# What is the measure of each interior angle of a regular pentagon?

Apr 18, 2017

${108}^{o}$

#### Explanation:

Consider this regular pentagon $A B C D E$.

Let us join vertices $A C$ and $E C$ as shown to form three triangles as shown. I have used letters $a , b , c , d , e , f , g , h , i$ to represent internal angles of triangles for sake of simplicity.

Since the sum of interior angles of a triangle is ${180}^{o}$,

In $\triangle A B C , b + c + d = {180}^{o}$
In $\triangle A C E , a + e + i = {180}^{o}$
In $\triangle E C D , h + f + g = {180}^{o}$

Sum of interior angles of the pentagon is

$a + b + c + d + e + f + g + h + i$
$= \left(b + c + d\right) + \left(a + e + i\right) + \left(h + f + g\right)$
$= {180}^{o} + {180}^{o} + 1 {80}^{o}$ [using the above three results]
$= {540}^{o}$

$i . e . \angle A + \angle B + \angle C + \angle D + \angle E = {540}^{o}$
Since it is a regular octagon, $\angle A = \angle B = \angle C = \angle D = \angle E$

$\implies \angle A + \angle A + \angle A + \angle A + \angle A = {540}^{o}$
$\implies 5 \cdot \angle A = {540}^{o}$
$\implies \angle A = \frac{540}{5} = {108}^{o} = \angle B = \angle C = \angle D = \angle E$

Hence internal angle of a regular pentagon is ${108}^{o}$.