Question

The magnitude of interior angle(in radians) of a regular pentagon is, while that of a regular octagon is

Answer 1

Interior angle of a $n$ sided regular polygon is $\frac{n-2}{n}\times {180}^{\circ }$
For a pentagon, $n=5$
$\Rightarrow$ Magnitude of Interior angle $=\frac{5-2}{5}\times {180}^{\circ }=\frac{3}{5}\times {180}^{\circ }={108}^{\circ }$
Now, Using the relation:
$\frac{D}{{180}^{\circ }}=\frac{R}{\pi }$
$\Rightarrow R=\pi \times \frac{D}{{180}^{\circ }}$

$\Rightarrow R=\pi \times \frac{{108}^{\circ }}{{180}^{\circ }}=\frac{3\pi }{5}$
Hence, magnitude of interior angle of a pentagon $=\frac{3\pi }{5}$

Similarly, for an octagon $n=8$
$\Rightarrow$ Magnitude of Interior angle $=\frac{8-2}{8}\times {180}^{\circ }=\frac{6}{8}\times {180}^{\circ }={135}^{\circ }$
Now, Using the relation:
$\frac{D}{{180}^{\circ }}=\frac{R}{\pi }$
$\Rightarrow R=\pi \times \frac{D}{{180}^{\circ }}$

$\Rightarrow R=\pi \times \frac{{135}^{\circ }}{{180}^{\circ }}=\frac{3\pi }{4}$
Hence, magnitude of interior angle of an octagon $=\frac{3\pi }{4}$