (c) The ratio of their perimeters is 2 × 3 : 5. (d) The ratio of their altitudes is 22 : 52. (iii) The shadow of a statue 8 m long has length 5 m. At the same time the shadow of a pole 5.6 m high is (a) 3 m (b) 3.5 m (c) 4 cm (d) 4.5 m (iv) For two similar polygons which of the following is not true? (a) They are not flipped horizontally. (b) They are dilated by a scale factor. (c) They cannot be translated down. (d) They are mirror images of each other. (v) Two similar triangles have a scale factor of 1 : 2. Then their corresponding altitudes have a ratio (a) 2 : 1 (b) 4 : 1 (c) 1 : 2 (d) 1 : 1 Triangles 115

Answers 1. (i) (c) (ii) (d) (iii) (b) (iv) (a) (v) (d) 2. (i) (c) (ii) (a) (iii) (d) (iv) (b) (v) (c) 3. (i) (c) (ii) (d) (iii) (b) (iv) (d) (v) (c) 4. (i) (a) (ii) (d) (iii) (a) (iv) (a) (v) (a) 5. (i) (c) (ii) (b) (iii) (b) (iv) (a) (v) (d) 6. (i) (b) (ii) (d) (iii) (a) (iv) (a) (v) (d) 7. (i) (a) (ii) (c) (iii) (b) (iv) (d) (v) (a) 8. (i) (a) (ii) (a) (iii) (b) (iv) (d) (v) (c) ASSERTION-REASON QUESTIONS The following questions consist of two statements—Assertion(A) and Reason(R). Answer these questions selecting the appropriate option given below: (a) Both A and R are true and R is the correct explanation for A. (b) Both A and R are true and R is not the correct explanation for A. (c) A is true but R is false. (d) A is false but R is true. 1. Assertion (A) : ∆ABC ~ ∆DEF such that ar(∆ABC) = 36 cm2 and ar(∆DEF) = 49 cm2 then, AB : DE = 6 : 7. Reason (R) : If ∆ABC ~ ∆DEF, then ar (DABC) = AB2 = BC2 = AC2 ar (DDEF) DE2 EF2 DF2 2. Assertion (A) : In DABC, DE || BC such that AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) cm and EC = 3x cm than x equal to 5. Reason (R) : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distant point, than the other two sides are divided in the same ratio. 3. Assertion (A) : DABC is an isosceles triangle right angled of C, then AB2 = 2AC2. Reason (R) : In right DABC, right angled at B, AC2 = AB2 + BC2. 4. Assertion (A) : In the DABC, AB = 24 cm, BC = 10 cm and AC = 26 cm, then DABC is a right angle triangle. Reason (R) : If in two triangles, their corresponding angles are equal, then the triangles are similar. Answers 1. (a) 2. (d) 3. (a) 4. (b) HINTS/SOLUTIONS OF SELECTED MCQs 1. a DABC is a right angled triangle A D ` The line drawn from right angled vertex to the hypotenuse divides the triangle into two similar triangles. So, DABD ~ DCAD ` BADD = AD & AD2 = BD× CD. CD B C Hence, option (c) is correct. 116 Mathematics–X: Term–1

3. Since, DABC ~ DEDF Therefore the ratio of their corresponding sides are equal. But BECF ! AB & BC.DE ! AB.EF DE Hence, option (c) is correct. 7. Since, TPRQ + TXYZ  The ratio of their corresponding sides are equal. PQ QR ⇒ XZ = YZ Hence, option (c) is correct. 9. In two triangles ABC and DEF +B = +E, +C = +F and AB= 3DE So DABC ~ DDEF (By AA similarity) But both triangles are not congruent. Since there is no information about sides. Hence, option (b) is correct. 12. We have, In DPAB and DPDC BP = AP PC PD ∠APB = ∠DPC = 50° (Vertically opposite angles) ∴ DPAB ~ DPDC (By SAS similarity criteria) ∴ ∠A = ∠D = 30° In DABP, we have ∠A + ∠B + ∠P = 180° ⇒ 30° + ∠B + 50° = 180° ⇒ ∠B = 100° ⇒ ∠PBA = 100° Hence, option (d) is correct. 13. Given, ∆ABC ∼ ∆QRP ∴ ar (TABC) = BC2 ⇒ ar (TQRP) RP2 9 = (15) 2 ⇒ 15 = 3 4 (RP) 2 RP 2 ⇒ RP = 15 × 2 = 5×2 = 10 ⇒ RP = 10 cm 3 Hence, option (a) is correct. 17. We have, ar (DXYZ) = XY2 = f XY 2 A ar (DPQR) PQ2 PQ p ar (DXYZ) = e 2 2 = 4 ar (DPQR) 5 25 o Hence, option (a) is correct. 19. a D is mid point of BC BD C Let BC = a. a 2 ` BD = BC/2 & BD = E Triangles 117

ar (DABC) = 3 a2 4 ar (DBDE) = 3 d a 2 = 3 e a2 o 4 2 4 4 n Now ar (DABC) = 3 a2 = a2 ×4 = 4 = 4 :1 ar (DBDE) 4 a2 1 3 a2 44 Hence, option (d) is correct. 20. a AC is the longest side. AC2 = AB2 + BC2 & 144 = (6 3)2 + (6)2 & 144 = 108 + 36 & 144 = 144 ` +B is right angle. [By converse of Pythagoras theorem] Hence, option (c) is correct. 21. In DABC A DE|| BC Then AD = AE [By BPT] 4 cm 6 cm BD EC D E 4 = 6   ⇒  4 × 3 = BD × 6 B 3 cm BD 3 C      ⇒  BD = 4×3 = 2cm 6 ` Option (a) is correct. 22. By Pythagoras theorem in DABC C AC2 = AB2 + BC2 = (3)2 + (4)2 = 9 + 16 North AC2 = 25 4 km AC = 5 ` Option (c) is correct. East 23. Given AB = 13 cm, BC = 12 cm and AC = 5 cm A 3 km B B C Here longest side is 13 cm, So, it is the hypotenuse, R AB2 = AC2 + BC2 169 = 25 + 144 13 cm 169 = 169 12 cm So ∠C is 90° [Angle opposite to hypotenuse is 90°] ` Option (c) is correct. 24. Given ar]TABCg = 9 A 5 cm ar^TPQRh 4 Since DABC ~ DPQR A ar]TABCg = AB2 P ar^TPQRh PQ2 4 4 9 = 42 sq. units 4 PQ2 9 sq. units PQ2 × 9 = 4 × 16 B CQ 118 Mathematics–X: Term–1

⇒ PQ2 = 4 ×16 9 4 ×16 2×4 8 PQ = 9 ⇒ PQ = 3 = 3 cm Hence, option (a) is correct. 25. In DABC and DEDF, AB = DE BC FD AE B CD F In given figure included angle is ∠B and ∠D So, ∠B = ∠D B Thus option (c) is correct. 26. In ∆ABC, 6√7 cm 24 cm AB2 + AC2 = ^6 7 h2 + ]18g2 = 252 + 324 = 576 = BC2 So, by Pythagoras theorem, it is a right angle. A 18 cm C Hence, option (c) is correct. 27. a B oth, the triangles are similar, so the ratio of their areas is equal to the ratio of square of their corresponding sides. ⇒ ar_3PRQ i = f QR 2 = d 3 2 = 9 or 9 ar^3 BCAh BC 1 1 p n Hence, option (a) is correct. 28. In ∆PQR, P PS = QS = RS (Given) ...(i) In ∆PSR, 2 PS = RS S 1 ⇒ ∠1 = ∠2 ... (ii) Similarly in ∆SQR, 4 3 QS = RS Q R ⇒ ∠3 = ∠4 [Corresponding angles of equal sides are equal] Now, in ∆PQR, ∠P + ∠Q + ∠R = 180° [Angle sum property] ⇒ ∠2 + ∠4 + ∠1 + ∠3 = 180° ⇒ ∠1 + ∠3 + ∠1 + ∠3 = 180° [From equation (i) and (ii)] ⇒ 2 [∠1 + ∠3] = 180° ⇒ ∠1 + ∠3 = 90° So, ∠R = 90° In ∆PQR, by Pythagoras theorem, PR2 + QR2 = PQ2 Hence, option (c) is correct. Triangles 119

