Question

A 6 sided regular polygon (hexagon) is inscribed in a circle of radius 10 cm, find area of the hexagon. [2 MARKS]

Answer 1

Solution: 2 Marks

$\angle AOB=\frac{{360}^{\circ }}{6}={60}^{\circ }$

Since OA = OB = 10 cm, triangle OAB is isosceles which gives

$\angle OAB=\angle OBA$

So, all three angles of the triangle are equal and therefore it is an equilateral triangle. Hence

AB = OA = OB = 10 cm.

Area of an equilateral triangle of side $a$ is $=\frac{\sqrt{3}}{4}\times (a{)}^2$

Therefore, Area of the equilateral triangle AOB
$=\frac{\sqrt{3}}{4}\times (10{)}^2$

$=43.30c{m}^2$

Area of hexagon
$=6\times areaofoneequilateraltriangle$

$=6\times 43.30$

$=259.8c{m}^2$