Question

Prove that the area of an equilateratl triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Answer 1

REF.Image

The two triangles

are $△APB$ and $△ABQ$

respectively

let side of square = a

$\Rightarrow AB=a\Rightarrow AP=BP=AB=a$

Area of $△APB=\frac{\sqrt{3}}{4}(side{)}^2=\frac{\sqrt{3}}{4}{a}^2$

Now, $B{D}^2=D{C}^2+B{C}^2$ (pythagorous theorem)

$\Rightarrow B{D}^2=a^2+a^2\Rightarrow BD=\sqrt{2}a$

$\Rightarrow$ area of $△BDQ=\frac{\sqrt{3}}{4}(\sqrt{2}a{)}^2=\frac{\sqrt{3}}{2}{a}^2$

$\Rightarrow △APB=\frac{1}{2}\left(ar△BDQ\right)$