29. Two triangles DEF and PQR, in which +D = +Q and +R = +E (Given) & DDEF ~ DQRP [By AA similarity] So, ratio of corresponding sides are equal. DE = DF = EF QR QP RP Hence, DE = EF is not true. PQ RP Hence, option (b) is correct. 30. a DABC ~ DDEF (Given) & DABE = BC & AB = 3 EF 2AB EF & 12 = 3 & EF = 6 cm EF Hence, option (d) is correct. B P1 31. In DAQM AQ2 = 4 + 1 & AQ = 5 In DPQN 12 O N PQ2 = 4 + 1 & PQ = 5 Q 12 In DBPO A 2M PB2 = 4 + 1 & PB = 5 AB = AQ + QP + PB = 5 + 5 + 5 = 3 5 units Hence, option (c) is correct. 32. In DABC and DPQR P +A = +R = 25° B +B = +P = 35° 35° ` DABC ~ DRPQ (By AA similarity) 16 units 35° 4units So, ar (DABC) = AB2 = 162 ar (RPQ) PR2 42 25° 120° 120° 25° A Q R = 256 = 16 C 16 1 & ar (DABC) =16 ar (DRPQ) Hence, option (d) is correct. A 33. As we know that altitude also acts as a median of an equilateral triangle. 12 BC 2 ` BM = CM = 2 = = 6 cm In DABM AM2 = AB2 – BM2 = 122 – 62 = 144 – 36 = 108 CM B 12 cm AM = 108 = 36×3 = 6 3 cm Hence, option (d) is correct. 120 Mathematics–X: Term–1

34. Since the ratio of corresponding sides of all equilateral triangles are same. So all equilateral triangles are similar. Hence, option (a) is correct. 35. It is given that ar (DPQR) = 4 ar (DABC) 1 We know that in similar triangles ratio of areas is equal to ratio of square of its corresponding sides. So, ar (DPQR) = PQ2 = 4 ⇒ PQ = 2 ar (DABC) AB2 1 AB 1 Again ratio of perimeters is equal to ratio of corresponding sides. So, Perimeter of DPQR = PQ = 2 i.e., 2 :1 Perimeter of DABC AB 1 Hence, option (a) is correct. B 36. Three sides are a, 3 a and 2 a , let AB = 2 a, AC = a, BC = 3 a Here, AB2 + AC2 = BC2 23 & ( 2 a)2 + (a)2 = ( 3 a)2 & 2a2 + a2 = 3a2 & 3a2 = 3a2 Here sum of squares of two sides is equal to square of third side. A C So, the triangle is right angled triangle. 10 cm P & Angle opposite to longest side is 90°. 9 cm Hence, option (b) is correct. 6 cm B 37. In DAOQ and DBOP, we have +AOQ = +BOP (Vertically opposite angle) +OAQ = +OBP (Both of 90°) A O ` DAOQ ~ DBOP (By AA similarity) So AO = AQ & 10 = AQ & AQ = 10×9 = 15 cm BO PB 6 9 6 Hence, option (a) is correct. Q 38. As we know that ratio of area of two triangles is equal to ratio of square of medians of both the triangle. So, area of triagle 1 = (median 1)2 area of triangle 2 (median 2)2 & 18414 = (16) 2 & (median 2)2 = 81× 256 = 144 (median) 2 144 ⇒ median 2 = 144 = 12 cm Hence, option (c) is correct. 39. In DPQR , PO = QO = RO (Given) In DPOR, OP = OR & +1 = +2 …(i) (Equal sides have equal opposite angles) Similarly in DROQ +3 = +4 …(ii) Now in DPQR, Triangles 121

+P + +Q + +R = 180° 2 R 4 P 1 Q & +2 + +4 + +1 + +3 = 180° & +1 + +3 + +1 + +3 = 180° 3 & 2 (+1 + +3) = 180° & +1 + +3 = 90° O & DPQR is right angled triangle. So, PQ2 = PR2 + QR2 Hence, option (b) is correct. 40. From the correspondence of sides we can clearly say that DABC ~ DDFE. Hence, option (d) is correct. 41. As we know ratio of areas of two similar triangles is equal to ratio of squares of its corresponding sSiod,e s.aarr (DDEF) = EF2 =d EF 2 = d 5 2 = 25 = 25 (DABC) BC2 BC n 1 1 n Hence, option (b) is correct. 42. a DABC ~ DDEF ` Ratio of corresponding sides are equal. So, AB = BC = AC & BC.DF = AC.EF is true DE EF DF or AB.DF = AC.DE is true or BC.DE = AB.EF is true But BC.DE = AB.FD is not correct. Hence, option (d) is correct. 43. Given that AD = AE and +ADE = 70° A BD EC By BPT, DE ; ; BC 70° E D & +AED = +ACB (Corresponding angles) = 50° In DADE +A + +ADE + +AED = 180° (By ASP) 50° BC & +A = 180° – 70° – 50° & +A = 60° & +BAC = 60° A Hence, option (d) is correct. 3 cm 5 cm 44. a CE = 4 = 0.8 and D AE 5 E CBFF = 2 = 20 = 4 = 0.8 2.5 25 5 4 cm & CAEE CF 4 cm BF & = EF ;; AB [By converse of BPT] Hence, option (c) is correct. B 2.5 cm F 2 cm C 45. In DAPB and DDPC , we have AD +A = +D (Both of 90°) P +APB = +DPC (Vertically opposite angles) ` DAPB ~ DDPC (By AA similarty) & DAPP = PB & AP× PC = DP × PB B C PC Hence, option (d) is correct. 122 Mathematics–X: Term–1

46. Given AB = 4BX and YC = 2 cm A XY a XY ; ; BC ` BAXB = AC = 4 & AC = 4 CY 1 CY 1 AC = 4 [a YC = 2 cm] 2 1 & AC = 8 cm B C A Now AY = AC – CY & (8 – 2) cm & 6 cm Hence, option (b) is correct. 47. a D, E and F are mid points of BC, AC and AB respectively. ` DE = 1 AB, EF = 1 BC and DF = 1 AC 2 2 2 and DE ; ; AB, EF ; ; BC, DF ; ; AC FE Therefore DDEF ~ DABC So, ar (DDEF) = EF2 = EF2 ar (DABC) BC2 (2EF) 2 BD C = EF2 = 1 or 1: 4. 4EF2 4 Hence, option (b) is correct. 48. In DABC a AB ; ; XY Z D Y C ` BXXC = AY (By BPT) …(i) A X YC In parallelogram ABCD, AB ; ; CD ; ; XZ In DACD, a CD ; ; YZ B & YACY = AZ (By BPT) …(ii) ZD From (i) and (ii), we get BX = AZ XC ZD Hence, option (c) is correct. 49. a Diagonals of rhombus bisect each other at 90°. ` DAOB, DAOD, DBOC and DCOD are right angled triangles. 1 1 Here, OA = OC = 2 AC and OB = OD = 2 BD In DAOB, AB AB2 = OA2 + OB2 O DC & AB2 = d AC 2 + d BD 2 2 2 Triangles 123 n n & AB2 = AC2 + BD2 4 4 & 4AB2 = AC2 + BD2

