5.2.4 Circles, PT3 Focus Practice

Question 10:
Diagram below shows two quadrants, AOC and EOD with centre O.


Sector AOB and sector BOC have the same area.
Calculate the area, in cm², of the coloured region.
$\left(\text{Use }\pi =\frac{22}{7}\right)$

Solution:
$\begin{array}{l}\text{Area of the sector }AOB=\text{Area of the sector }BOC\\ \text{Therefore, }\angle AOB=\angle BOC\\ ={90}^o÷2\\ ={45}^o\\ \text{Area of the coloured region}\\ =\frac{{45}^o}{{360}^o}\times \frac{22}{7}\times {16}^2\\ =100\frac{4}{7}{\text{ cm}}^2\end{array}$

Question 11:
The Town Council plans to build an equilateral triangle platform in the middle of a roundabout. The diameter of circle RST is 24 m and the perpendicular distance from R to the line ST is 18 m. as shown in Diagram below.
Find the perimeter of the platform.


Solution:

Given diameter = 24 m
hence radius = 12 m
O is the centre of the circle.
Using Pythagoras’ theorem:
$\begin{array}{l}{x}^2={12}^2-6^2\\ x=\sqrt{144-36}\\ =10.39\text{ m}\\ TS=RS=RT\\ =10.39\text{ m }\times 2\\ =20.78\text{ m}\\ \text{Perimeter of the platform}\\ TS+RS+RT\\ =20.78\times 3\\ =63.34\text{ m}\end{array}$