Triangles Exercise 6.4 – class 10

1. Let ∆ABC∼ ∆PQR and their areas be, respectively, 64 cm² and 121 cm² . If EF = 15.4 cm, find BC.

Solution:

2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Solution:

Since AB || CD,

∴ ∠OAB = ∠OCD and ∠OBA = ∠ODC (Alternate interior angles)

In ∆AOB and ∆COD, ∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD (Alternate interior angles)

∠OBA = ∠ODC (Alternate interior angles)

∴ ∆AOB ∼ ∆COD (By AAA similarity criterion)

3. In the following figure, ABC and DBC are two triangles on the same base BC. If AD intersects  BC at O, such that, [area(∆ABC)]/[area(∆DBC)] = AO/DO

Solution:

Let us draw two perpendiculars AP and DM on line BC.

We know that area of a triangle = 1/2 x height x base

In ∆APO and ∆DMO,

∠APO = ∠DMO (Each = 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ∆APO ∼ ∆DMO (By AA similarity criterion)

4. If the areas of two similar triangles are equal, prove that they are congruent.

Solution:

Let us assume two similar triangles as ∆ABC ∼ ∆PQR.

5. D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ABC. Find the ratio of the area of ∆DEF and ∆ABC.

Solution:

D and E are the mid-points of ∆ABC.

6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Solution:

Let us assume two similar triangles as ∆ABC ∼ ∆PQR.

Let AD and PS be the medians of these triangles.

∆ABC ∼ ∆PQR

∴ AB/PQ = AC/PR = BC/QR ————(1)

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ————(2)

Since AD and PS are medians,

BD = DC = BC/2 and QS = SR = QR/2

Equation (1) becomes

AB/PQ = BD/QS = BC/QR  —————(3)

In ∆ABD and ∆PQS,

∠B = ∠Q [Using equation (2)]

and AB/PQ = BD/QS [using (3)]

∴ ∆ABD ∼ ∆PQS (SAS similarity criterion) Therefore, it can be said that

7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Solution:

Let ABCD be a square of side a.

Therefore, its diagonal = √2a

Two desired equilateral triangles are formed as ∆ABE and ∆DBF.

Side of an equilateral triangle, ∆ABE, described on one of its sides = a

Side of an equilateral triangle, ∆DBF, described on one of its diagonals√2a

We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

8. ABC and BDE 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) 4 : 1

(D) 1 : 4

Solution:

We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other.

Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

Let side of ∆ABC = x

Therefore, side of ΔBDE = x/2

Hence, the correct answer is (C).

9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio (A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81

Solution:

If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles. It is given that the sides are in the ratio 4:9.

Therefore, ratio between areas of these triangles = [4/9]² = 16/81

Hence, the correct answer is (D).