& AB2 + AB2 + AB2 + AB2 = AC2 + BD2 & AB2 + BC2 + CD2 + AD2 = AC2 + BD2 Hence, option (c) is correct. C 50. Here AD = BC In DABD AB2 = AD2 + BD2 …(i) In DACD AC2 = AD2 + CD2 … (ii) D A B From (i) and (ii) AB2 = (AC2 – CD2) + BD2 [Put value of AD2] & AB2 + CD2 = AC2 + BD2 Hence, option (c) is correct. 51. All right angle triangles should satisfy the Pythagoras Theorem So, h2 = p2 + b2 (34)2 = (30)2 + (16)2 ⇒ 1156 = 900 + 256 & 1156 = 1156 So, sides of a right triangle are 34 cm, 30 cm and 16 cm Hence, option (b) is correct. 52. We know from theorem that if two triangles are similar then ratio of area of these triangles is equal to ratio of square of their corresponding sides. Hence, option (c) is correct. 53. As we know that in two similar triangles ratio of area of two triangles is equal to ratio of square of their corresponding medians So, Area of triangle 1 = (median1) 2 Area of triangle 2 (median2) 2 & 16444 = (12) 2 & (median 2)2 = 64 ×144 (median 2)2 144 median 2 = 8 cm Hence, option (b) is correct. 54. In DABC, AD is the bisectors of +A . AB BD ` AC = DC … (i) A LD From A draw AL = BC 1 ` aarr (DABD) = 2 BD × AL = BD (DACD) CD 1 CD×AL 2 (DABD) & aarr (DACD) = AB [From (i)] AC B C Hence, option (d) is correct. 55. ABCD is a square and APC is a triangle described on diagonal and BCQ, is a triangle described on a side of square. Let each side of square be of length a unit. 124 Mathematics–X: Term–1

Diagonal AC = a2 + a2 = a 2 units P C a Both triangles are equilateral triangles. D ` aarr((DDBACPQC)) = 3 × (a 2)2 = a2 ×2 = 2 i.e., 2 :1 4 a2 1 3 Q 4 a2 C Hence, option (c) is correct. 56. a Perpendicular of an equilateral triangle act as a median. A B BC B A ` BD = CD = 2 D In DABD, A AB2 = AD2 + BD2 2 BC2 BC 4 AB2 = AD2 +d 2 n & AB2 = AD2 + & AB2 = AD2 + AB2 [a AB = BC = AC] 4 4AD2 + AB2 & AB2 = 4 & 3AB2 = 4AD2 Hence, option (c) is correct. 57. a Perpendicular of an equilateral triangle act as a median. ` BD = CD = BC/2 In DADC AC2 = AD2 + CD2 B D C & BC2 = AD2 + CD2 [a AB = BC = AC] B & (2CD)2 = AD2 + CD2 & 4CD2 = AD2 + CD2 & AD2 = 3CD2 Hence, option (c) is correct. 58. As we know that the perpendicular from the right angle vertex on hypotenuse divides the triangle into two similar triangles. ` DABD ~ DACD D C Ratio of corresponding sides are equal. BD AB CD = AC Hence, option (b) is correct. A 59. Since we know that all the congruent figures are similar. Therefore only Vatsal is correct. ` Option (a) is correct. 60. We know that all circles are similar, all squares are also similar because their angles are equal but all right triagles are not congruent because different right triangles can have different sides. Also, all equilateral triangles are not congruent due to same reason. ` Option (a) is correct. 61. We have, AB = 3 m, BC = 2 m, CD = 5 m, DA = 1 m and PQ = 6 m, QR = 4 m, RS = 10 m, SP = 2 m ` AB = BC = CD = DA = 1 PQ QR RS SP 2 Triangles 125

` Quadrilateral ABCD ~ 4PQRS ` +A = +P and +D = +S (Corresponding angles) & 130° = y and 58° = x ` x = 58° and y =130° ` Option (a) is correct. 62. For two quadrilaterals to be similar whose diagonals bisect each other, the diagonals should bisected at right angles and their adjacent angles should be equal. ` Option (c) is correct. 63. In DAOB and DDOC, we have AO = 1.6 = 2 and OOBC = 4.8 =2 OD 0.8 2.4 ` AO = OB OD OC +AOB = +COD (Vertically opposite angles are equal) ` DAOB ~ DDOC (By SAS similarity criteria) Thus, option (d) is correct. 64. Statement (i), +B = 70 ` +A = 180° – (60° + 70°) = 50° & 7x + 1 = 50 & 7x = 49 & x = 7 ` +D = 3x + 29 = 3×7 + 29 = 21 + 29 = 50 ` +A = +D and +AOB = +COD = 60° Also third angle +B = +C = 70° DAOB~DDCO (By AAA similarity criteria) Now, if we take statement (ii), +C = 70°, then also same result will be found i.e., DAOB ~ DDOC Thus, option (c) is correct, i.e; each statement alone is sufficient. 65. We have In DABC PQ ; ; BC ` AQ = AP & 1Q.B3 = 2.4 = 24 QB PC 3.3 33 & 1.3 = 8 & QB = 1.3×11 = 1.78 = 1.8 cm QB 11 8 ` Option (c) is correct. 66. It is given that QR ; ; AB, RP ; ; BD and, CQ = x + 2, QA = x, CP = 5x + 4, PD = 3x, Now, In DABC, QR ; ; AB ` CQ = CR … (i) AQ RB In DBCD, RP ; ; BD ` CR = CP …(ii) RB PD 126 Mathematics–X: Term–1

From (i) and (ii), we have CQ = CP & x +2 = 5x + 4 & x+2 = 5x + 4 AQ PD x 3x 1 3 & 3x + 6 = 5x + 4 & 2 = 2x & x = 1 ` Option (a) is correct. 67. We have, PQ = 3.2, PR = 4 and +P = y Also, EF = 4.8, DE = 6 and +E = y In DPQR and DEFD, we have PQ = 3.2 = 2 , PR = 4 = 2 EF 4.8 3 DE 6 3 and +P = +E = y & PQ = PR and +P = +E EF DE ` Both triangles are similar by SAS similarity criteria. ` Option (b) is correct. 68. In DPQR we have PR = QR & +Q = +P = 40° +R = 180° – (+P + +Q) = 180° – (40° + 40°) = 100° ` +R = 100° 5 1 AC 5 1 AB AC 1 AB 10 2 QR 10 2 PR QR 2 We have, PR = = , = = & = = Condition for both be triangles to be similar is +A = +R = 100° ` Option (b) is correct. 69. Let AB be position of Rohit and PQ be the pole. Now, at the same instant +C = +R +B = +Q = 90° P Q +C = +R A ` DABC ~ DPQR (By AA similarity criteria) E ` AB = BC (Corresponding sides) 6 ft PQ QR B 5 ft & 6 = 5 & PQ = 36 ft . C 30 ft R PQ 30 D ` Height of the pole be 36 ft F ` Option (d) is correct. Triangles 127 70. Let h feet be the height of the building. Since both the triangle formed by Ankit and a building, are similar. 5 h 2 = 48 & h = 24 ×5 = 120 feet A ` Option (b) is correct. 71. Let DABC and DDEF are two similar triangles. ` ar (DABC) = AB2 = BC2 = AC2 ar (DDEF) DE2 EF2 DF2 & a = AB2 B C k2 a DE2

