Question

ABCDEF is a regular hexagon. Find each angle of $△$ABE.

Answer 1

Given that $ABCDEF$ is a regular hexagon, so all sides of it are equal in length.

Sum of the angles of a $n-$ sided polygon $=(n-2)\times 180$

Hence,

Sum of all angles of a hexagon

$=(6-2)\times 180$

$=720$

And,

Each angle of a regular hexagon

$=\frac{720}{6}$

$=120$.......(1)

Quadrilateral $ABEF$ contains half the angle measure of the hexagon.

Hence, its interior angle measure

$=720/2=360$

$\angle AFE=120$.......(ii) [ from (1)]

In $△AFE$,

$AF=EF$ [ since, they are a part of regular hexagon].......(iii)

$\Rightarrow \angle FAE=\angle FEA$

[ Angles opposite to equal sides are equal]

In $△AFE$,

$\angle FAE+\angle FEA+\angle AFE=180$

[ sum of all angles of a triangle is ${180}^o$]

substituting (ii) & (iii) in above equation,

$\angle FAE=\angle FEA={30}^o$

$\angle FAE+\angle EAB={120}^o$ [ Angle of a regular hexagon]

$\therefore \angle EAB={90}^o$.......(iv)

$\angle ABE=1/2\times {120}^o$ $[\because \angle ABE$ bisects $\angle ABC]$

$={60}^o$

In $△ABE$,

$\angle AEB+\angle ABE+\angle EBA={180}^o$

[ sum of all angles of a triangle]

$\angle AEB={180}^o-{60}^o-{90}^o={30}^o$ [ from (iv) & (v) ]

$\angle ABE={60}^o\angle EAB={90}^o\angle AEB={30}^o$