Question

If a regular pentagon and regular decagon have the same perimeter, then their areas are as 2 to \[\sqrt{5}\].
A. true
B. false

Answer 1

Hint: We will suppose the sides of a regular pentagon and regular decagon. Then we will find the relation between the sides. Also, we will use the formula to find the area of polygon which is given as follows:
Area of polygon \[=\dfrac{n{{a}^{2}}}{4}\cot \left( \dfrac{\pi }{n} \right)\] where ‘n’ is the number of sides of the polygon.


Complete step-by-step answer:
We have been given that a regular pentagon and regular decagon have the same perimeter.
Now, let us suppose each side of a regular pentagon to be ‘a’ unit and that of a regular decagon to be ‘b’ unit.
We know that the perimeter of a regular polygon is given by na, where ‘n’ is the number of sides and ‘a’ is the length of a side. So a pentagon has 5 sides, so we get n = 5 and for a decagon, we get n = 10 as it has 10 sides.
So the perimeter of pentagon = 5a and perimeter of decagon = 10b.
\[\begin{align}
  & 5a=10b \\
 & a=2b......(1) \\
\end{align}\]
Hence the side of a pentagon is twice the side of a decagon.
As we know that the area of a polygon is as follows:
Area of polygon \[=\dfrac{n{{a}^{2}}}{4}\cot \left( \dfrac{\pi }{n} \right)\] where ‘n’ is the number of sides of the polygon and ‘a’ is measurement of its side.
So area of pentagon \[=\dfrac{5{{a}^{2}}}{4}\cot \left( \dfrac{\pi }{5} \right)\]
Area of decagon \[=\dfrac{10{{b}^{2}}}{4}\cot \left( \dfrac{\pi }{10} \right)\]
Now, the ratio of the area of a pentagon to the area of a decagon is given by as follows:
Ratio \[=\dfrac{Area\text{ }of\text{ }pentagon}{Area\text{ }of\text{ }decagon}\] \[=\dfrac{\dfrac{5{{a}^{2}}}{4}\cot \left( \dfrac{\pi }{5} \right)}{\dfrac{10{{b}^{2}}}{4}\cot \left( \dfrac{\pi }{10} \right)}\].
As we know that from equation (1) a = 2b.
So, by substituting the value of a = 2b in the ratio, we will get as follows:
Ratio \[=\dfrac{\dfrac{5}{4}{{\left( 2b \right)}^{2}}\cot {{36}^{\circ }}}{\dfrac{10{{b}^{2}}}{4}\cot {{18}^{\circ }}}=\dfrac{5\times 4{{b}^{2}}\cot {{36}^{\circ }}}{10{{b}^{2}}\cot {{18}^{\circ }}}=\dfrac{2\cot {{36}^{\circ }}}{\cot {{18}^{\circ }}}\].
As we know that \[\cot {{36}^{\circ }}=\dfrac{\cos {{36}^{\circ }}}{\sin {{36}^{\circ }}}\] and \[\cot {{18}^{\circ }}=\dfrac{\cos {{18}^{\circ }}}{\sin {{18}^{\circ }}}\].
 So by substituting these values in the ratio, we will get as follows:
     Ratio \[=\dfrac{2\cos {{36}^{\circ }}\times \sin {{18}^{\circ }}}{\sin {{36}^{\circ }}\times \cos {{18}^{\circ }}}\].
As we know that \[\sin 2\theta =2\sin \theta \cos \theta \].
Hence, \[sin{{36}^{\circ }}=2\sin {{18}^{\circ }}\cos {{18}^{\circ }}\].
So, by substituting the value of \[sin{{36}^{\circ }}\], in the above ratio, we will get as follows:
\[\Rightarrow \dfrac{2\cos {{36}^{\circ }}\times \sin {{18}^{\circ }}}{2\sin {{18}^{\circ }}\times \cos {{18}^{\circ }}\times \cos {{18}^{\circ }}}=\dfrac{2\cos {{36}^{\circ }}}{2{{\cos }^{2}}{{18}^{\circ }}}\]
On using the identity \[2{{\cos }^{2}}\theta =1+\cos 2\theta \], we will get as follows:
Ratio \[=\dfrac{2\cos {{36}^{\circ }}}{1+\cos {{36}^{\circ }}}\].
As we already know that \[\cos {{36}^{\circ }}=\dfrac{\sqrt{5}+1}{4}\].
Ratio \[=\dfrac{2\left( \dfrac{\sqrt{5}+1}{4} \right)}{1+\dfrac{\sqrt{5}+1}{4}}=2\left( \dfrac{\sqrt{5}+1}{4} \right)\times \dfrac{4}{5+\sqrt{5}}\]
\[\begin{align}
  & =\dfrac{2\left( \sqrt{5}+1 \right)}{\sqrt{5}\left( \sqrt{5}+1 \right)}=\dfrac{2}{\sqrt{5}} \\
 & =2:\sqrt{5} \\
\end{align}\]
Therefore, the correct answer of the given question is option A.

Note: We can easily calculate the ratio if we remember the value of \[\cot {{36}^{\circ }}\] and \[\cot {{18}^{\circ }}\]. Also, be careful while substituting the values of \[\cos {{36}^{\circ }}\] as there is a chance of calculation mistake here.