Question

The difference in the areas of the regular hexagon circumscribing a circle of radius 10 cm and the regular hexagon inscribed in the circle is ___________.

Answer 1

In the figure, O is the centre of circle of radius 10 cm. ABCDEF is a regular hexagon circumscribing the circle and PQRSTU is a regular hexagon inscribed in the circle.

Here, ∆OPQ is an equilateral triangle.

Length of each side of equilateral ∆OPQ = Radius of the circle = 10 cm

Now,

Area of the regular hexagon PQRSTU

= 6 × Area of equilateral ∆OPQ

$=6\times \frac{\sqrt{3}}{4}\times {\left(10\right)}^2$ $\left[\mathrm{Area}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}=\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^2\right]$

$=150\sqrt{3}{\mathrm{cm}}^2.....\left(1\right)$

Also,

∆OAB is an equilateral triangle.

Length of an altitude of equilateral ∆OAB = Radius of the circle = 10 cm

We know

Length of an altitude of an equilateral triangle = $\frac{\sqrt{3}}{2}\left(\mathrm{Side}\right)$

$$\begin{gathered} \therefore \frac{\sqrt{3}}{2}\times \mathrm{AB}=10\mathrm{cm} \\ \Rightarrow \mathrm{AB}=\frac{20}{\sqrt{3}}\mathrm{cm} \end{gathered}$$

Area of the regular hexagon ABCDEF

= 6 × Area of equilateral ∆OAB

$=6\times \frac{\sqrt{3}}{4}\times {\left(\frac{20}{\sqrt{3}}\right)}^2$ $\left[\mathrm{Area}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}=\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^2\right]$

$=200\sqrt{3}{\mathrm{cm}}^2.....\left(2\right)$

∴ Required difference in areas

= Area of the regular hexagon ABCDEF − Area of the regular hexagon PQRSTU

$$\begin{gathered} =200\sqrt{3}-150\sqrt{3}\left[\mathrm{Using}\left(1\right)\mathrm{and}\left(2\right)\right] \\ =50\sqrt{3}{\mathrm{cm}}^2 \end{gathered}$$

The difference in the areas of the regular hexagon circumscribing a circle of radius 10 cm and the regular hexagon inscribed in the circle is $\underline{50\sqrt{3}{\mathrm{cm}}^2}$.