Question

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

(i) AD = a\(\sqrt3\)

(ii) Area (ΔABC) = \(\sqrt{3a^2}\)

Answer 1

(i) In ΔABD and ΔACD 

ΔADB = ΔADC = 90º

AB = AC (given) 

AD = AD (common)

ΔADB ~ ΔACD 

BD = CD = a (By c.p.c.t) 

In ΔADB 

AD² + BD² = AB² 

AD² + a² = (2a)² 

AD² = 4a² - a² 

AD² = 3a² 

AD = a \(\sqrt3\)

(ii) Area of ABC = 1/2 x BC x AD 

= 1/2 x 2a x a\(\sqrt3\)

= \(\sqrt{3a^2}\)