& 1 = AB2 & AB = 1 k2 DE2 DE k ` AB : DE =1: k ` Option (a) is correct. 72. In DABC, PQ ; ; BC ` DAPQ ~ DABC (By AAA similarity criteria) ` Perimeter of DABC = 3 & AB = 3 … (i) Perimeter of DAPQ 1 AP 1 Now, area of DABC = AB2 = d 3 2 & ar (DABC) = 9 area of DAPQ AP2 1 ar (DAPQ) n & ar (DABC) = 9× ar (DAPQ) a ar (DAPQ) is whole number. ` ar (DABC) is a multiple of 9. In the given option it is only 99. ` Option (c) is correct. 73. We have DABC ~ DPQR and ar (DABC) = 25 & AM2 = 25 = 52 ar (DPQR) 144 PN2 144 122 AM = 5 PN 12 ` Option (a) is correct. 74. Let x cm be length of corresponding sides of larger triangle. ` (15) 2 = 9 & 15 = 3 & x = 60 = 20 cm x2 16 x 4 3 Difference between both sides = 20 – 15 = 5 cm ` Option (d) is correct. 75. We have In DAPB and DABC +APB = +ABC = 90° +BAP = +BAC (Common) ` DAPB ~ DABC (By AA similarity criteria) ` AB = AP & AB2 = AC $ AP & 25 = (2x + 5) . x AC AB & 25 = 2x2 + 5x & 2x2 + 5x – 25 = 0 & 2x2 + 10x – 5x – 25 = 0 & 2x (x + 5) – 5 (x + 5) = 0 & (x + 5) (2x – 5) = 0 & 2x – 5 = 0 (a x + 5 ! 0 & x ! – 5 and length cannot be negative) ` x = 5 = 2.5 2 Length of PC = x + 5 = 2.5 + 5 = 7.5 cm ` Option (d) is correct. 76. We have In DAPB and DABC +APB = +ABC = 90 128 Mathematics–X: Term–1

+BAP = +BAC (Common) ` DAPB ~ DABC (By AA similarity criteria) …(i) Similarly, DBPC ~ DABC (BY AA similarity criteria) ...(ii) BC = AC & BC2 = CP $ AC PC BC Also DAPB ~ DBPC (From (i) and (ii)) So, all the option (a), (b) and (c) are correct. The only option (d) is not true. Thus, option (d) is correct. 77. To prove AB2 + BC2 = AC2 In right DABC, we have to prove first DAPB ~ DABC and again DBPC ~ DABC. Mr Shah follows steps to prove DAPB ~ DABC and AP = AB & AB2 = AP.AC AB AC So the next step be to prove DBPC ~ DABC . ` Option (c) is correct. W N Z X 112 m 78. In right DXYZ, we have 15 m XZ2 = XY2 + YZ2 YE & XZ2 = (112)2 + (15)2 = 12544 + 225 & XZ2 = 12769 ` XZ = 12769 = 113 m ` Required distance = 113 m ` Option (a) is correct. S 79. Option (b) is correct. Because its given length of sides are 7 cm, 24 cm, 25 cm ` (7)2 + (24)2 = (25)2 It satisfies Pythagoras Theorem. 80. We have DABC with sides m, n and r. A mn So, claim (i): ABC is a right triangle provided n2 – &m2 = r2 n2 = r2 + m2 (Pythagoras theorem) ` Claim (i) would be correct. Also, triangle with side lengths 2m, 2n and 2r is a right angle triangle, if B r C DABC is right triangle. ` Claim (iii) would be correct. ` Option (c) is correct. 81. We have, ar (DAEF) = e 1 2 = 1 ⇒ ar (DDEF) = 1 [ ar(DAEF) = ar(DDEF)] ar (DABC) 2 4 ar (DABC) 4 o Hence, option (b) is correct. 82. Since D ABC ∼ D DEF ∴ ar (TABC) = AB2 = d AB 2 = e 1.2 2 ar (TDEF) DE2 DE 1.4 n o   = c 12 2 = d 6 2 = 36 14 7 49 m n ∴ Option (d) is correct. Triangles 129

84. We have, A 2.7 cm In D ABC, DE || BC DE ⇒ AD = AE ⇒ 3 = 2.7 DB EC 2 EC BC ⇒ EC = 2.7 × 2 = 1.8 cm 3 ∴ Option (b) is correct. 85. Given, DABC ~ DDFE, \\ ∠A = ∠D, ∠B = ∠F and ∠C = ∠E ⇒ ∠D = 30°, ∠E = 50° ⇒ ∠F =180° – (30° + 50°) = 100°  (Using angle sum property) Also, AB = AC ⇒ 5 = 8 ⇒ DE = 12 cm DF DE 7.5 DE \\ DE = 12 cm and ∠F = 100° Hence, option (b) is correct. 86. Given, AB = AC AE EF DE For two triangles (ABC and EFD) to be similar when ∠A = ∠E (Corresponding angles) ∴ Option (b) is correct. B CF D SOLUTIONS OF CASE-BASED QUESTIONS 1. (i) Let h m be the height of tower, therefore using property of similar triangle between two triangle i.e. for Vijay’s house and for tower with shadows. We have, 20 = 10 ⇒ h = 100 m h 50 ∴ Height of tower is 100 m. ∴ Option (c) is correct. (ii) When Vijay’s house casts a shadow of 12 m, we have 20 = 12 , where x is the length of shadow of tower. h x ⇒ 20 = 12 ⇒ x = 60 m 100 x ∴ Option (d) is correct. (iii) Let H m be the height of Ajay’s house. ∴ 20 = H ⇒ H = 40 m 10 20 ∴ Option (b) is correct. (iv) When the tower casts a shadow of 40 m. ∴ 100 = Height of Ajay's house 40 Length of shadow of Ajay's house ⇒ 5 = 40 2 Shadow length 130 Mathematics–X: Term–1

⇒ Length of shadow of Ajay house = 16 m. ∴ Option (a) is correct. (v) We have, 100 = Height of Vijay's house ⇒ 5 = Length 20 shadow 40 Length of its shadow 2 of its ⇒ Length of shadow of Vijay house = 8 m ∴ Option (d) is correct. 2. (i) Pythagoras Theorem will be used to find the distance AC in right angle ∆ACD. ∴ Option (c) is correct. (ii) In right angle ∆ACD, we have AC2 = AD2 + CD2 (Pythagoras Theorem) ⇒ AC2 = (30)2 + (40)2 = 900 + 1600 = 2500 ⇒ AC = 50 m ∴ Option (a) is correct. (iii) Here, (21, 20, 28) is not a Pythagorean triplet because (21)2 + (20)2 ≠ (28)2 ⇒ 441 + 400 ≠ 784 ⇒ 841 ≠ 784 ∴ Option (d) is correct. (iv) AB = AC – BC = 50 – 12 = 38 m ∴ Option (b) is correct. (v) Total length of rope used = BC + CD + DA = 12 + 40 + 30 = 82 m ∴ Option (c) is correct. 3. (i) S­ince scale is 1 : 4. Let the width of the scale model is x cm. So 4x = 60 ⇒ x = 60 =15 cm 4 ∴ Option (c) is correct. (ii) They are not the mirror image of one another. ∴ Option (d) is correct. (iii) Since the scale factor is a : b then their altitudes have the ratio a : b. ∴ Option (b) is correct. (iv) By basic proportionality theorem 5 = 2 ⇒ x = 25 =5 m 12.5 x 5 ∴ Option (d) is correct. (v) E is the mid point of AT i.e; ET = AT =6 m 2 F is the mid point of BT i.e; FT = BT =6 m 2 Now in ∆TEF and ∆TAB a Both are similar. (By AA similarity) ∴ ET = EF ⇒ 6 = EF ⇒ EF = 6 m­ AT AB 12 12 ∴ Option (c) is correct. Triangles 131

