Question

Exterior angle of a regular polygon having n sides is more than that of the polygon having ${{n}^{2}}$ side by ${{50}^{{}^\circ }}$. Find the number of sides of each polygon.

Answer 1

Hint: We know that exterior angle of a regular polygon having n-side $=\dfrac{{{360}^{{}^\circ }}}{n}$ also, exterior angle of regular polygon having ${{n}^{2}}$ sides = $\dfrac{{{360}^{{}^\circ }}}{{{n}^{2}}}$ also $\dfrac{360}{n}-\dfrac{360}{{{n}^{2}}}=50$. We will solve this quadratic equation to find the number of sides of a polygon.

Complete step-by-step solution -
It is given in the equation that exterior angle of a regular polygon having n-side is more than the exterior angle of a regular polygon having ${{n}^{2}}$ side by ${{50}^{{}^\circ }}$then we have to find the number of side of each polygon. We know that exterior angle of n side regular polygon is 50 greater than ${{n}^{2}}$ side regular polygon, therefore we get, $\dfrac{360}{n}-\dfrac{360}{{{n}^{2}}}=50$.
 Solving this equation by taking LCM as ${{n}^{2}}$, we get $\dfrac{360n-360}{{{n}^{2}}}=50$ solving further, we get $360n-360=50{{n}^{2}}$ or $50{{n}^{2}}-360n+360=0$ taking 10 as common factor and transposing it to RHS we get reduced equation as \[5{{n}^{2}}-36n+36=0\].
Solving the equation by splitting the middle term, we get \[5{{n}^{2}}-30n-6n+36=0\] solving further, we get \[5n\left( n-6 \right)-6\left( n-6 \right)=0\] therefore we get \[\left( 5n-6 \right)\left( n-6 \right)=0\]. Hence the value of n we get are $n=6,n=\dfrac{6}{5}$. Since the number of sides cannot be a fraction, and has to be an integer, therefore neglecting $n=\dfrac{6}{5}$, we get $n=6$.
That is, first polygon has 6 sides and exterior angle as $\dfrac{360}{6}={{60}^{{}^\circ }}$ and the second polygon has sides ${{n}^{2}}$ that is ${{6}^{2}}=36$ and exterior angle as $\dfrac{360}{36}={{10}^{{}^\circ }}$. Also, according to question we get ${{60}^{{}^\circ }}-{{10}^{{}^\circ }}={{50}^{{}^\circ }}$.

Note: Many students confuse in last step of the solution that why we are considering only 6 as the value of n and not $\dfrac{6}{5}$ as the value of n. This is because the number of sides is counted in natural numbers only and $\dfrac{6}{5}$ is not a natural number, instead a fraction. It is very true that both 6 and $\dfrac{6}{5}$ are the roots of the quadratic equation so formed, but simply number of sides cannot be a rational number, thus neglecting $\dfrac{6}{5}$.