Question

Prove that the area of an equilateral 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

Given:
$ABCD$ is a Square,
$DB$ is a diagonal of square,
$△DEB$ and $△CBF$ are Equilateral Triangles.
To Prove:
$\frac{A\left(△CBF\right)}{A\left(△DEB\right)}=\frac{1}{2}$
Proof:

Since, $△DEB$ and $△CBF$ are Equilateral Triangles.

$\therefore$ Their corresponding sides are in equal ratios.

In a Square ABCD, $DB=BC\sqrt{2}$ $.....\left(1\right)$

$\therefore$ $\frac{A\left(△CBF\right)}{A\left(△DEB\right)}=$$\frac{\frac{\sqrt{3}}{4}\times (BC{)}^2}{\frac{\sqrt{3}}{4}\times (DB{)}^2}$

$\therefore$ $\frac{A\left(△CBF\right)}{A\left(△DEB\right)}=$$\frac{\frac{\sqrt{3}}{4}\times (BC{)}^2}{\frac{\sqrt{3}}{4}\times (BC\sqrt{2}{)}^2}$ (From 1)

$\therefore$ $\frac{A\left(△CBF\right)}{A\left(△DEB\right)}=\frac{1}{2}$