Question

In a hexagon ABCDEF, side AB is parallel to side FE and $\angle B:\angle C:\angle D:\angle E=6:4:2:3.$ Find $\angle B,\angle C,\angle D$ and $\angle E$.

• A. $\angle B={216}^{\circ },\angle C={144}^{\circ },\angle D={72}^{\circ },\angle E={108}^{\circ }$
• B. $\angle B={236}^{\circ },\angle C={144}^{\circ },\angle D={72}^{\circ },\angle E={120}^{\circ }$
• C. $\angle B={226}^{\circ },\angle C={144}^{\circ },\angle D={72}^{\circ },\angle E={108}^{\circ }$
• D. $\angle B={296}^{\circ },\angle C={180}^{\circ },\angle D={108}^{\circ },\angle E={144}^{\circ }$

Answer 1

The correct option is A $\angle B={216}^{\circ },\angle C={144}^{\circ },\angle D={72}^{\circ },\angle E={108}^{\circ }$

In a hexagon ABCDEF, side AB is parallel to side FE.
$\angle A+\angle F={180}^o$ (Sum of interior angles on the same side of the transversal is supplementary).
$\angle B:\angle C:\angle D:\angle E=6:4:2:3$
Let $\angle B=6x,\angle C=4x,\angle D=2x,\angle E=3x$
Sum of interior angles of hexagon = ${180}^o(n-2)={180}^o(6-2)={720}^o$ (where n = number of sides of polygon )
Now,
$\angle A+\angle F+\angle B+\angle C+\angle D+\angle E={720}^o$
${180}^o+6x+4x+2x+3x={720}^o$
$15x={540}^o$
$x={36}^o$
$\angle B=6x=6\times {36}^o={216}^o$
$\angle D=2x=2\times {36}^o={72}^o$
$\angle C=4x=4\times {36}^o={144}^o$
$\angle E=3x=3\times {36}^o={108}^o$