Question

ABCD is a trapezium in which AB is parallel to DC. If $\angle A$ = $\angle B$ = ${40}^{\circ }$, what are the measures of the other two angles?

• A. $\angle BCD$= ${140}^{\circ }$ and $\angle ADC$ = ${40}^{\circ }$
• B. $\angle BCD$= ${140}^{\circ }$ and $\angle ADC$ = ${140}^{\circ }$
• C. $\angle BCD$= ${40}^{\circ }$ and $\angle ADC$ = ${140}^{\circ }$
• D. $\angle BCD$= ${40}^{\circ }$ and $\angle ADC$ = ${40}^{\circ }$

Answer 1

The correct option is B $\angle BCD$= ${140}^{\circ }$ and $\angle ADC$ = ${140}^{\circ }$

Given: ABCD is a trapezium.
AB||DC and $\angle A$ = $\angle B$ = ${40}^{\circ }$

$\angle BAD$ + $\angle ADC$ = ${180}^{\circ }$ $\because \text{(Sum of interior angles on the same}\text{side of the transversal is}{180}^{\circ })$
${40}^{\circ }$ + $\angle ADC$ = ${180}^{\circ }$
$\angle ADC$ = ${180}^{\circ }$ - ${40}^{\circ }$ = ${140}^{\circ }$
$\angle ADC$ = ${140}^{\circ }$

$\angle ABC$ + $\angle BCD$ = ${180}^{\circ }$ $\because \text{(Sum of interior angles on the same}\text{side of the transversal is}{180}^{\circ })$
${40}^{\circ }$ + $\angle BCD$ = ${180}^{\circ }$
$\angle BCD$ = ${180}^{\circ }$ - ${40}^{\circ }$ = ${140}^{\circ }$
$\angle BCD$ = ${140}^{\circ }$

So, both $\angle BCD$ and $\angle ADC$ are ${140}^{\circ }$.