Question

In the given figure, $AB$ || $ED$ and angle B, C and D are in the ratio $5:6:7$. Find the measure of angles B, C and D.

• A. $\angle B={100}^{\circ },\angle C={120}^{\circ }$ and $\angle D={140}^{\circ }$
• B. $\angle B={110}^{\circ },\angle C={120}^{\circ }$ and $\angle D={140}^{\circ }$
• C. $\angle B={100}^{\circ },\angle C={120}^{\circ }$ and $\angle D={110}^{\circ }$
• D. none of the above

Answer 1

The correct option is A $\angle B={100}^{\circ },\angle C={120}^{\circ }$ and $\angle D={140}^{\circ }$

A pentagon ABCDE with sides AB and ED parallel to each other.
and $\angle B:\angle C:\angle D=5:6:7.$

Then, AE behaves as a transversal.

$\angle BAE+\angle AED={180}^o$ (Sum of interior angles of the same side of transversal is supplementary.

So,$\angle A+\angle E={180}^o$

Sum of all interior angles of pentagon are = ${180}^o(n-2)={180}^o(5-2)={540}^o$

where n = number of sides of polygon

Let $\angle B=5x,\angle C=6x,\angle D=7x$

Now,

$\angle A+\angle E+\angle B+\angle C+\angle D={540}^o$

${180}^o+5x+6x+7x={540}^o$

$x={20}^o$

$\angle B=5\times {20}^o={100}^o$

$\angle C=6\times {20}^o={120}^o$

$\angle D=7\times {20}^o={140}^o$