Question

In quadrilateral ABCD, side AB is parallel to side DC. If $\angle A:\angle D=1:2and\angle C:\angle B=4:5$

(i) Calculate each angle of the quadrilateral.

(ii) Assign a special name to quadrilateral ABCD.

Answer 1

$\because \angle A:\angle D=1:2Let\angle A=xand\angle B=2x\because \angle C:\angle B=4:5Let\angle C=4yand\angle B=5y\because AB||DC\therefore \angle A+\angle D={180}^{o}x+2x={180}^{o}3x={180}^{o}x={60}^{o}A={60}^o\angle D=2x=2\times 60={120}^{o}again\angle B+\angle C={180}^{o}5y+4y={180}^{o}9y={180}^{o}y={20}^o\therefore \angle B=5y=5\times 20={100}^o\angle C=4y=4\times 20={80}^{o}Hence\angle A={60}^o;\angle B={100}^o;\angle C={80}^{o}and\angle D={120}^o$.

(ii) Quadrilateral ABCD is a trapezium because one pair of opposite side is parallel