1.4.1 Circular Measure SPM Practice (Paper 1 Question 1 – 10)

Question 1:

Solution:
(a)
$\begin{array}{l}s=r\theta \\ 5.2=13\left(\angle COB\right)\\ \angle COB=0.4\text{ radian}\end{array}$

(b)
$\begin{array}{l}\cos \angle COB=\frac{OA}{OC}\\ \cos 0.4=\frac{OA}{13}\text{ (change calculator to Rad mode)}\\ OA=11.97\text{ cm}\\ \therefore AB=13-11.97=1.03\text{ cm}\\ \\ CA=\sqrt{{13}^2-{11.97}^2}\\ CA=5.07\text{ cm}\end{array}$

Question 2:

Solution:
(a) Arc AB = 6cm
$\begin{array}{l}\text{Area of sector }AOB\\ =\frac{1}{2}{r}^2\theta =\frac{1}{2}{\left(5\right)}^2\left(\frac{6}{5}\right)=15c{m}^2\end{array}$
$\begin{array}{l}\text{Area of shaded region}\\ \text{=}\frac{1}{2}{r}^2\left(\theta -\sin \theta \right)\left(\text{change calculator to Rad mode}\right)\\ \text{=}\frac{1}{2}{\left(5\right)}^2\left(\frac{6}{5}-\sin \frac{6}{5}\right)\\ =3.35{\text{ cm}}^2\end{array}$

Question 3:
Solution:

Question 4:
Solution:

Question 5:
Solution:
(a) s = r θ
$=\frac{1}{2}{\left(8\right)}^2\left(0.75\right)-\frac{1}{2}{\left(3\right)}^2\left(1.2\right)$

Question 6:
Solution:

Question 7 (SPM 2017 – 3 marks):
Diagram 7 shows two sectors AOD and BOC of two concentric circles with centre O.

Diagram 7

The angle subtended at the centre O by the major arc AD is 7α radians and the perimeter of the whole diagram is 50 cm.
Given OB = r cm, OA = 2OB and ∠BOC = 2α, express r in terms of α.


Solution:

$\begin{array}{l}\text{Length of major arc }AOD\\ =2r\times 7\alpha \\ =14r\alpha \\ \\ \text{Length of minor arc }BOC\\ =r\times 2\alpha \\ =2r\alpha \\ \\ \text{Perimeter of the whole diagram}\\ =50\text{ cm}\\ 14r\alpha +2r\alpha +r+r=50\\ 16r\alpha +2r=50\\ 8r\alpha +r=25\\ r\left(8\alpha +1\right)=25\\ r=\frac{25}{8\alpha +1}\end{array}$

Question 8 (SPM 2018 – 4 marks):
Diagram 5 shows a circle with centre O.

Diagram 5

PR and QR are tangents to the circle at points P and Q respectively. It is given that the length of minor arc PQ is 4 cm and $OR=\frac{5}{\alpha }\text{ cm}\text{.}$  
Express in terms of α,
(a) the radius, r, of the circle,
(b) the area, A, of the shaded region.


Solution:
(a)
$\begin{array}{l}\text{Given }{s}_{PQ}=4\\ r\alpha =4\\ r=\frac{4}{\alpha }\text{ cm}\end{array}$

(b)

$\begin{array}{l}PR=\sqrt{{\left(\frac{5}{\alpha }\right)}^2-{\left(\frac{4}{\alpha }\right)}^2}\\ PR=\sqrt{\frac{9}{{\alpha }^2}}\\ PR=\frac{3}{\alpha }\\ \\ A=\text{ Area of shaded region}\\ A=\text{ Area of quadrilateral }OPRQ-\\ \text{Area of sector }OPQ\\ =2\left(\text{Area of }△OPR\right)-\frac{1}{2}{r}^2\theta \\ =2\left[\frac{1}{2}\times \frac{3}{\alpha }\times \frac{4}{\alpha }\right]-\left[\frac{1}{2}\times {\left(\frac{4}{\alpha }\right)}^2\times \alpha \right]\\ =\frac{12}{{\alpha }^2}-\frac{8}{\alpha }\\ =\frac{12-8\alpha }{{\alpha }^2}{\text{ cm}}^2\end{array}$

Question 9 (SPM 2005):
Diagram 6 shows a circle with centre O. 

The length of the minor arc AB is 16 cm and the angle of the major sector AOB is 290^o.
Using π = 3.142, find
(a) the value of θ, in radians.
(Give your answer correct to four significant figures.) 

(b) the length, in cm , of the radius of the circle.
[3 marks]


Answer:
(a)
$$ \begin{aligned} \theta & =360^{\circ}-290^{\circ} \\ & =70^{\circ} \\ & =70 \times \frac{3.142}{180} \\ & =1.222 \text { radians } \end{aligned} $$

(b)
$$ \begin{aligned} 16 & =r \times 1.222 \\ r & =\frac{16}{1.222} \\ & =13.09 \mathrm{~cm} \end{aligned} $$

Question 10 (SPM 2006):
Diagram 7 shows sector OAB with centre O and sector AXY with centre A. 

Given OB = 10 cm, AY = 4 cm, ∠XAY = 1.1 radians and the length of AB = 7 cm.
Calculate
(a) the value of θ, in radian.
(b) the area, in cm², of the shaded region.
[4 marks]


Answer:
(a)
$$ \begin{aligned} 10 \theta & =7 \\ \theta & =0.7 \mathrm{rad} \end{aligned} $$

(b)
$$ \begin{aligned} & \text { Area of the shaded region } \\ & =\text { Area of sector } O A B \text { – Area of sector } A X Y \\ & =\frac{1}{2}\left(10^2\right)(0.7)-\frac{1}{2}\left(4^2\right)(1.1) \\ & =35-8.8 \\ & =26.2 \mathrm{~cm}^2 \end{aligned} $$