Question

The number of sides of two regular polygons are as 5:4 and the difference between their angles is \[{{9}^{\circ }}\]. Find the number of sides of the polygons.

Answer 1

Hint: First of all we will suppose the number of sides of the polygon to be a variable according to the question. Then we will use the formula of each angle of a regular polygon having ‘n’ sides is given by as follows:
Each angle \[=\left( \dfrac{n-2}{n} \right)\times {{180}^{\circ }}\]

Complete step-by-step answer:
Let us suppose the first polygon and second polygon to be \[{{P}_{1}}\] and \[{{P}_{2}}\] respectively and their sides are \[{{N}_{1}}\] and \[{{N}_{2}}\] respectively.
According to the question,
\[\dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{5}{4}\]
On cross multiplication, we get as follows:
\[4{{N}_{1}}=5{{N}_{2}}\]
On dividing the equation by 4 we get as follows:
\[\begin{align}
  & \dfrac{4}{4}\times {{N}_{1}}=\dfrac{5}{4}\times {{N}_{2}} \\
 & \Rightarrow {{N}_{1}}=\dfrac{5}{4}{{N}_{2}}.....(1) \\
\end{align}\]
We already know that each interior angle of n sided regular polygon is given as follows:
Each angle \[=\left( \dfrac{n-2}{n} \right)\times {{180}^{\circ }}\]
So, for \[{{P}_{1}}\], each angle \[=\left( \dfrac{{{N}_{1}}-2}{{{N}_{1}}} \right)\times {{180}^{\circ }}\]
So, for \[{{P}_{2}}\], each angle \[=\left( \dfrac{{{N}_{2}}-2}{{{N}_{2}}} \right)\times {{180}^{\circ }}\]
We have been given that the difference between their angles is \[{{9}^{\circ }}\].
\[\Rightarrow \left( \dfrac{{{N}_{1}}-2}{{{N}_{1}}} \right)\times {{180}^{\circ }}-\left( \dfrac{{{N}_{2}}-2}{{{N}_{2}}} \right)\times {{180}^{\circ }}={{9}^{\circ }}\]
On dividing the equation by \[{{180}^{\circ }}\], we get as follows:
\[\begin{align}
  & \Rightarrow \left( \dfrac{{{N}_{1}}-2}{{{N}_{1}}} \right)\times \dfrac{{{180}^{\circ }}}{{{180}^{\circ }}}-\left( \dfrac{{{N}_{2}}-2}{{{N}_{2}}} \right)\times \dfrac{{{180}^{\circ }}}{{{180}^{\circ }}}=\dfrac{{{9}^{\circ }}}{{{180}^{\circ }}} \\
 & \Rightarrow \dfrac{{{N}_{1}}-2}{{{N}_{1}}}-\dfrac{{{N}_{2}}-2}{{{N}_{2}}}={{\left( \dfrac{1}{20} \right)}^{\circ }} \\
\end{align}\]
On substituting \[{{N}_{1}}=\dfrac{5}{4}{{N}_{2}}\] using equation (1) we get as follows:
\[\Rightarrow \dfrac{\dfrac{5}{4}{{N}_{2}}-2}{\dfrac{5}{4}{{N}_{2}}}-\dfrac{{{N}_{2}}-2}{{{N}_{2}}}={{\dfrac{1}{20}}^{\circ }}\]
Taking \[\dfrac{1}{{{N}_{2}}}\] as common we get as follows:
\[\begin{align}
  & \Rightarrow \dfrac{1}{{{N}_{2}}}\left[ \dfrac{\dfrac{5}{4}{{N}_{2}}-2}{\dfrac{5}{4}}-{{N}_{2}}+2 \right]={{\dfrac{1}{20}}^{\circ }} \\
 & \Rightarrow \dfrac{1}{{{N}_{2}}}\left[ \dfrac{\dfrac{5}{4}{{N}_{2}}-2-\dfrac{5}{4}{{N}_{2}}+\dfrac{5}{4}\times 2}{\dfrac{5}{4}} \right]=\dfrac{1}{20} \\
 & \Rightarrow \dfrac{1}{{{N}_{2}}}\left[ \dfrac{-2+\dfrac{5}{2}}{\dfrac{5}{4}} \right]=\dfrac{1}{20} \\
 & \Rightarrow \dfrac{1}{{{N}_{2}}}\left[ \dfrac{\left( -4+5 \right)\times 4}{2\times 5} \right]=\dfrac{1}{20} \\
 & \Rightarrow \dfrac{1}{{{N}_{2}}}\left[ \dfrac{4}{10} \right]=\dfrac{1}{20} \\
\end{align}\]
On cross multiplication, we get as follows:
\[\begin{align}
  & 10{{N}_{2}}=10\times 4 \\
 & \Rightarrow {{N}_{2}}=\dfrac{20\times 4}{10}=8 \\
\end{align}\]
On substituting \[{{N}_{2}}=8\] in equation (1) we get as follows:
\[\Rightarrow {{N}_{1}}=\dfrac{5}{4}\times 8=5\times 2=10\]
Therefore, the number of sides of the polygons is 8 and 10.

Note: Be careful while solving the equation and take care of the sign also. Don’t get confused while formulating the equations and solving them as it might lead to errors. Use the formula of each angle of regular polygon \[=\left( \dfrac{n-2}{n} \right)\times {{180}^{\circ }}\] where n is the number of sides of regular polygon.