If the area of the circle is $A_{1}$ and the area of the regular pentagon inscribed in the circle is $A_{2}$ then the ratio $A_{1} ∣ A_{2}$ be $\frac{π}{k} sec (\frac{π}{h})$ .Find $k * h$ ?

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Solution

In $Δ O A B, O A = O B = r$
$∠ A O B = \frac{360^{\circ}}{5} = 72^{\circ}$
$∴ A r e a o f Δ A O B = \frac{1}{2} r \times r sin 72^{\circ} = \frac{1}{2} r^{2} cos 18^{\circ}$
Area of pentagon $= (A_{2}) = 5 (A r e a o f Δ A O B) = 5 (\frac{1}{2} r^{2} cos 18^{\circ})$ (i)
Also,Area of the circle $(A_{1}) = π r^{2}$ (ii)
Hence, $= \frac{A_{1}}{A_{2}} = \frac{π r^{2}}{\frac{5}{2} r^{2} cos 18^{\circ}} = \frac{2 π}{5} sec (\frac{π}{10})$ [from Eqs.(i) and (ii)]

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