Question

An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is $12d{m}^2$ then the difference of their areas (in $d{m}^2$ ) is :

• A. $2$
• B. $4$
• C. $6$
• D. 1$8$

Answer 1

Answer: (D)

$18$

Let, side of triangle is $x$ and side of hexagon is $a$.

So,
Perimeter are equal
$3x=6a$
$x=2a$
Area of eq. $△=\frac{\sqrt{3}}{4}{x}^2=12d{m}^2$ .....(1)

$=\dfrac{\sqrt{3}}{4}
Area of hexagon $=6\frac{\sqrt{3}}{4}\times a^2=6\frac{\sqrt{3}}{4}{\left(\frac{x}{2}\right)}^2$
$=\frac{6}{4}(\frac{\sqrt{3}}{4}\times x^2)$ $=\frac{6}{4}\times 12$ .......From (1)

Now difference betwwen them is,

$\Rightarrow 12-\frac{6}{4}\times 12=18d{m}^2$

Hence, the difference of their areas is $18d{m}^2$.