4. (i) In DAEB and DCED, ∠E = ∠E (Common) ∠EDC = ∠EBA (Corresponding angles) DAEB ~ DCED (By AA similarity) ∴ Option (a) is correct. (ii) AE = BE (By BPT) CE DE ∴ Option (d) is correct. (iii) We have, TAEB + TCED & AB = AE & 3 = 6 & CE = 6 =2m CD CE 1 CE 3 ∴ Option (a) is correct. (iv) In right triangle ABE BE = AB2 + AE2 (Pythagoras theorem) = 82 + 62 = 64 + 36 = 10 m ∴ Option (a) is correct. ar (DABC) AB2 (v) DABC ~ DPQR & ar (DPQR) = PQ2 & 4 = AB2 & AB2 = 4 ×100 = 16 & AB2 = 16 cm & AB= 4 cm 25 102 25 ∴ Option (a) is correct. 5. (i) Since the ladder placed against wall forms a right angled triangle, so length of ladder can be found using Pythagoras theorem. ∴ Option (c) is correct. (ii) In DABC, AB2 = 42 + 32 = 16 + 9 = 25 & Length of ladder (AB) = 5 m ∴ Option (b) is correct. (iii) Let C be the window, CE be the ladder. C 5m 3m ∴ In ∆CDE ED2 = CE2 – CD2 = 25 – 9 = 16 ED = 4 m ∴ Option (b) is correct. ED (iv) In right DPQR R RQ2 = RP2 + PQ2 (By Pythagoras theorem) = PQ2 + PQ2 (a RP = PQ) = 2PQ2 ∴ Option (a) is correct. PQ (v) OA2 = OB2 + BA2 O B A 132 Mathematics–X: Term–1

&   DOAB is a right triangle, right angled at B (Converse of Pythagoras theorem) ∴ Option (d) is correct. 6. (i) Since a right triangle is formed, so distance between two buildings can be calculated using Pythagoras theorem. ∴ Option (b) is correct. (ii) EF = BF – BE = 41 – 32 = 9 m ∴ Option (d) is correct. (iii) DF = DE2 + EF2 (Using Pythagoras theorem) = 122 + 92 = 144 + 81 = 225 = 15 m Option (a) is correct. (iv) Longest side = QR (25 cm) & +P is a right angle. [ a Angle opposite to longest side is equal to 90°.] ∴ Option (a) is correct. A 2a (v) Altitude, AD = AB2 – BD2 = (2a)2 – a2 = 4a2 – a2 = 3 a units Option (d) is correct. B a DaC 7. (i) AD = AE [By Thales theorem] A DB EC E Option (a) is correct C (ii) Since DE ;; BC, AD = AE [By Thales Theorem] D AB AC 20 AE 1 & 80 = 100 & AE = 4 ×100 = 25 cm B ∴ Option (c) is correct. (iii) We have AD = AE (By Thales Theorem) DB EC x+1 x+3 & 3x – 1 = 3x + 4 & (x + 1) (3x + 4) = (x + 3) (3x – 1) & 3x2 + 4x + 3x + 4 = 3x2 – x + 9x – 3 & 7x – 8x = –3 – 4 & x = 7 ∴ Option (b) is correct. (iv) Out of the given options, BD ! AE AD EC Option (d) is correct. (v) P and Q are midpoints of sides YZ and XZ respectively. Z ` YPPZ = 1, XQ = 1 & YP = XQ QZ PZ QZ & PQ ; ; YX ∴ Option (a) is correct. PQ 8. (i) Given scale = 1 : 8 Let actual length of car = x cm. Then 40 = 1 & x = 320 cm YX x 8 ∴ Option (a) is correct. Triangles 133

(ii) Ratio of medians = 2 : 5, since two similar triangles have ratio 2 : 5. ∴ Option (a) is correct (iii) Length of statue = Length of pole Length of shadow of statue Length of its shadow & 8 = 5.6 & x= 5.6 × 5 = 3.5 m 5 x 8 ∴ Option (b) is correct. (iv) Two similar polygons are not the mirror images of each other. So, (d) is correct option. (v) The ratio of corresponding altitude of two similar triangles is equal to their scale factor. So, (c) is correct option. SOLUTIONS OF ASSERTION-REASON QUESTIONS 1. ar]DABCg = AB2 & 36 = AB2 ar (DDEF) DE2 & 49 DE2 & AB = 6 AB : DE = 6 : 7 DE 7 So, both A and R are correct and R explain A. Hence, option (a) is correct. 2. We have, AD = AE & 7x – 4 = 5x – 2 DB EC 3x + 4 3x ⇒ 21x2 – 12x = 15x2 + 20x – 6x – 8 ⇒ 6x2 – 26x + 8 = 0 3x2 – 12x – x + 4 = 0 ⇒ 3x2 – 13x + 4 = 0 ⇒ ⇒ 3x(x – 4) –1 (x – 4) = 0 ⇒ (x – 4)(3x –1) = 0 ⇒ x = 4, 1 3 So, A is incorrect but R is correct. Hence, option (d) is correct. 3. In an isosceles DABC, right angled at C is AB2 = AC2 + BC2 ⇒ AB2 = AC2 + AC2 ⇒ AB2 = 2AC2 (AC = BC) So, both A and R are correct and R explains A. Hence, option (a) is correct. 4. We have, AB2 + BC2 = (24)2 + (10)2 = 576 + 100 = 676 = AC2 AB2 + BC2 = AC2 ∴ ABC is a right angled triangle. Also, two triangles are similar if their corresponding angles are equal. So, both A and R are correct but R does not explain A. Hence, option (b) is correct. zzz 134 Mathematics–X: Term–1

6 INTRODUCTION TO TRIGONOMETRY BASIC CONCEPTS & FORMULAE 1. Trigonometry is the branch of mathematics which deals with the measurement of sides and angles of the triangles. Trigonometric ratios: C Let ABC be a right-triangle, right angled at B. Let ∠CAB = q. Then, sin i = BC cos i = AB tan i = BC θ B AC AC AB A cot i = AB sec i = AC cosec i = AC BC AB BC Relation between trigonometric ratios: (i) Reciprocal Relations sin i = 1 ⇒ cosec i = 1 ⇒ sin q . cosec q = 1 cosec i sin i cos i = 1 ⇒ sec i = 1 ⇒ cos q . sec q = 1 sec i cos i tan i = 1 ⇒ cot i = 1 ⇒ tan q . cot q = 1 cot i tan i (ii) Quotient Relations tan i = sin i and cot i = cos i cos i sin i 2. An expression having equal to sign (=) is called an equation. 3. An equation which involves trigonometric ratios of an angle and is true for all values of the angle is called a trigonometric identity. Some common trigonometric identites are (i) sin2 q + cos2 q = 1 for 0° ≤ q ≤ 90° (ii) sec2 q = 1 + tan2 q for 0° ≤ q ≤ 90° (iii) cosec2 q = 1 + cot2 q for 0° ≤ q ≤ 90° 4. Values of Trigonometric Ratios of Standard Angles 30° 45° 60° sin q 1 13 2 22 cos q 3 1 1 2 2 2 Introduction to Trigonometry 135

