Question

In a ΔABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that

(i) AD = a`sqrt3`

(ii) Area (ΔABC) = `sqrt3` a²

Answer 1

(i) In ΔABD and ΔACD

∠ADB = ∠ADC                          [Each 90°]

AB = AC                                  [Given]

AD = AD                                  [Common]

Then, ΔABD ≅ ΔACD                 [By RHS condition]

∴ BD = CD = a                         [By c.p.c.t]

In ΔADB, by Pythagoras theorem

AD² + BD² = AB²

⇒ AD² + (a)² = (2a)²

⇒ AD² + a² = 4a²

⇒ AD² = 4a² − a² = 3a²

⇒ AD = a`sqrt3`

(ii) Area of ΔABC = `1/2xxBCxxAD`

`=1/2xx2axxasqrt3`

`=sqrt3` a²