Question

In figure, if $ABCD$ is a parallelogram, $\angle ADE={40}^{\circ }$, and $\angle C={110}^{\circ }$, calculate the measure of $\angle BED$.

• A. ${30}^{\circ }$
• B. ${40}^{\circ }$
• C. ${70}^{\circ }$
• D. ${110}^{\circ }$
• E. ${150}^{\circ }$

Answer 1

The correct option is E ${150}^{\circ }$

Given, $\angle$ $C$ $=$ ${110}^o,\angle ADE={40}^o$

$\angle$ $D=\angle B$ {Opposite angle}

Also,

$\angle D+\angle C={180}^o$ {because these interior angles lie on the same side of the transversal}

$\angle D={180}^o-{110}^o={70}^o$

Therefore,

$\angle$ $EDC=\angle D-\angle ADE={70}^o-{40}^o={30}^o$

$\angle$ $EBC=\angle B={70}^o$, $\angle$ $DCB={110}^o$

Therefore,

$\angle$ $BED={360}^o-{110}^o-{70}^o-{30}^o={150}^o$ {Sum of angle of quadrilateral EBCD = ${360}^o$}