# Ex.3.2 Q5 Understanding Quadrilaterals Solution - NCERT Maths Class 8

## Question

(a) Is it possible to have a regular polygon with measure of each exterior angle as \(22^\circ\)?

(b)Can it be an interior angle of a regular polygon? Why?

## Text Solution

**What is Known?**

Measure of an exterior angle is \(22^\circ\).

**What is Unknown?**

To find whether a regular polygon with exterior angle \(= 22^\circ\) is possible or not.

**Reasoning:**

\[\begin{align}{\text{Number of sides of any polygon}}\,{\text{ }}&= \,\frac{{{\rm{36}}{{\rm{0}}^{\rm{\circ}}}}}{{{\text{Exterior angle}}}}\\{\text{where the answer is whole number}}{\text{.}}

\end{align}\]

**Steps:**

(a) Is it possible to have a regular polygon with measure of each exterior angle as \(22^\circ\)?

Total measure of all exterior angles \(= {\rm{36}}0^\circ \)

Let number of sides be \(= \rm n\).Measure of each exterior angle \(= {\rm{22}}^\circ \)

\[\begin{align}{\text{Therefore, the number of sides }}&=\,\frac{{{\text{Sum of exterior angles}}}}{{{\text{Each exterior angle}}}}\\&= \frac{{{{360}^{\rm{o}}}}}{{{{22}^{\rm{o}}}}}\\&= 16.36\end{align}\]We cannot have regular polygon with each exterior angle = \(22^\circ \) as the number of sides is not a whole number [ \(22\) is not a perfect divisor of \(360^\circ \)].

(b) Can it be an interior angle of a regular polygon? Why?

\[\begin{align}{\text{Measure of each interior angle}}\,{\rm{ }}&= 2{{\rm{2}}^{\rm{\circ}}}\\

{\text{Measure of each exterior angle }}&= 180^\circ - {\rm{22}}^\circ {\rm{ }}\\&= {\rm{158}}^\circ

\end{align}\]

\[\begin{align}{\text{Number of sides}}\,{\rm{ }}&= \frac{{{\text{Sum of exterior angles}}}}{{{\text{Each exterior angle}}}}\\&= \frac{{{\rm{36}}0^\circ }}{{{\rm{158}}^\circ }}\\&= 2.27\end{align}\]

We cannot have regular polygon with each interior angle as \(22^\circ \) because the number of sides is not a whole number[ \(22\) is not a perfect divisor of \(360^\circ \)].