Question

A square and an equilateral triangle have equal perimeters. Ifthe diagonal of the square is $6\sqrt{2}$ cm, then the area of the triangle is

• A. $16\sqrt{2}c{m}^2$
• B. $16\sqrt{3}c{m}^2$
• C. $12\sqrt{2}c{m}^2$
• D. $12\sqrt{3}c{m}^2$

Answer 1

Answer: (B)

$16\sqrt{3}c{m}^2$

Given that,

Diagonal of a square

$=6\sqrt{2}$

$=side\sqrt{2}$

So,

$side=6cm$

Now,

Perimeter of square $=6\times 4=24$

Perimeter of triangle

$=3\times a=24$

$a=8cm$

Area of equilateral triangle

$=\frac{\sqrt{3}}{4}{a}^2$

$=\frac{\sqrt{3}}{4}\times 8^2$

$=\frac{\sqrt{3}}{4}\times 64$

$=16\sqrt{3}c{m}^2$

Hence,

This the required solution.