Question

In equilaterak triangle with 'a'.prove that area of triangle Is root 3÷4a^

Answer 1

let, ∆ABC be the equalitaral triangle with sides 'a'
now draw a median to base BC and vertex A
such that median bisect the base BC at D
so that BD=a/2,. DC =a/2. ^=square
now in ∆ABD
AB^=BD^+AD^
AB^-BD^=AD^
(a)^-(a/2)^=AD^
a^-a^/4=AD^
3a^/4=AD^
√3a/2=AD
ar(∆ABC)=1/2bh
=1/2×BC×AD
=1/2×a×√3a/2 (BC=a )
=√3a^/4
hence proved
it may help you