Question

Find the area of the hexagon given below:

• A. $51\sqrt{3}c{m}^2$
• B. $52\sqrt{3}c{m}^2$
• C. $53\sqrt{3}c{m}^2$
• D. $54\sqrt{3}c{m}^2$

Answer 1

The correct option is D

$54\sqrt{3}c{m}^2$

Area of hexagon ABCDEF $=6\times$ Area $\left(\Delta ABO\right)$

(All the triangles are equilateral)

Proof: Each internal angle in a regular hexagon with $n$ sides = $=\frac{(n-2)\times {180}^{\circ }}{n}$

$=\frac{(6-2)\times {180}^{\circ }}{6}=\frac{4\times {180}^{\circ }}{6}={120}^{\circ }$

$\angle ABC={120}^{\circ }$

$\therefore \angle ABO=\angle OBC={60}^{\circ }$

Hence all the triangle are equilateral

Area $\left(\Delta ABO\right)=\frac{\sqrt{3}}{4}{a}^2$

$=\frac{\sqrt{3}}{4}\times 6^2=9\sqrt{3}c{m}^2$

$\therefore$ Area (hexagon ABCDEF) $=6\times 9\sqrt{3}=54\sqrt{3}c{m}^2$