The problem asks for the area of a regular hexagon inscribed inside a circle with a given radius.
A regular hexagon can be divided into 6 identical equilateral triangles. The side length ($s$) of each equilateral triangle is equal to the radius ($r$) of the circle in which the hexagon is inscribed.
$s = r = 14 \text{ cm}$
$\text{Area}_{\triangle} = \frac{\sqrt{3}}{4} s^2$
Substitute $s = 14$ cm:$\text{Area}_{\triangle} = \frac{\sqrt{3}}{4} (14 \text{ cm})^2 = \frac{\sqrt{3}}{4} \times 196 \text{ cm}^2 = 49\sqrt{3} \text{ cm}^2$
$\text{Area}_{\text{Hexagon}} = 6 \times \text{Area}_{\triangle} = 6 \times 49\sqrt{3} \text{ cm}^2 = 294\sqrt{3} \text{ cm}^2$
$\text{Area}_{\text{Hexagon}} \approx 294 \times 1.73205 \text{ cm}^2 \approx 509.2237 \text{ cm}^2$
The calculated area is approximately $509.22 \text{ cm}^2$. Comparing this with the given options, Option B ($509.21 \text{ cm}^2$) is the closest value.
Calculate the area of the triangle whose sides are 8 cm, 9 cm and 13 cm. (Rounded up to two decimal places)