Question

In the given figure, ABC is a right angled triangle with ∠BAC = 90°.

1. Prove that : ΔADB ∼ ΔCDA.
2. If BD = 18 cm and CD = 8 cm, find AD.
3. Find the ratio of the area of ΔADB is to area of ΔCDA.

Answer 1

i. Let ∠CAD = x

`=>` m ∠DAB = 90° – x

`=>` m ∠DBA = 180° – (90° + 90° – x) = x

`=>` ∠CDA = ∠DBA   ...(1)

In ΔADB and ΔCDA,

∠ADB = ∠CDA   ...[Each 90°]

∠ABD = ∠CAD   ...[From (1)]

∴ ΔADB ∼ ΔCDA  ...[By A.A]

ii. Since the corresponding sides of similar triangles are proportional, we have.

`(BD)/(AD) = (AD)/(CD)`

`=> (18)/(AD) = (AD)/(8)`

`=>` AD² = 18 × 8 = 144

`=>` AD = 12 cm

iii. The ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides. 

`=> (Ar(ΔADB))/(Ar(ΔCDA)) = (AD^2)/(CD^2)`

= `12^2/8^2`

= `144/64`

= `9/4`

= 9 : 4