Question

In a regular pentagon $ABCDE$, draw a diagonal $BE$ and then find the meausre of:
$\left(iii\right)$ $\angle BED$

Answer 1

Step 1: Finding the measure of $\angle BAE$


We know,
Number of sides in the pentagon $=5$
Then each interior angle $=\frac{(n-2)\times {180}^o}{n}$
$=\frac{(5-2)\times {180}^o}{5}$

$=\frac{3\times {180}^o}{5}$

$={108}^o$

Hence, $\angle BAE={108}^o$. ...(1)

Step 2: Finding the measure of $\angle AEB$ in $\Delta ABE$,

$AB=AE$ [$\because$ it is a regular polygon]

$\Rightarrow \angle AEB=\angle ABE$ ... (2)

Therefore, $\angle BAE+\angle ABE+\angle AEB={180}^o$
[Angle sum property]

$\Rightarrow {108}^o+2\angle AEB={180}^o$ [from 1 and 2]

$\Rightarrow 2\angle AEB={180}^o-{108}^o$

$\Rightarrow 2\angle AEB={72}^o$

$\Rightarrow \angle AEB=\frac{{72}^o}{2}$

$\Rightarrow \angle AEB={36}^o$ ...(3)

Step 3: Finding the measure of $\angle BED$

$\angle AED={108}^o$

[$\because$ In a regular polygon all interior angles are equal]

$\angle AED=\angle AEB+\angle BED={108}^o$

$\Rightarrow {36}^o+\angle BED={108}^o$ [from 3]

$\Rightarrow \angle BED={108}^o-{36}^o$

$\Rightarrow \angle BED={72}^o$

Hence, the value of $\angle BED$ is ${72}^o$