Question

A regular hexagon is inscribed in a circle of radius r By how much is the area of the circle is more than the area of the hexagon?

• A. $(\pi -2\sqrt{3})r^2$
• B. $(\pi -\sqrt{3})r^2$
• C. $(\pi -\frac{\sqrt{3}}{2})r^2$
• D. $(\pi -\frac{3\sqrt{3}}{2})r^2$

Answer 1

The correct option is D $(\pi -\frac{3\sqrt{3}}{2})r^2$

When a regular hexagon is inscribed in a circle of radius , $r$, its length of side will also be $r$
Area of the hexagon of side "r" $=\frac{3\sqrt{3}}{2}{r}^2$
And Area of a circle of radius r $=\pi r^2$
So, Area of the circle is more than the area of hexagon by $\pi r^2-\frac{3\sqrt{3}}{2}{r}^2=(\pi -\frac{3\sqrt{3}}{2})r^2$