Question

In a hexagon ABCDEF, side AB is parallel to side FE and $\angle B:\angle C:\angle D:\angle E=6:4:2:3$. Find $\angle Band\angle D$.

Answer 1

Given : Hexagon ABCDEF in which AB || EF and $\angle B:\angle C:\angle D:\angle E=6:4:2:3$.

To find : $\angle Band\angle D$

Proof : No. of sides n = 6

$\therefore \text{Sum of interior angles}=(n-2)\times {180}^o=(6-2)\times {180}^o={720}^o\because AB||EF\left(Given\right)\angle A+\angle F={180}^{o}But\angle A+\angle B+\angle C+\angle D+\angle E+\angle F={720}^o\left(Proved\right)\angle B+\angle C+\angle D+\angle E+\angle {180}^o={720}^o\therefore \angle B+\angle C+\angle D+\angle E={720}^o-{180}^o={540}^{o}Ratio=6:4:2:3\text{Sum of parts}=6+4+2+3=15\angle B=\frac{6}{15}\times 540={216}^o\angle D=\frac{2}{15}\times {540}^o={72}^{o}Hence\angle B={216}^o;\angle D={72}^o$.