Question

The figure, given below, shows a pentagon ABCDE with sides AB and ED parallel to each other, and $\angle B:\angle C|\angle D=5:6:7$.

(i) Using formula, find the sum of interior angles of the pentagon.

(ii) Write the value of $\angle A+\angle E$

(iii) Find angles B, C and D.

Answer 1

(i) Sum of interior angles of the pentagon $=(5-2)\times {180}^o=3\times {180}^o={540}^o[\because \text{sum for a polygon of x sides}=(x-2)\times {180}^o]$

(ii) Since AB||ED

$\therefore \angle A+\angle E={180}^o$

(iii) Let $\angle B=5x\angle C=6x\angle D=7x\therefore 5x+6x+7x+{180}^o={540}^o(\angle A+\angle E={180}^o)$ Proved in (ii)

$18x={540}^o-{180}^o\Rightarrow 18x={360}^o\Rightarrow x={20}^o\therefore \angle B=5\times {20}^o={100}^o,\angle C=6\times 20={120}^o\angle D=7\times 20={140}^o$