Question

Prove that the area of an equilateral triangle is equal to √3÷4×a²  where a is the side of the triangle. 

Answer 1

base of equilateral triangle = b

height = square root of(side² - (side/2)²   by pythagoras theorem

= √s²-s²÷4

= √(4s²-s²)/4

=√3s²/4

= √3  x s/2

area = 1/2 x b x h

     = 1/2 x s x √3  x s/²

     = √3/4 s²

Answer 2

A line perpendicular to the the opposite base which will form 2 right angle triangles. Each right triangle leg length x/2. 

according to the pythagorean theorem : $x^2=(x/2)^2+h^2$

$x^2=(x/2)^2+h^2$ 
$x^2-(x/2)^2=h^2$ 
$x^2- x^2/4=h^2$ 
$4x^2- x^2=4h^2$ 
$3x^2/4=h^2$ 
$(√3/2) x=h$ 

Each right angle triangle area is  $(1/2)*(x/2)*h$. And the area of the equilateral triangle is double of that.

$A = 2*(1/2)*(x/2)*h$ 
$A = (x/2)*h$ 
$A = (x/2)*(√3/2)x$ 
$A = (√3/4) x^2$  ( a can be x) (  $√$ means √)