tan q 1 1 3 cot q 3 1 1 3 3 sec q 2 22 3 cosec q 2 2 2 3 MULTIPLE CHOICE QUESTIONS Choose and write the correct option in the following questions. 1. The value of (sin 30° + cos 30°) – (sin 60° + cos 60°) is  [NCERT Exemplar] [NCERT Exemplar] (a) – 1 (b) 0 (c) 1 (d) 2 2. The value of tan 30° is cot 60° 1 1 (a) 2 (b) 3 (c) 3 (d) 1 3. The value of (sin 45° + cos 45°) is [NCERT Exemplar] (a) 1 (b) 2 (c) 3 (d) 1 2 2 4. If 6cot θ + 2cosec θ = cot θ + 5cosec θ, then cos θ is (a) 4 (b) 5 (c) 3 (d) 5 5 3 5 4 2 tan 30° 5. The value of 1 – tan2 30° is equal to (a) cos 60° (b) sin 60° (c) tan 60° (d) sin 30° 6. If sin A = 1 , then the value of cot A is [NCERT Exemplar] 2 (a) 3 (b) 1 (c) 3 (d) 1 3 2 7. Given that sin i = a , then cos θ is equal to  [NCERT Exemplar] b (a) b (b) b (c) b2 – a2 (d) a b2 – a2 a b b2 – a2 8. If sin A + sin2 A = 1, then the value of the expression (cos2A + cos4A) is  [NCERT Exemplar] (a) 1 (b) 1 (c) 2 (d) 3 2 9. If 4 tan θ = 3, then d 4 sin i – cos i n is equal to  [NCERT Exemplar] 4 sin i + cos i (a) 2 (b) 1 (c) 1 (d) 3 3 3 2 4 136 Mathematics–X: Term–1

10. Value of sin 30° tan 45° is (a) 3 (b) 2 (c) 1 (d) 0 2 2 11. If cos A = 4 , then the value of tan A is [NCERT Exemplar] 5 3 3 5 (a) 5 (b) 4 (c) 4 (d) 3 3 12. The value of tan2 60° – sin2 30° is tan2 45° + cos2 30° 7 11 13 11 (a) 11 (b) 13 (c) 11 (d) 7 13. If x = r sin q and y = r cos q then the value of x2 + y2 is (a) r (b) r2 (c) 1 (d) 1 r 14. If 3 sec i – 5= 0 then cot i is equal to (a) 5 (b) 4 (c) 3 (d) 3 3 5 4 5 15. If a + b = 90° and a = 2b, then cos2 a + sin2 b is equal to (a) 1 (b) 1 (c) 0 (d) 2 2 16. If tan x + sin x = m and tan x – sin x = n then m2 – n2 is equal to (a) 4 mn (b) mn (c) 2 mn (d) none of them 17. If sec i + tan i = x, the value of sec i is (a) 1 d x – 1 n (b) x2 – 1 (c) 1 dx + 1 n (d) none of them 2 x x2 + 1 2 x 18. The value of 1 + cos i is 1 – cos i (a) cot i – cosec i (b) cosec i + cot i (c) cosec2 i + cot2 i (d) (cot i + cosec i)2 19. sin i + 1 cos i is equal to 1 – cot i – tan i (a) 0 (b) 1 (c) sin i + cos i (d) sin i – cos i 20. Given that sin i = a , then tan i is equal to b (a) b (b) b2 – a2 (c) a (d) b2 – a2 b2 – b b2 – a a2 a2 21. Considering the diagram below. P x M y Q R Introduction to Trigonometry 137

Which of the following statements is true? [CBSE Question Bank] (a) Side PR is adjacent to ∠y in triangle PMR and side QR is adjacent to ∠y in triangle PQR (b) Side MR is adjacent to ∠y in triangle PMR and side PR is adjacent to ∠y in triangle PQR (c) Side PR is adjacent to ∠y in triangle PMR and side MR is adjacent to ∠y in triangle PQR (d) Side PR is adjacent to ∠y in triangle PMR and triangle PQR 22. Consider the triangle shown below. X 15 Y θ 8 17 Z What are the values of tan q, cosec q and sec q? [CBSE Question Bank] (a) tan i = 8 , cosec i = 17 , sec i = 17 (b) tan i = 8 , cosec i = 17 , sec i = 17 15 8 15 15 15 8 (c) tan i = 17 , cosec i = 8 , sec i = 17 (d) tan i = 8 , cosec i = 17 , sec i = 8 15 15 8 15 15 17 23. If sin q = 7 , what are the values of tan q, cos q and cosec q? [CBSE Question Bank] 85 (a) tan i = 6 , cos i = 7 and cosec i = 85 (b) tan i= 7 , cos i= 7 and cosec i = 85 7 85 7 6 85 7 (c) tan i = 7 , cos i = 6 and cosec i = 85 (d) tan i= 7 , cos i= 6 and cosec i = 85 6 85 7 6 85 6 24. The two legs AB and BC of right triangle ABC are in a ratio 1 : 3. What will be the value of sin C? [CBSE Question Bank] (a) 10 (b) 1 (c) 3 (d) 1 10 10 2 25. What is the value of 3 – sin2 60° ? [CBSE Question Bank] tan 30° tan 60° (a) 2 1 (b) 3 1 (c) 2 3 (d) 3 3 4 4 4 4 4 – sin245° 26. The value of cot k tan 60° is 3.5. [CBSE Question Bank] What is the value of k? (a) 30° (b) 45° (c) 60° (d) 90° [CBSE Question Bank] 27. Which of these is equivalent to 2 tan x (sec2x – 1) ? cos3x (d) 2 cot3 x sec3 x (a) 2 tan3 x cosec x (b) 2 cot3 x cosec3 x (c) 2 tan3 x sec3 x 28. Which of the following option makes the statement below true? 1 sec x + sec x x [CBSE Question Bank] cos2 x – 1 – tan2 (d) −sec x cot x (a) −cosec x tan x (b) −sec x tan x (c) −cosec x cot x 138 Mathematics–X: Term–1

29. If sin i = 1 , the value of (2cot2 q + 2) is 3 (a) 16 (b) 20 (c) 12 (d) 18 30. If x = a cos q and y = b sin q, then the value of b2x2 + a2y2 is (a) a2 + b2 (b) a2 (c) a2b2 (d) None of these b2 31. If 3 tan q = 3 sin q, (q ≠ 0) then the value of sin2 q – cos2 q is (a) 1 (b) 3 (c) 2 (d) 3 3 3 2 32. (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) is equal to (a) 0 (b) 1 (c) –1 (d) none of these 33. (sec A + tan A) (1 – sin A) is equal to (a) sec A (b) sin A (c) cosec A (d) cos A 34. Given that sin a = 1 and cos b= 1 , then the value of (a + b) is 2 2 (a) 0° (b) 30° (c) 60° (d) 90° 35. If sin q + cos q = 2 cos q, (q ≠ 90) then the value of tan q is [CBSE Sample Question Paper 2020] (a) 2 – 1 (b) 2 +1 (c) 2 (d) – 2 36. If 2x = sec θ and 2 = tan i, then the value of d x2 – 1 n is x x2 (a) 4 (b) 1 (c) 2 (d) 1 4 2 37. The value of (1 + cot q – cosec q)(1 + tan q + sec q) is (a) 1 (b) 2 (c) –1 (d) None of these 38. If sin 77° = x, than the value of tan 77° is (a) 1 (b) x (c) x x2 . (d) None of these 1+ x2 1 + x2 1– 39. sec4 A – sec2 A is equal to (a) tan2A – tan4A (b) tan4A – tan2A (c) tan4A + tan2A (d) None of these 40. If a cos θ + b sin θ = m and a sin θ – b cos θ = n, then a2 + b2 is equal to (a) m2 – n2 (b) m2n2 (c) n2 – m2 (d) m2 + n2 41. If cos A + cos2A = 1, then sin2A + sin4A is equal to (a) –1 (b) 0 (c) 1 (d) None of these 42. The value of the expression sin 60° is cos 30° (a) 3 (b) 1 (c) 1 (d) 2 2 2 43. If 4 tan x = 3, than cos x + sin x is equal to cos x – sin x (a) 7 (b) 1 (c) – 7 (d) – 1 7 7 44. cos4 A – sin4 A is equal to (a) 2cos2A + 1 (b) 2cos2A – 1 (c) 2sin2A – 1 (d) 2sin2 A + 1 Introduction to Trigonometry 139

