If a regular pentagon and a regular decagon h | vedclass

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(B) Let the perimeter of both the regular pentagon and the regular decagon be $10x$.For the regular pentagon,the number of sides $n_1 = 5$. The side l

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$2:\sqrt{5}$

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(B) Let the perimeter of both the regular pentagon and the regular decagon be $10x$.For the regular pentagon,the number of sides $n_1 = 5$. The side length $s_1 = \frac{10x}{5} = 2x$.The area of a regular polygon with $n$ sides and side length $s$ is given by $A = \frac{n s^2}{4 	an(\frac{\pi}{n})}$.Area of the pentagon $A_1 = \frac{5 (2x)^2}{4 	an(\frac{\pi}{5})} = \frac{5 \cdot 4x^2}{4 	an(36^{\circ})} = 5x^2 \cot(36^{\circ})$.For the regular decagon,the number of sides $n_2 = 10$. The side length $s_2 = \frac{10x}{10} = x$.Area of the decagon $A_2 = \frac{10 x^2}{4 	an(\frac{\pi}{10})} = \frac{5}{2} x^2 \cot(18^{\circ})$.The ratio of the area of the pentagon to the area of the decagon is:$\frac{A_1}{A_2} = \frac{5x^2 \cot(36^{\circ})}{\frac{5}{2} x^2 \cot(18^{\circ})} = 2 \cdot \frac{\cot(36^{\circ})}{\cot(18^{\circ})} = 2 \cdot \frac{\cos(36^{\circ})}{\sin(36^{\circ})} \cdot \frac{\sin(18^{\circ})}{\cos(18^{\circ})}$Using $\sin(36^{\circ}) = 2 \sin(18^{\circ}) \cos(18^{\circ})$:$\frac{A_1}{A_2} = 2 \cdot \frac{\cos(36^{\circ})}{2 \sin(18^{\circ}) \cos(18^{\circ})} \cdot \frac{\sin(18^{\circ})}{\cos(18^{\circ})} = \frac{\cos(36^{\circ})}{\cos^2(18^{\circ})}$Using $\cos^2(18^{\circ}) = \frac{1 + \cos(36^{\circ})}{2}$:$\frac{A_1}{A_2} = \frac{\cos(36^{\circ})}{\frac{1 + \cos(36^{\circ})}{2}} = \frac{2 \cos(36^{\circ})}{1 + \cos(36^{\circ})}$Since $\cos(36^{\circ}) = \frac{\sqrt{5} + 1}{4}$:$\frac{A_1}{A_2} = \frac{2(\frac{\sqrt{5} + 1}{4})}{1 + \frac{\sqrt{5} + 1}{4}} = \frac{\frac{\sqrt{5} + 1}{2}}{\frac{5 + \sqrt{5}}{4}} = \frac{\sqrt{5} + 1}{2} \cdot \frac{4}{\sqrt{5}(\sqrt{5} + 1)} = \frac{2}{\sqrt{5}}$.

If a regular pentagon and a regular decagon have the same perimeter,then the ratio of their areas is

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