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

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## 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 $$360^\circ$$].

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