45. If tan A= 3 , then the value of cos A is 2 (a) 3 (b) 2 (c) 2 (d) 13 13 13 3 2 46. 1 + tan2 A is equal to 1 + cot2 A (a) sec2A (b) –1 (c) cot2 A (d) tan2 A sin i 47. 1 + cos i is equal to 1 + cos i 1 + cos i 1– cos i 1– sin i (a) sin i (b) cos i (c) sin i (d) cos i Answers 2. (d) 3. (b) 4. (c) 5. (c) 6. (a) 8. (a) 9. (c) 10. (c) 11. (b) 12. (d) 1. (b) 14. (c) 15. (b) 16. (a) 17. (c) 18. (b) 7. (c) 20. (c) 21. (b) 22. (a) 23. (c) 24. (b) 13. (b) 26. (c) 27. (c) 28. (c) 29. (d) 30. (c) 19. (c) 32. (b) 33. (d) 34. (d) 35. (a) 36. (b) 25. (a) 38. (c) 39. (c) 40. (d) 41. (c) 42. (c) 31. (a) 44. (b) 45. (b) 46. (d) 47. (c) 37. (b) 43. (a) ASSERTION-REASON QUESTIONS The following questions consist of two statements—Assertion(A) and Reason(R). Answer these questions selecting the appropriate option given below: (a) Both A and R are true and R is the correct explanation for A. (b) Both A and R are true and R is not the correct explanation for A. (c) A is true but R is false. (d) A is false but R is true. 1. Assertion (A) : sin2 67° + cos2 67° = 1 Reason (R) : For any value of θ, sin2 θ + cos2 θ = 1. 2. Assertion (A) : If cos A + cos2 A = 1 then sin2 A + sin4 A = 2. Reason (R) : 1 – sin2 A = cos2 A, for any value of A. 3. Assertion (A) : In a right angled triangle, if tan θ = 3 then sin i = 3 . 4 5 1 Reason (R) : sin 60° = 2 4. Assertion (A) : In a right angled triangle, if cos i = 1 and sin i = 3 , then tan θ = 3. 2 2 sin i Reason (R) : tan i= cos i Answers 1. (a) 2. (d) 3. (c) 4. (a) 140 Mathematics–X: Term–1

HINTS/SOLUTIONS OF SELECTED MCQs 2. We have, 1 tan 30° = 3 =1 cot 60° 1 3 ` Option (d) is correct. 3. We have, sin 45° + cos 45° = 1+ 1= 2= 2 2 2 2 ∴ Option (b) is correct. 6. Given, sin A= 1 2 ⇒ sin A= sin 30° ⇒ A = 30° ∴ cot A= cot 30° = 3 ∴ Option (a) is correct. 8. Given sin A + sin2 A = 1 ...(i) ⇒ sin2 A = 1 – sin A ⇒ 1 – cos2 A = 1 – sin A ...(ii) Now, we have to find value of cos2A + cos4A Put value from (ii), we get sin A + sin2 A = 1 [From equation (i)] ∴ Option (a) is correct. 9. If 4 tan θ =3⇒ tan θ = 3 4 4 sin i – cos i 4 sin i –1 (Divide Nr and Dr by cos θ) 4 sin i + cos i cos i Now, = sin i 4 cos i +1 = 4 tan i – 1 = 4× 3 –1 4 tan i + 1 4 3 4× 4 + 1 = 3–1 = 2 = 1 3+1 4 2 Hence, option (c) is correct. 10. sin 30° tan 45° = 1 ×1 = 1 2 2 Hence option (c) is correct. 11. cos A = 4 , (Given) 5 sin A= 1 – cos2 A = 1 – c 4 2 = 1– 16 = 9 = 3 5 25 25 5 m Introduction to Trigonometry 141

3 tan A = sin A & tan A = 5 = 3 cos A 4 4 5 ` Option (b) is correct. 12. Given expression is tan2 60° – sin2 30° tan2 45° + cos2 30° ^ 3 h2 – c 1 2 3– 1 2 1+ 4 m 3 4 = 32 = 2o (1)2 + e 12 – 1 = 4 = 11 4+3 7 4 ` Option (d) is correct. 13. Given x = r sin i and y = r cos i then x2 + y2 = (r sin i)2 + (r cos i)2 = r2 sin2 i + r2 cos2 i = r2 (sin2 i + cos2 i) = r2 ` Option (b) is correct. 14. 3 sec i – 5 = 0 & sec i = 5 3 But sec i = Hypotenuse & h = 5 and b = 3 A base 5 By Pythagoras theorem, q h2 = p2 + b2 & (5)2 = p2 + (3)2 B3 ⇒ p2 = 25 – 9 = 16 & p = 4 So cot i = base = b = 3 C perpendicular p 4 ` Option (c) is correct. 15. Given a + b = 90° and a = 2b & 2b + b = 90° & b = 30° So, a = 2b = 2×30° = 60° Now value of cos2 a + sin2 b Put the values of a and b, we get & cos2 (60°) + sin2 30° & d 1 2 + d 1 2 = 1 + 1 = 2 = 1 2 2 4 4 4 2 n n ` Option (b) is correct. 16. Given tan x + sin x = m and tan x – sin x = n Now m2 – n2 = (tan x + sin x)2 – (tan x – sin x)2 = tan2 x + sin2 x + 2 sin x tan x – tan2 x – sin2 x + 2 sin x tan x 142 Mathematics–X: Term–1

= 4 sin x tan x = 4 sin2 x tan2 x =4 sin2 x × sin2 x =4 sin2 x (1 – cos2 x) cos2 x cos2 x =4 sin2 x – sin2 x cos2 x = 4 tan2 x – sin2 x cos2 x = 4 (tan x – sin x) (tan x + sin x) = 4 mn ` Option (a) is correct. 17. Given sec i + tan i = x ...(i) From identity sec2 i – tan2 i = 1 & (sec i – tan i) (sec i + tan i) = 1 & (sec i – tan i) x = 1 1 & sec i – tan i = x ...(ii) By adding (i) and (ii), we get 2 sec i = x + 1 & sec i = 1 dx + 1 n x 2 x ` Option (c) is correct. 18. 1 + cos i 1 – cos i By rationalisation, we get 1 + cos i × 1 + cos i & (1 + cos i)2 1 – cos i 1 + cos i (1 – cos2 i) & 1 + cos i & 1 + cos i & sin1 i + cos i & cosec i + cot i sin2 i sin i sin i ` Option (b) is correct. 19. sin i + cos i & sin i + cos i 1 – cot i – tan 1 i 1– cos i 1 – sin i sin i cos i & sin sin i + cos i & sin2 i + cos2 i i i – cos cos i – sin i sin i – cos i cos i – sin sin i cos i sin2 i cos2 i sin2 i – cos2 i & sin i – cos – sin i – cos i & sin i – cos i = (sin i – cos i) (sin i + cos i) (sin i – cos i) i & sin i + cos i ` Option (c) is correct. 20. a sin i = a & sin2 i = a2 b b2 cos2 i = 1 – sin2 i = 1– a2 = b2 – a2 & cos i = b2 – a2 a b2 b2 b tan i = sin i = b = a cos i b2 – a2 b2 – a2 b ` Option (c) is correct. Introduction to Trigonometry 143

