Question 6:
In diagram below, PQRSTU is a regular hexagon QUV is a straight line.



Find the value of x.

Solution:
$\begin{array}{l}\text{Each interior angle}\\ =\frac{\left(6-2\right)\times {180}^o}{6}\\ ={120}^o\\ \angle PUQ=\frac{{180}^o-{120}^o}{2}={30}^o\text{ (}△PUQ\text{ is a isosceles triangle)}\\ x^o={180}^o-{30}^o\\ ={150}^o\\ x=150\end{array}$

Question 7:
In diagram below, PQRSTU is a hexagon. UPE, PQF, QRG, RSH and UTJ are straight lines.



Find the value of x.

Solution:
$\begin{array}{l}\text{Sum of all the exterior angles of any polygon}={360}^o\\ {25}^o+3x^o+2x^o+{40}^o+{75}^o+{55}^o={360}^o\\ 5x^o+{195}^o={360}^o\\ 5x^o={360}^o-{195}^o\\ ={165}^o\\ x^o={165}^o÷5\\ ={33}^o\\ x=33\end{array}$

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:
$\begin{array}{l}\text{Exterior angle }=\frac{{360}^o}{9}={40}^o\\ \text{Interior angle}={180}^o-{40}^o={140}^o\\ \\ \text{Sum of interior angles of pentagon }ABCDE\\ =\left(5-2\right)\times {180}^o\\ ={540}^o\\ \\ x^o+{140}^o+{140}^o+{140}^o+x^o={540}^o\\ 2x^o={540}^o-{420}^o\\ ={120}^o\\ x^o={60}^o\\ x=60\end{array}$

In Diagram below, PQRSTU is a hexagon. APQ and BTS are straight lines.

Solution:
$\begin{array}{l}\angle QPU={180}^o-{160}^o={20}^o\\ \text{Reflex}\angle PUT={360}^o-{80}^o={280}^o\\ \angle UTS={180}^o-{120}^o={60}^o\\ \angle TSR={180}^o-{35}^o={145}^o\\ \\ \text{Sum of interior angles of a hexagon}\\ =\left(6-2\right)\times {180}^o\\ ={720}^o\\ \\ x^o+y^o+{145}^o+{60}^o+{280}^o+{20}^o={720}^o\\ x^o+y^o={720}^o-{505}^o\\ ={215}^o\\ x+y=215\end{array}$

Diagram below shows a regular hexagon PQRSTU. PUV is a straight line.

In the diagram below, KLMNP is a regular pentagon. LKS and MNQ are straight lines.

In Diagram below, PQR is an isosceles triangle and PRU is a straight line.