Question

In a pentagon ABCDE, AB is parallel to DC and $\angle A:\angle E:\angle D=3:4:5$. Find the measure of $\angle E$.

• A. ${142}^{\circ }$
• B. ${111}^{\circ }$
• C. ${136}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is D ${120}^{\circ }$

In a pentagon ABCDE, AB is parallel to DC and $\angle A:\angle E:\angle D=3:4:5$
So, $\angle B+\angle C={180}^o$ (Sum of interior angles on the same side of transversal are supplementary.)
Let $\angle A=3x,\angle E=4x,\angle D=5x$
Sum of all interior angles of pentagon = ${180}^o(n-2)={180}^o(5-2)={540}^o$
Now,
$\angle B+\angle C+\angle A+\angle E+\angle D={540}^o$
$=>{180}^o+3x+4x+5x={540}^o$
$=>12x={360}^o$
$=>x={30}^o$
$\angle E=4x=4\times {30}^o={120}^o$