Question

In a ∆ABC, ∠A = 90°, AB = 5 cm and AC = 12 cm. If AD ⊥ BC, then AD =

(a) $\frac{13}{2}\mathrm{cm}$
(b) $\frac{60}{13}\mathrm{cm}$
(c) $\frac{13}{60}\mathrm{cm}$
(d) $\frac{2\sqrt{15}}{13}\mathrm{cm}$

Answer 1

Given: In ΔABC,, AC = 12cm, and AB = 5cm.

To find: AD

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

In ∆ACB and ∆ADC,

$\angle \mathrm{C}=\angle \mathrm{C}$ (Common)

$\angle \mathrm{A}=\angle \mathrm{ADC}=90°$

∴ ∆ACB $~$∆ADC (AA Similarity)

We got the result as .