A regular pentagon and regular decagon have the same perimeter than ratio of their areas is:

A regular pentagon and regular decagon have the same perimeter than ratio of their areas is:

1. A
$1:\sqrt{5}$
2. B
$2:\sqrt{5}$
3. C
$3:\sqrt{5}$
4. D
$4:\sqrt{5}$

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Solution:

Concept- We will first let each side of the pentagon as  and decagon be $x$. Use the given condition to find the measure of each side of the decagon.
Let each side be of .
Hence, the perimeter of the pentagon is .
Then the sum of 10 sides is $10x$
Now, the measure of each side is $\frac{10x}{10}=x$
We know that the area of the pentagon is $5{x}^{2}\mathit{cot}\frac{\pi }{5}$
Similarly, the area of the decagon is $\frac{5}{2}{x}^{2}\mathit{cot}\frac{\pi }{10}$
$5{x}^{2}\mathit{cot}\frac{\pi }{5}:\frac{5}{2}{x}^{2}\mathit{cot}\frac{\pi }{10}$
Represent the ratio as a fraction.
$\frac{5{x}^{2}\mathit{cot}\frac{\pi }{5}}{\frac{5}{2}{x}^{2}\mathit{cot}\frac{\pi }{10}}=\frac{\mathit{cot}\frac{\pi }{5}}{\frac{1}{2}\mathit{cot}\frac{\pi }{10}}$ $⇒\frac{\mathit{cot}\frac{\pi }{5}}{\frac{1}{2}\mathit{cot}\frac{\pi }{10}}=\frac{2\mathit{cot}\frac{\pi }{5}}{\mathit{cot}\frac{\pi }{10}}$ We know that $\mathit{cot\theta }=\frac{\mathit{cos\theta }}{\mathit{sin\theta }}$
Then,
$\frac{2\frac{\mathit{cos}\frac{\pi }{5}}{\mathit{sin}\frac{\pi }{5}}}{\frac{\mathit{cos}\frac{\pi }{10}}{\mathit{sin}\frac{\pi }{10}}}=\frac{2\mathit{cos}\frac{\pi }{5}}{\mathit{sin}\frac{\pi }{5}}×\frac{\mathit{sin}\frac{\pi }{10}}{\mathit{cos}\frac{\pi }{10}}$ Now, $\mathit{sin}2\theta =2\mathit{sin\theta cos\theta }$
$\frac{2\mathit{cos}\frac{\pi }{5}}{\mathit{sin}\frac{\pi }{5}}×\frac{\mathit{sin}\frac{\pi }{10}}{\mathit{cos}\frac{\pi }{10}}=\frac{2\mathit{cos}\frac{\pi }{5}}{2\mathit{sin}\frac{\pi }{10}\mathit{cos}\frac{\pi }{10}}×\frac{\mathit{sin}\frac{\pi }{10}}{\mathit{cos}\frac{\pi }{10}}$ On simplifying the above expression, we get
$\frac{2\mathit{cos}\frac{\pi }{5}}{2\mathit{sin}\frac{\pi }{10}\mathit{cos}\frac{\pi }{10}}×\frac{\mathit{sin}\frac{\pi }{10}}{\mathit{cos}\frac{\pi }{10}}=\frac{\mathit{cos}\frac{\pi }{5}}{{\mathit{cos}}^{2}\frac{\pi }{10}}$ Also, $2{\mathit{cos}}^{2}\theta =1+\mathit{cos}2\theta$
Then the above expression is simplified as,
Now, substitute the value of $\mathit{cos}\frac{\pi }{5}=\frac{\sqrt{5}+1}{4}$
Then,
Multiply numerator and denominator by $5-\sqrt{5}$ to rationalise the denominator
Then,
Now, multiply numerator and denominator by $\sqrt{5}$
$\frac{8\sqrt{5}}{20}×\frac{\sqrt{5}}{\sqrt{5}}=\frac{2}{\sqrt{5}}$ So, the ratio of the area of the pentagon to the area of the decagon is $2:\sqrt{5}.$
Hence, the correct option is 2.

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