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

STEP 1 : Construction

$ABCD$ is a square whose one diagonal is $AC$.

$△APC$ and $△BQC$ are two equilateral triangles described on the diagonals $AC$ and side $BC$ of the square $ABCD$.

STEP 2 : Finding the relation between the diagonal and side of the squares

Let the side of the square be $a$

i.e. $AB=BC=CD=DA=a$

Diagonal $AC=\sqrt{A{B}^2+B{C}^2}$

$AC=\sqrt{a^2+a^2}$

$AC=\sqrt{2a^2}$

$AC=\sqrt{2}a$

STEP 3 : Finding the area of the triangles $\mathit{B}\mathit{Q}\mathit{C}$ and $\mathit{A}\mathit{P}\mathit{C}$

$ar\left(△BQC\right)=\frac{\sqrt{3}}{4}\times sid{e}^2=\frac{\sqrt{3}}{4}{a}^2$

$ar\left(△APC\right)=\frac{\sqrt{3}}{4}\times sid{e}^2=\frac{\sqrt{3}}{4}{\left(\sqrt{2}\times a\right)}^2$

$ar\left(△APC\right)=\frac{\sqrt{3}}{4}{\left(\sqrt{2}\right)}^2\times a^2=\frac{2\sqrt{3}}{4}{a}^2$

$ar\left(△APC\right)=2\times \frac{\sqrt{3}}{4}{a}^2$

$ar\left(△APC\right)=2\times ar\left(△BQC\right)$

$\Rightarrow ar\left(△BQC\right)=\frac{1}{2}\times ar\left(△APC\right)$

Hence, 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.