Question

A square and an equilateral triangle have the same perimeter. If the diagonal of the square is 12$\sqrt{2}$ cm, then the area of the triangle is:

• A. $24\sqrt{3}c{m}^2$
• B. $24\sqrt{2}c{m}^2$
• C. $64\sqrt{3}c{m}^2$
• D. $32\sqrt{3}c{m}^2$

Answer 1

The correct option is C $64\sqrt{3}c{m}^2$

Let the side of square be $s$ and equilateral triangle be $a$

Given that $4\times s=3\times a$ (1)

Also diagonal of square $=12\sqrt{2}cm$

Applying Pythagoras theorem, $s^2+s^2=(12\sqrt{2}{)}^2$

on solving we get s $=$12 cm

now, putting $s$ in (1)

$4\times 12=3\times a\Rightarrow a=16cm$

Area of an equilateral triangle $=\frac{\sqrt{3}}{4}{a}^2$

$=64\sqrt{3}c{m}^2$