21. Clearly side MR is adjacent to ∠y in DPMR and side PR is adjacent to ∠y in DPQR. P x M y Q R ` Option (b) is correct. 8 P YZ 15 22. We have, tan i = B & tan i = XY = and cosec i = H = XZ = 17 P YZ 8 Also sec i = H = XZ = 17 B XY 15 ` Option (a) is correct. 23. Given, sin i = 7 & cos i = 1–e 72 85 85 o & cos i = 1– 49 = 36 & cos i = 6 85 85 85 tan i = sin i = 7/ 85 = 7 and cosec i = 1 = 85 cos i 6 85 6 sin i 7 ` Option (c) is correct. 24. Let AB = x, BC = 3x A ` AC = AB2 + AC2 = x2 + 9x2 = 10x2 = 10 x 10x 3x C ` sin C = AB = x = 1 x AC 10 x 10 ` Option (b) is correct. B 25. We have, 3 – sin2 60° 3–e 32 3 1 tan 30° . tan 60° 2o 4 4 = 1 = 3– = 2 3 × 3 ` Option (a) is correct. 26. Given 4 – sin2 45° = 3.5 cot k tan 60° & 4–e 12 & 4 – 1 2o 2 = 3.5 = 3.5 cot k 3 3 cot k & 2 7 = cot k & 7 = cot k & 1 = cot k & cot 60° = cot k 3 ×3.5 73 3 & k = 60° ` Option (c) is correct. 144 Mathematics–X: Term–1

27. We have 2 tan x (sec2 x – 1) cos3 x 2 tan x × tan2 x = cos3 x = 2 tan3 x sec3 x ` Option (c) is correct. 28. 1 x + sec x = (cos x + sec x) sec cos2 x – (1 + tan2 x) cos2 x – 1 – tan2 x = (cos x + sec x) = (cos x cos x + sec x x) cos2 x – sec2 x – sec x) (cos x + sec = cos 1 x = – sec 1 x x – sec x – cos = – 1 = – 1 ×cos x cos2 1 x – cos x 1 – x cos cos x cos x 1 = – sin2 x = – sin x × sin x = – cosec x × cot x ` Option (c) is correct. 29. sin q = 1 = Perpendicular 3 Hypotenuse ∴ (3)2 = (Perpendicular)2 + (Base)2 (Base)2 = 9 – 1 = 8 Base = 2 2 cot2θ = Base = 22 Perpendicular 1 ∴ 2cot2 θ + 2 = 2 (2 2)2 + 2 = 2(8) + 2 = 16 + 2 = 18 Hence, option (d) is correct. 30. We have, b2 x2 + a2 y2 = b2 × a2 cos2 i + a2 × b2 sin2 i = a2 b2 (cos2 i + sin2 i) = a2b2 ×1= a2b2 ` Option (c) is correct. 31. We have, 3 tan q = 3 sin q ⇒ 3 sin i = 3 sin q cos i ⇒ 3 = 3 ⇒ 3 = cos q ⇒ 1 = cos i cos i 3 3 ∴ cos q = 1 3 Now, sin2 q – cos2 q = 1 – cos2 q – cos2 q = 1 – 2 cos2 q 12 =1–2× e 3o =1– 2 = 1 3 3 ` Option (a) is correct. Introduction to Trigonometry 145

32. We have, (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = d 1 i – sin i n d 1 i – cos in d sin i + cos i n sin cos cos i sin i = e 1 – sin2 i o e 1 – cos2 i o e sin2 i + cos2 i o sin i cos i sin i cos i = cos2 i × sin2 i × sin 1 i =1 sin i cos i i cos ` Option (b) is correct. 33. We have, (sec A + tan A) (1 – sin A) = c 1 A + sin A m (1 – sin A) = (1+ sin A) (1 – sin A) cos cos A cos A = 1 – sin2 A = cos2 A = cos A cos A cos A ` Option (d) is correct. 34. We have, sin a = 1 = sin 30° & a = 30° 2 and, cos b = 1 = cos 60° & b = 60° 2 ` a + b = 30° + 60° = 90° Hence, option (d) is correct. 35. Given, sin i + cos i = 2 cos i Dividing both sides by cos θ, we get sin i +1 = 2 & tan i = 2 –1 cos i ` Option (a) is correct. 36. We have, (2x) 2 – c 2 2 = sec2 i – tan2 i x m ⇒ 4x2 – 4 =1 ⇒ 4d x2 – 1 n =1 x2 x2 1 ∴ d x2 – x2 n = 1 4 ∴ Option (b) is correct. 37. We have, (1 + cot q – cosec q) (1 + tan q + sec q) = d1 + cos i – 1 i n d1 + sin i + 1 n sin i sin cos i cos i = d sin i + cos i – 1 nd sin i + cos i + 1 n sin i cos i = (sin i + cos i)2 – (1)2 p= sin2 i + cos2 i + 2 sin cos i –1 f sin i cos i sin i cos i = 1+ 2 sin i cos i – 1 = 2 sin i cos i =2 sin i cos i sin i cos i ∴ Option (b) is correct. 146 Mathematics–X: Term–1

38. We have, sin 77° = x A x cos 77° = 1 – x2 [a cos2 i = 1 – sin2 i] x tan 77° = 1 – x2 ` Option (c) is correct. 39. We have, B 77° sec4A – sec2A = sec2A (sec2A –1) C = sec2A × tan2A = (1 + tan2A) tan2A = tan2A + tan4A = tan4A + tan2A ∴ Option (c) is correct. 40. We have, m2 + n2 = (a cos i + b sin i)2 + (a sin i – b cos i)2 = a2 cos2 i + b2 sin2 i + 2ab cos i sin i + a2 sin2 i + b2 cos2 i – 2ab sin i cos i = a2 (cos2 i + sin2 i) + b2 (sin2 i + cos2 i) ⇒ m2 + n2 = a2 ×1+ b2 ×1= a2 + b2 ∴ a2 + b2 = m2 + n2 ∴ Option (d) is correct. 41. Given, cosA + cos2A = 1 ⇒ cos A = 1 – cos2A = sin2A ⇒ sin2 A = cos A ⇒ sin4 A = cos2A ∴ sin2 A + sin4 A = cos A + cos2 A = 1 ∴ Option (c) is correct. 3 42. sin 60° = 2 =1 cos 30° 3 2 ∴ Option (c) is correct. 43. We have, 4 tan x = 3 ⇒ tan x = 3 4 ∴ cos x + sin x = 1 + tan x (Divide Nr and Dr by cos x) cos x – sin x 1 – tan x = 1+ 3 =7 1– 4 3 4 ∴ Option (a) is correct. 44. cos4 A – sin4 A = (cos2 A – sin2 A) (cos2 A + sin2 A) = (cos2 A – sin2 A) ×1 = cos2 A – (1 – cos2 A) = 2 cos2 A – 1 ∴ Option (b) is correct. Introduction to Trigonometry 147