Question

The corresponding sides of ΔABC and ΔPQR are in the ratio 3 : 5. AD ⊥ BC and PS ⊥ QR, as shown in the following figures:

1. Prove that ΔADC ~ ΔPSR.
2. If AD = 4 cm, find the length of PS.
3. Using (ii) find ar(ΔABC) : ar(ΔPQR).

Answer 1

(i) In ΔADC and ΔPSR

∠ADC = ∠PSR  ....(Each 90°)

∠ACD = ∠PRS   .....(∵ ΔАBC ~ ΔPQR) (Given)

∴ ΔADC ~ ΔPSR   ....(By AA similarity)

(ii) Given, AD = 4 cm

∴ `(AD)/(PS) = (AC)/(PR) = 3/5`   ...(Given)

`4/(PS) = 3/5`

3PS = 20

∴ PS = `20/3` cm

(iii) The areas of two similar triangles are in the ratio of the squares of their corresponding sides.

`("ar"(ΔABC))/("ar"(ΔPQR)) = ((AC)/(PR))^2`

= `(3/5)^2`

= `9/25`