Question

$H_1$ is a regular hexagon circumscribing a circle. $H_2$ is a regular hexagon inscribed in the circle. Find the ratio of areas of $H_1$ and $H_2$.

• A. $4:3$
• B. $2:1$
• C. $3:1$
• D. $3:2$

Answer 1

Answer: (A)

Let $H_1$ be $PQRSTU$ and $H_2$ be $ABCDEF$.
Let $R$ be the radius of the circle.
$\therefore R=\frac{\sqrt{3}(PU)}{2}\Rightarrow PU=\frac{2R}{\sqrt{3}}$
$AB=R(\because$ A hexagon inscribed in a circle must have its side equal to the radius of the circle).
Area of $H_1=\frac{3\sqrt{3}}{2}{\left(\frac{2}{\sqrt{3}}R\right)}^2$
And area of $H_2=\frac{3\sqrt{3}}{2}(R{)}^2$
$\left(\because \text{ Area of hexagon }=\frac{3\sqrt{3}}{2}\times (\text{ Its side }{)}^2\right)$
$\text{ Required ratio }=\frac{{\left(\frac{2}{\sqrt{3}}R\right)}^2}{R^2}=\frac{4}{3}$