Question

In a regular pentagon ABCDE, draw a diagonal BE and then find the measure of $\angle BED$

• A. ${72}^{\circ }$
• B. ${36}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is A ${72}^{\circ }$

In a regular pentagon ABCDE, BE is a diagonal.
$\angle BAE$ is the interior angle of the polygon.
Each interior angle of a regular polygon is = $\frac{{180}^o(n-2)}{n}$
where $n=$ number of sides of polygon, $n=5$
$\angle BAE=\frac{{180}^o(5-2)}{5}={108}^o$
So, each angle of polygon is ${108}^o$ also $\angle AED={108}^o$
Now, $△ABE$ is an isosceles triangle in which AB = AE.
So, $\angle ABE=\angle AEB$
$\angle ABE+\angle AEB+\angle BAE={180}^o$
$=>2\angle AEB={180}^o-{108}^o$
$=>\angle AEB=\frac{{72}^o}{2}={36}^o$
$\angle BED=\angle AED-\angle AEB={108}^o-{36}^o$
$\angle BED={72}^o$