These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6 Other Exercises Question 1. Triangles ABC and DEF are similar. (i) If area (∆ABC) = 16 cm², area (∆DEF) = 25 cm² and BC = 2.3 cm, find EF. (C.B.S.E. 1992) (ii) If area (∆ABC) = 9 cm², area (∆DEF) = 64 cm² and DE = 5.1 cm, find AB.(iii) If AC = 19 cm and DF = 8 cm, find the ratio of the area of two triangles. (C.B.S.E. 1992C) (iv) If area (∆ABC) = 36 cm², area (∆DEF) = 64 cm² and DE = 6.2 cm, find AB. (C.B.S.E. 1992) (v) If AB = 1.2 cm and DE = 1.4 cm, find the ratio of the areas of ∆ABC and ∆DEF. (C.B.S.E. 1991C) Solution: Question 2. In the figure, ∆ACB ~ ∆APQ. If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. Also, find the area (∆ACB) : area (∆APQ).
Solution: Question 3. The areas of two similar triangles are 81 cm² and 49 cm² respectively. Find the ratio of their corresponding heights, what is the ratio of their corresponding medians ? Solution: Areas of two similar triangles are 81 cm² and 49 cm² The ratio of the areas of two similar triangles are proportion to the square of their corresponding altitudes and also squares of their corresponding medians Ratio in their altitudes = √81 : √49 = 9 : 7Similarly, the ratio in their medians = √81 : √49 = 9 : 7 Question 4. The areas of two similar triangles are 169 cm² and 121 cm² respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle. Solution: Triangles are similar Area of larger triangle = 169 cm² and area of the smaller triangle =121 cm² Length of longest sides of the larger triangles = 26 cm Let the length of longest side of the smaller triangle = x
Question 5. The areas of two similar triangles are 25 cm² and 36 cm² respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other. Solution: Area of first triangle = 25 cm² Area of second = 36 cm² Altitude of the first triangle = 2.4 cm Let altitude of the second triangle = x The triangles are similar
Question 6. The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas. Solution: Length of the corresponding altitude of two triangles are 6 cm and 9 cm triangles are similar
Question 7. ABC is a triangle in which ∠A = 90°, AN ⊥ BC, BC = 12 cm and AC = 5 cm. Find the ratio of the areas of the ∆ANC and ∆ABC. Solution:
Question 8. In the figure, DE || BC (i) If DE = 4 cm, BC = 6 cm and area (∆ADE) = 16 cm², find the area of ∆ABC. (ii) If DE = 4 cm, BC = 8 cm and area of (∆ADE) = 25 cm², find the area of ∆ABC. (C.B.S.E. 1991) (iii) If DE : BC = 3 : 5, calculate the ratio of the areas of ∆ADE and the trapezium BCED
Solution: Question 9. In ∆ABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ∆ADE and ∆ABC. Solution: In ∆ABC, D and E are the mid points of AB and AC respectively
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Question 12. In ∆ABC ~ ∆DEF such that AB = 5 cm and (∆ABC) = 20 cm² and area (∆DEF) = 45 cm², determine DE. Solution:
Question 13. In ∆ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides ∆ABC into two parts equal in area. Find \(\frac { BP }{ AB }\). Solution: In ∆ABC, PQ || BC and PQ divides ∆ABC in two parts ∆APQ and trap. BPQC of equal in area i.e., area ∆APQ = area BPQC
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Question 18. Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights. Solution: Two isosceles triangles have equal vertical angles So their base angles will also be the equal to each other Triangles will be similar Now, ratio in their areas = 36 : 25
Question 19. In the figure, ∆ABC and ∆DBC are on the same base BC. If AD and BC intersect
Solution: Given : Two ∆ABC and ∆DBC are on the same base BC as shown in the figure AC and BD intersect eachother at O
Question 20. ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove that (i) ∆AOB ~ ∆COD (ii) If OA = 6 cm, OC = 8 cm, find
Solution: Given : ABCD is a trapezium in which AB || CD Diagonals AC and BD intersect each other at O
Question 21. In ∆ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ∆APQ and trapezium BPQC. Solution: In ∆ABC, P is a point on AB such that AP : PQ = 1 : 2 PQ || BC Now we have to find the ratio between area ∆APQ and area trap BPQC
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Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6 are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. |