Question

In a ΔABC, AB = BC = CA = 2a and AD⟂ BC. Prove that (i) AD =a√3 (ii) area (ΔABC) = √3a^2

Answer 1

in ACD and ABD
angle ADC =angle ADB(AD is perpendicular to BC)
AC=AB (given)
AD=AD(common)
triangle ACD is confident to triangle ABD (by RHS,)
CD=BD(by cpct)

in triangle ACD use Pythagoras theorem
H^2=P^2+B^2
AC=BC=2a
BC=2a
BD+CD=2a
BD=CD(proved above)
CD+CD=2a
2CD=2a
CD =a
AC^2=CD^2+AD^2
(2a)^2=a^2+AD^2
4a^2=a^2+AD^2
4a^2-a^2=AD^2
3a^2=AD^2
squaring root both side
√3a=AD

Area of triangle ABC=1/2×base×height
=1/2×2a×√3a
=√3a^2