A regular pentagon and regular decagon have the same perimeter than ratio of their areas is: - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

Concept- We will first let each side of the pentagon as

2 x units

and decagon be

x

. Use the given condition to find the measure of each side of the decagon.
Let each side be of

2 x units

.
Hence, the perimeter of the pentagon is

5 (2 x) = 10 x

.
Then the sum of 10 sides is

10 x

Now, the measure of each side is

\frac{10 x}{10} = x

We know that the area of the pentagon is

5 x^{2} \cot \frac{π}{5}

Similarly, the area of the decagon is

\frac{5}{2} x^{2} \cot \frac{π}{10}

5 x^{2} \cot \frac{π}{5} : \frac{5}{2} x^{2} \cot \frac{π}{10}

Represent the ratio as a fraction.

\frac{5 x^{2} \cot \frac{π}{5}}{\frac{5}{2} x^{2} \cot \frac{π}{10}} = \frac{\cot \frac{π}{5}}{\frac{1}{2} \cot \frac{π}{10}}

\Rightarrow \frac{\cot \frac{π}{5}}{\frac{1}{2} \cot \frac{π}{10}} = \frac{2 \cot \frac{π}{5}}{\cot \frac{π}{10}}

We know that

cotθ = \frac{cosθ}{sinθ}

Then,

\frac{2 \frac{\cos \frac{π}{5}}{\sin \frac{π}{5}}}{\frac{\cos \frac{π}{10}}{\sin \frac{π}{10}}} = \frac{2 \cos \frac{π}{5}}{\sin \frac{π}{5}} \times \frac{\sin \frac{π}{10}}{\cos \frac{π}{10}}

Now,

\sin 2 θ = 2 sinθcosθ

\frac{2 \cos \frac{π}{5}}{\sin \frac{π}{5}} \times \frac{\sin \frac{π}{10}}{\cos \frac{π}{10}} = \frac{2 \cos \frac{π}{5}}{2 \sin \frac{π}{10} \cos \frac{π}{10}} \times \frac{\sin \frac{π}{10}}{\cos \frac{π}{10}}

On simplifying the above expression, we get

\frac{2 \cos \frac{π}{5}}{2 \sin \frac{π}{10} \cos \frac{π}{10}} \times \frac{\sin \frac{π}{10}}{\cos \frac{π}{10}} = \frac{\cos \frac{π}{5}}{\cos^{2} \frac{π}{10}}

Also,

2 \cos^{2} θ = 1 + \cos 2 θ

Then the above expression is simplified as,

\frac{2 \cos \frac{π}{5}}{1 + \cos (\frac{π}{5})}

Now, substitute the value of

\cos \frac{π}{5} = \frac{\sqrt{5} + 1}{4}

Then,

\frac{2 (\frac{\sqrt{5} + 1}{4})}{1 + (\frac{\sqrt{5} + 1}{4})} = \frac{2 (\sqrt{5} + 1)}{\sqrt{5} + 5}

Multiply numerator and denominator by

5 - \sqrt{5}

to rationalise the denominator
Then,

\frac{2 (\sqrt{5} + 1)}{\sqrt{5} + 5} \times \frac{5 - \sqrt{5}}{5 - \sqrt{5}} = \frac{8 \sqrt{5}}{20}

Now, multiply numerator and denominator by

\sqrt{5}

\frac{8 \sqrt{5}}{20} \times \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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