**Question 6:**In diagram below,

*PQRSTU*is a regular hexagon

*QUV*is a straight line.

Find the value of

*x*.

*Solution*:$\begin{array}{l}\text{Eachinteriorangle}\\ =\frac{\left(6-2\right)\times {180}^{o}}{6}\\ ={120}^{o}\\ \angle PUQ=\frac{{180}^{o}-{120}^{o}}{2}={30}^{o}\text{(}\u25b3PUQ\text{isaisoscelestriangle)}\\ {x}^{o}={180}^{o}-{30}^{o}\\ \text{}={150}^{o}\\ x=150\end{array}$

**Question 7:**In diagram below,

*PQRSTU*is a hexagon.

*UPE*,

*PQF*,

*QRG*,

*RSH*and

*UTJ*are straight lines.

Find the value of

*x*.

*Solution*:$\begin{array}{l}\text{Sumofalltheexterioranglesofanypolygon}={360}^{o}\\ {25}^{o}+3{x}^{o}+2{x}^{o}+{40}^{o}+{75}^{o}+{55}^{o}={360}^{o}\\ 5{x}^{o}+{195}^{o}={360}^{o}\\ \text{}5{x}^{o}={360}^{o}-{195}^{o}\\ \text{}={165}^{o}\\ \text{}{x}^{o}={165}^{o}\xf75\\ \text{}={33}^{o}\\ \text{}x=33\end{array}$

**Question 8:**In Diagram below,

*A*,

*B*,

*C*,

*D*and

*E*are vertices of a 9 sided regular polygon.

Find the value of

*x*.

*Solution*:$\begin{array}{l}\text{Exteriorangle}=\frac{{360}^{o}}{9}={40}^{o}\\ \text{Interiorangle}={180}^{o}-{40}^{o}={140}^{o}\\ \\ \text{Sumofinterioranglesofpentagon}ABCDE\\ =\left(5-2\right)\times {180}^{o}\\ ={540}^{o}\\ \\ {x}^{o}+{140}^{o}+{140}^{o}+{140}^{o}+{x}^{o}={540}^{o}\\ 2{x}^{o}={540}^{o}-{420}^{o}\\ \text{}={120}^{o}\\ \text{}{x}^{o}={60}^{o}\\ \text{}x=60\end{array}$