**2.2.1 Polygons II, PT3 Practice**

**Question 1:**

Diagram below shows a pentagon

*PQRST.**TPU*and*RSV*are straight lines.Find the value of

*x*.

Solution:Solution:

$\begin{array}{l}\text{Sumofinterioranglesofapentagon}\\ =\left(5-2\right)\times {180}^{o}\\ =3\times {180}^{o}\\ ={540}^{o}\\ \angle TSR={180}^{o}-{70}^{o}\\ \text{}={110}^{o}\\ \angle TPQ={180}^{o}-{85}^{o}\\ \text{}={95}^{o}\\ {x}^{o}={540}^{o}-\left({110}^{o}+{105}^{o}+{115}^{o}+{95}^{o}\right)\\ \text{}={540}^{o}-{425}^{o}\\ \text{}={115}^{o}\\ \text{}x=115\end{array}$

**Question 2:**

*PQRSTU*is a hexagon.

*APQ*and

*BTS*are straight lines.

Solution:Solution:

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

**Question 3:**

*PQRSTU. PUV*is a straight line.

Find the value of

*x*+*y*.

Solution:Solution:

$\begin{array}{l}\text{Size of each interior angle of a regular hexagon}\\ =\frac{\left(6-2\right)\times {180}^{o}}{6}\\ ={120}^{o}\\ {x}^{o}=\frac{{180}^{o}-{120}^{o}}{2}={30}^{o}\\ {y}^{o}={180}^{o}-{120}^{o}\\ \text{}={60}^{o}\\ {x}^{o}+{y}^{o}={30}^{o}+{60}^{o}\\ \text{}={90}^{o}\\ \text{}x+y=90\end{array}$

**Question 4:**

*KLMNP*is a regular pentagon.

*LKS*and

*MNQ*are straight lines

*.*

*Solution:*
$\begin{array}{l}\text{Size of each interior angle of a regular pentagon}\\ =\frac{\left(5-2\right)\times {180}^{o}}{5}\\ ={108}^{o}\\ \angle PKS=\angle PNQ={180}^{o}-{108}^{o}={72}^{o}\\ \text{Reflex angle}\angle KPN={360}^{o}-{108}^{o}={252}^{o}\end{array}$

$\begin{array}{l}\text{Sum of interior angles of a hexagon}\\ =\left(6-2\right)\times {180}^{o}\\ ={720}^{o}\\ \therefore {x}^{o}+{y}^{o}+{72}^{o}+{252}^{o}+{72}^{o}+{100}^{o}={720}^{o}\\ {x}^{o}+{y}^{o}={720}^{o}-{496}^{o}\\ \text{}={224}^{o}\\ \text{}x+y=224\end{array}$

**Question 5:**

*PQR*is an isosceles triangle and

*PRU*is a straight line.

Find the value of

*x*+*y.*

Solution:Solution:

$\begin{array}{l}{x}^{o}={180}^{o}-{20}^{o}-{20}^{o}={140}^{o}\\ \angle PRS={180}^{o}-{110}^{o}={70}^{o}\\ {y}^{o}+{85}^{o}+{75}^{o}+{70}^{o}={360}^{o}\\ {y}^{o}+{230}^{o}={360}^{o}\\ \text{}{y}^{o}={130}^{o}\\ {x}^{o}+{y}^{o}={140}^{o}+{130}^{o}\\ \text{}={270}^{o}\\ \text{}x+y=270\end{array}$