Solved 2024 Specimen Paper ICSE Class 10 Mathematics — Question 3

(a) In △ADF,

⇒ ∠ADF = 90°

By angle sum property of triangle,

⇒ ∠ADF + ∠AFD + ∠A = 180°

⇒ 90° + ∠AFD + ∠A = 180°

⇒ ∠AFD + ∠A = 180° - 90°

⇒ ∠AFD + ∠A = 90° .........(1)

In △ ABC,

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + ∠C = 180°

⇒ ∠A + ∠C = 180° - 90°

⇒ ∠A + ∠C = 90° .........(2)

From equation (1) and (2), we get :

⇒ ∠AFD + ∠A = ∠A + ∠C

⇒ ∠AFD = ∠C

In △ ADF and △ FEC,

⇒ ∠ADF = ∠FEC (Both equal to 90°)

⇒ ∠AFD = ∠C (Proved above)

∴ △ ADF ~ △ FEC [By A.A. axiom]

Hence, proved that △ ADF ~ △ FEC.

(b) In △ ADF and △ ABC,

⇒ ∠DAF = ∠BAC (Common angle)

⇒ ∠ADF = ∠ABC (Both equal to 90°)

∴ △ ADF ~ △ ABC [By A.A. axiom]

Hence, proved that △ ADF ~ △ ABC.

(c) From figure,

⇒ DB = FE = x (let)

⇒ AB = AD + DB = (6 + x) cm

⇒ DF = BE = BC - CE = 12 - 4 = 8 cm.

△ ADF ~ △ ABC [proved above]

We know that,

Corresponding sides of similar triangle are proportional.

$\Rightarrow \dfrac{AD}{AB} = \dfrac{DF}{BC} \\[1em] \Rightarrow \dfrac{6}{6 + x} = \dfrac{8}{12} \\[1em] \Rightarrow 8(6 + x) = 6 \times 12 \\[1em] \Rightarrow 48 + 8x = 72 \\[1em] \Rightarrow 8x = 72 - 48 \\[1em] \Rightarrow 8x = 24 \\[1em] \Rightarrow x = \dfrac{24}{8} = 3 \text{ cm}.$

Hence, FE = 3 cm.

(d) We know that,

The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles.

$\Rightarrow \dfrac{\text{Area of △ ADF}}{\text{Area of △ ABC}} = \dfrac{AD^2}{AB^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ ADF}}{\text{Area of △ ABC}} = \dfrac{6^2}{(6 + 3)^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ ADF}}{\text{Area of △ ABC}} = \dfrac{6^2}{9^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ ADF}}{\text{Area of △ ABC}} = \dfrac{36}{81} = \dfrac{4}{9}.$

Hence, area △ ADF : area △ ABC = 4 : 9.