Question

A square and an equilateral triangle have equal perimeters If the 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

The correct option is B $16\sqrt{3}$ $c{m}^2$

Given diagonal of square is $6\sqrt{2}$ cm

As per Pythagoras theorem

$(side{)}^2+(side{)}^2=(digonal{)}^2\Rightarrow 2(side{)}^2=(digoonal{)}^2$$2a^2=(d{)}^2\Rightarrow (a{)}^2=\frac{(6\sqrt{2}{)}^2}{2}=6$

Then side of square=6 cm

70dm

Then perimeter of square=$4\left(side\right)=4\left(6\right)=24cm$

But perimeter of square=perimeter of equilateral triangle =24 cm

Then side of equilateral triangle = $\frac{24}{3}=8$

Then area of equilateral triangle =$\frac{\sqrt{3}}{4}\times (8{)}^2=\frac{\sqrt{3}}{4}\times 64=16\sqrt{3}c{m}^3$