Question

A regular pentagon and regular decagon have the same perimeter than ratio of their areas is
A. $1:\sqrt 5 $
B. $2:\sqrt 5 $
C. $3:\sqrt 5 $
D. $4:\sqrt 5 $

Answer 1

Hint: We will first let each side of the pentagon as $2x$ units and decagon be x. Use the given condition to find the measure of each side of the decagon. Then, use the sides to find the area of the pentagon and the area of the decagon. Simplify the expression to find the ratio of the area of the pentagon and the area of the decagon.

Complete step by step Answer:

A regular pentagon is a pentagon with 5 equal sides.
Let each side be of $2x$ units.
Also, the perimeter of any shape is the sum of all its sides.
Hence, the perimeter of the pentagon is $5\left( {2x} \right) = 10x$.
We are given that the perimeter of the regular decagon is the same as the perimeter of the regular pentagon.
In a regular decagon, the polygon has 10 equal sides.
Then the sum of 10 sides is $10x$
Now, the measure of each side is $\dfrac{{10x}}{{10}} = x$
We know that the area of the pentagon is $5{x^2}\cot \left( {\dfrac{\pi }{5}} \right)$
Similarly, the area of the decagon is $\dfrac{5}{2}{x^2}\cot \left( {\dfrac{\pi }{{10}}} \right)$
But, we have to find the ratio of area of pentagon and area of decagon.
$5{x^2}\cot \left( {\dfrac{\pi }{5}} \right):\dfrac{5}{2}{x^2}\cot \left( {\dfrac{\pi }{{10}}} \right)$
Represent the ratio as a fraction.
$
  \dfrac{{5{x^2}\cot \left( {\dfrac{\pi }{5}} \right)}}{{\dfrac{5}{2}{x^2}\cot \left( {\dfrac{\pi }{{10}}} \right)}} = \dfrac{{\cot \left( {\dfrac{\pi }{5}} \right)}}{{\dfrac{1}{2}\cot \left( {\dfrac{\pi }{{10}}} \right)}} \\
   \Rightarrow \dfrac{{\cot \left( {\dfrac{\pi }{5}} \right)}}{{\dfrac{1}{2}\cot \left( {\dfrac{\pi }{{10}}} \right)}} = \dfrac{{2\cot \left( {\dfrac{\pi }{5}} \right)}}{{\cot \left( {\dfrac{\pi }{{10}}} \right)}} \\
$
We know that $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
Then,
 \[\dfrac{{\dfrac{{2\cos \left( {\dfrac{\pi }{5}} \right)}}{{\sin \left( {\dfrac{\pi }{5}} \right)}}}}{{\dfrac{{\cos \left( {\dfrac{\pi }{{10}}} \right)}}{{\sin \left( {\dfrac{\pi }{{10}}} \right)}}}} = \dfrac{{2\cos \left( {\dfrac{\pi }{5}} \right)}}{{\sin \left( {\dfrac{\pi }{5}} \right)}} \times \dfrac{{\sin \left( {\dfrac{\pi }{{10}}} \right)}}{{\cos \left( {\dfrac{\pi }{{10}}} \right)}}\]
Now, $\sin 2\theta = 2\sin \theta \cos \theta $
\[\dfrac{{2\cos \left( {\dfrac{\pi }{5}} \right)}}{{\sin \left( {\dfrac{\pi }{5}} \right)}} \times \dfrac{{\sin \left( {\dfrac{\pi }{{10}}} \right)}}{{\cos \left( {\dfrac{\pi }{{10}}} \right)}} = \dfrac{{2\cos \left( {\dfrac{\pi }{5}} \right)}}{{2\sin \left( {\dfrac{\pi }{{10}}} \right)\cos \left( {\dfrac{\pi }{{10}}} \right)}} \times \dfrac{{\sin \left( {\dfrac{\pi }{{10}}} \right)}}{{\cos \left( {\dfrac{\pi }{{10}}} \right)}}\]
On simplifying the above expression, we get
\[\dfrac{{2\cos \left( {\dfrac{\pi }{5}} \right)}}{{2\sin \left( {\dfrac{\pi }{{10}}} \right)\cos \left( {\dfrac{\pi }{{10}}} \right)}} \times \dfrac{{\sin \left( {\dfrac{\pi }{{10}}} \right)}}{{\cos \left( {\dfrac{\pi }{{10}}} \right)}} = \dfrac{{\cos \left( {\dfrac{\pi }{5}} \right)}}{{{{\cos }^2}\left( {\dfrac{\pi }{{10}}} \right)}}\]
Also, \[2{\cos ^2}\theta = 1 + \cos 2\theta \]
Then the above expression is simplified as,
\[\dfrac{{2\cos \left( {\dfrac{\pi }{5}} \right)}}{{1 + \cos \left( {\dfrac{\pi }{5}} \right)}}\]
Now, substitute the value of $\cos \dfrac{\pi }{5} = \dfrac{{\sqrt 5 + 1}}{4}$
Then, \[\dfrac{{2\left( {\dfrac{{\sqrt 5 + 1}}{4}} \right)}}{{1 + \left( {\dfrac{{\sqrt 5 + 1}}{4}} \right)}} = \dfrac{{2\left( {\sqrt 5 + 1} \right)}}{{5 + \sqrt 5 }}\]
Multiply numerator and denominator by $5 - \sqrt 5 $ to rationalise the denominator
Then, \[\dfrac{{2\left( {\sqrt 5 + 1} \right)}}{{5 + \sqrt 5 }} \times \dfrac{{5 - \sqrt 5 }}{{5 - \sqrt 5 }} = \dfrac{{8\sqrt 5 }}{{20}}\]
Now, multiply numerator and denominator by $\sqrt 5 $
\[\dfrac{{8\sqrt 5 }}{{20}} \times \dfrac{{\sqrt 5 }}{{\sqrt 5 }} = \dfrac{2}{{\sqrt 5 }}\]
Hence, the ratio of the area of the pentagon to the area of the decagon is $2:\sqrt 5 $
Therefore, option B is the correct answer.

Note: Perimeter is the sum of all the boundaries of the shape. Also, in a regular polygon, the measure of all sides are equal. Students should also know the formulas of trigonometry to simplify the questions.