# 2.2.1 Polygons II, PT3 Practice

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:
$\begin{array}{l}\text{Sum of interior angles of a pentagon}\\ =\left(5-2\right)×{180}^{o}\\ =3×{180}^{o}\\ ={540}^{o}\\ \angle TSR={180}^{o}-{70}^{o}\\ \text{}={110}^{o}\\ \angle TPQ={180}^{o}-{85}^{o}\\ \text{}={95}^{o}\\ {x}^{o}={540}^{o}-\left({110}^{o}+{105}^{o}+{115}^{o}+{95}^{o}\right)\\ \text{}={540}^{o}-{430}^{o}\\ \text{}={110}^{o}\\ \text{}x=110\end{array}$

Question 2:
In Diagram below, PQRSTU is a hexagon. APQ and BTS are straight lines.
Find the value of x + y.

Solution:

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

Question 3:
Diagram below shows a regular hexagon PQRSTU. PUV is a straight line.
Find the value of x + y.

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

Question 4:
In the diagram below, KLMNP is a regular pentagon. LKS and MNQ are straight lines.
Find the value of x + y.

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

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

Question 5:
In Diagram below, PQR is an isosceles triangle and PRU is a straight line.
Find the value of x + y.

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