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:
$\begin{array}{l}{x}^2={12}^2-6^2\\ x=\sqrt{144-36}\\ =10.39\text{ m}\\ TS=RS=RT\\ =10.39\text{ m }\times 2\\ =20.78\text{ m}\\ \text{Perimeter of the platform}\\ TS+RS+RT\\ =20.78\times 3\\ =63.34\text{ m}\end{array}$

Question 12:
Amy will place a ball on top of a pillar in Diagram below. Table below shows the diameters of three balls X, Y and Z.


Which ball X, Y or Z, can fit perfectly on the top of the pillar? Show the calculation to support Amy’s choice.

Solution:

$\begin{array}{l}\text{Let the radius of the top of the pillar}=r\text{ cm}\text{.}\\ O\text{ is the centre of the circle}\text{.}\\ \text{In }\Delta OQR,\\ r^2={\left(r-4\right)}^2+8^2\left(\text{using Pythagoras' theorem}\right)\\ r^2=r^2-8r+16+64\\ r^2=r^2-8r+80\\ r^2-r^2+8r=80\\ 8r=80\\ r=\frac{80}{8}\\ r=10\text{ cm}\\ \\ \text{Therefore, diameter}\\ =2\times 10\\ =20\text{ cm}\\ \\ \text{Ball }Y\text{ with diameter 20 cm can fit perfectly }\\ \text{on top of the pillar}\text{.}\end{array}$

Question 13:
Diagram below shows a rim of a bicycle wheel with a diameter of 26 cm. Kenny intends to build a holder for the rim.

Which of the rim holder, X, Y or Z, can fit the bicycle rim perfectly? Show the calculation to support your answer.

Solution:

$\begin{array}{l}\text{Let the radius of the rim holder}=r\text{ cm}\text{.}\\ O\text{ is the centre of the circle}\text{.}\\ \text{In }\Delta OQR,\\ r^2={\left(r-8\right)}^2+{12}^2\left(\text{using Pythagoras' theorem}\right)\\ r^2=r^2-16r+64+144\\ r^2=r^2-16r+208\\ r^2-r^2+16r=208\\ 16r=208\\ r=\frac{208}{16}\\ r=13\text{ cm}\\ \\ \text{Therefore, diameter}\\ =2\times 13\\ =26\text{ cm}\\ \\ \text{Rim holder }Z\text{ with diameter 26 cm can fit the bicycle perfectly}\text{.}\end{array}$

Question 6:
In the diagram below, CD is an arc of a circle with centre O.


Determine the area of the shaded region.
$\left(\text{Use }\pi =\frac{22}{7}\right)$

Solution:
$\begin{array}{l}\text{Area of sector}=\text{Area of circle}\times \frac{{72}^o}{{360}^o}\\ =\frac{22}{7}\times {\left(10\right)}^2\times \frac{{72}^o}{{360}^o}\\ =\frac{440}{7}{\text{ cm}}^2\\ \text{Area of }\Delta OBD=\frac{1}{2}\times 6\times 8\\ =24{\text{ cm}}^2\\ \text{Area of shaded region}=\frac{440}{7}-24\\ =38\frac{6}{7}{\text{ cm}}^2\end{array}$

Question 7:
In diagram below, ABC is a semicircle with centre O.

Calculate the area, in cm² , of the shaded region.
$\left(\text{Use }\pi =\frac{22}{7}\right)$

Solution:
$\begin{array}{l}\angle ACB={90}^o\\ AB=\sqrt{6^2+8^2}\\ =\sqrt{100}\\ =10\text{ cm}\\ \text{Radius}=10÷2\\ =5\text{ cm}\\ \\ \text{The shaded region}\\ =\left(\frac{1}{2}\times \frac{22}{7}\times 5\times 5\right)-\left(\frac{1}{2}\times 6\times 8\right)\\ =39\frac{2}{7}-24\\ =15\frac{2}{7}{\text{ cm}}^2\end{array}$

Question 8:
In diagram below, ABC is an arc of a circle centre O

The radius of the circle is 14 cm and AD = 2 DE.
Calculate the perimeter, in cm, of the whole diagram.
$\left(\text{Use }\pi =\frac{22}{7}\right)$

Solution:
$\begin{array}{l}\text{Length of arc }ABC\\ =\frac{3}{4}\times 2\pi r\\ =\frac{3}{4}\times 2\times \frac{22}{7}\times 14\\ =66\text{ cm}\\ \\ \text{Perimeter of the whole diagram}\\ =16+8+8+66\\ =98\text{ cm}\end{array}$

Question 9:
In diagram below, KLMN is a square and KLON is a quadrant of a circle with centre K.



Calculate the area, in cm², of the coloured region.
$\left(\text{Use }\pi =\frac{22}{7}\right)$

Solution:
$\begin{array}{l}\text{Area of the coloured region}\\ =\frac{{45}^o}{{360}^o}\times \pi r^2\\ =\frac{{45}^o}{{360}^o}\times \frac{22}{7}\times {14}^2\\ =77{\text{ cm}}^{\text{2}}\end{array}$

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

$\begin{array}{l}\text{Length of major arc}AB\\ =\frac{{216}^o}{{360}^o}\times 2\pi r\\ =\frac{{216}^o}{{360}^o}\times 2\times \frac{22}{7}\times 35\\ =132\text{cm}\end{array}$

$\begin{array}{l}\text{The area of sector }CBED\\ =\frac{{60}^o}{{360}^o}\times \pi r^2\\ =\frac{{60}^o}{{360}^o}\times \frac{22}{7}\times {14}^2\\ =102\frac{2}{3}{\text{ cm}}^{\text{2}}\end{array}$

= 62.90 cm

$$\boxed{\begin{array}{l}\text{ circumference}=\pi d,\text{ where }d=diameter\\ =2\pi r,\text{ where }r=radius\\ \\ \pi (pi)=\frac{22}{7}\text{ or }3.142\end{array}}$$
$\begin{array}{l}\text{Circumference}=\pi \times \text{Diameter}\\ =\frac{22}{7}\times 14\\ =44\text{cm}\end{array}$

$$\boxed{\begin{array}{l}\\ \frac{\text{Length of arc}}{\text{Circumference}}=\frac{\text{Angle at centre}}{{360}^o}\\ \end{array}}$$