Question 7:
Solution:

Question 8:
Solution:

Question 5 (4 marks):
Diagram shows a circle with centre O.

Diagram

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 6 (3 marks):
Diagram shows two sectors AOD and BOC of two concentric circles with centre O.

Diagram

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 10 (8 marks):
Diagram shows a circle and a sector of a circle with a common centre O. The radius of the circle is r cm.


It is given that the length of arc PQ and arc RS are 2 cm and 7 cm respectively. QR = 10 cm.
[Use θ = 3.142]
Find
(a) the value of r and of θ,
(b) the area, in cm², of the shaded region.


Solution:
(a)
$\begin{array}{l}\text{Length of arc }PQ=2\text{ cm}\\ r\theta =2\mathrm{.................}\left(1\right)\\ \\ \text{Length of arc }RS=7\text{ cm}\\ \left(r+10\right)\theta =7\\ r\theta +10\theta =7\mathrm{.................}\left(2\right)\\ \\ \text{Substitute }\left(1\right)\text{ into }\left(2\right):\\ 2+10\theta =7\\ 10\theta =5\\ \theta =\frac{5}{10}\\ \theta =0.5\text{ rad}\\ \\ \text{From}\left(1\right):\\ \text{When }\theta =0.5\text{ rad,}\\ r\times 0.5=2\\ r=4\end{array}$

(b)
$\begin{array}{l}OS=OR=4+10=14\text{ cm}\\ \text{Area of shaded region}\\ =\text{area of }\Delta ORS-\text{ area of sector }OPQ\\ =\left(\frac{1}{2}\times {14}^2\times \sin 0.5\text{ rad}\right)-\left(\frac{1}{2}\times 4^2\times 0.5\right)\\ =42.981{\text{ cm}}^2\end{array}$

Question 9 (7 marks):
Mathematics Society of SMK Mulia organized a competition to design a logo for the society.


Diagram shows the circular logo designed by Adrian. The three blue coloured regions are congruent. It is given that the perimeter of the blue coloured region is 20π cm.
[Use π = 3.142]
Find
(a) the radius, in cm, of the logo to the nearest integer,
(b) the area, in cm², of the yellow coloured region.

Solution:
(a)
$\begin{array}{l}6\text{ arcs }=20\pi \\ 6r\theta =20\pi \\ 6r\left[\cancel{{60}^o}\times \frac{\pi }{{\cancel{{180}^o}}^3}\right]=20\pi \\ 2\pi r=20\pi \\ r=10\text{ cm}\end{array}$

(b)

$\begin{array}{l}\text{Area of yellow coloured region}\\ =3\left[\text{area of triangle }OAB\right]-6\left[\text{area of segment}\right]\\ =3\left[\frac{1}{2}ab\sin C\right]-6\left[\frac{1}{2}{r}^2\left(\theta -\sin \theta \right)\right]\\ =3\left[\frac{1}{2}\left(10\right)\left(10\right)\sin {120}^o\right]-6\left[\frac{1}{2}{\left(10\right)}^2\left(\theta -\sin \theta \right)\right]\\ =3\left(43.3013\right)-6\left[50\left(1.0473-\sin 1.0473\right)\right]\leftarrow \boxed{\begin{array}{l}\text{change to rad mode}\\ \theta ={60}^o\times \frac{3.142}{{180}^o}=1.0473\end{array}}\\ =129.9039-6\left(9.0612\right)\\ =129.9039-54.3672\\ =75.54{\text{ cm}}^2\end{array}$

Question 8:


Solution:
(a)
$\begin{array}{l}\frac{1}{2}AB=\sin 0.41\times 10\to \left(\text{Change the calculator to Rad mode}\right)\\ \frac{1}{2}AB=3.99\\ \therefore \text{The length of chord }AB=3.99\times 2=7.98cm.\end{array}$

(b)
$\begin{array}{l}\text{Let }\frac{1}{2}\theta =\alpha ,\theta =2\alpha \\ \sin \alpha =\frac{3.99}{5}\\ \alpha =0.924\text{ rad}\\ \therefore \theta =0.924\times 2=1.848\text{ radian}\text{.}\end{array}$

(c)

Question 7:


Solution:
$\begin{array}{l}\text{Area of shaded region}\\ \text{=}\frac{1}{2}{r}^2\left(\theta -\sin \theta \right)\left(\text{change calculator to Rad mode}\right)\\ \text{=}\frac{1}{2}{\left(10\right)}^2\left(\frac{2\pi }{3}-\sin \frac{2\pi }{3}\right)\\ =50\left(2.094-0.866\right)\\ =61.40{\text{ cm}}^2\end{array}$

Question 6:
Diagram below shows a circle PQT with centre O and radius 7 cm.

QS is a tangent to the circle at point Q and QSR is a quadrant of a circle with centre Q. Q is the midpoint of OR and QP is a chord. OQR and SOP are straight lines.
[Use π = 3.142]
Calculate
(a) angle θ, in radians,
(b) the perimeter, in cm ,of the shaded region,
(c) the area, in cm² ,of the shaded region.


Solution:
(a)
$\begin{array}{l}OQ=QR=QS=7\text{ cm}\\ \tan \theta =1\\ \theta ={45}^o\\ ={45}^o\times \frac{\pi }{{180}^o}\\ =0.7855\text{ rad}\end{array}$

(b)
$\begin{array}{l}\text{Length of arc }RS\\ =7\times \left(0.7855\times 2\right)\to \boxed{\begin{array}{l}\pi \text{ rad}={180}^o\\ {45}^o=0.7855\text{ rad}\\ {90}^o=0.7855\times \text{2 rad}\end{array}}\\ =7\times 1.571\\ =10.997\text{ cm}\\ \\ \text{Length of arc }QP\\ =7\times \left(0.7855\times 3\right)\\ =7\times 2.3565\\ =16.496\text{ cm}\\ \\ \text{Length of chord }QP\\ =\sqrt{7^2+7^2-2\left(7\right)\left(7\right)\cos {135}^o}\leftarrow \boxed{\begin{array}{l}\text{refer form 4 chapter 10}\\ \left(\text{solution of triangle}\right)\text{for cosine rule}\end{array}}\\ =\sqrt{167.30}\\ =12.934\text{ cm}\\ \\ \text{Perimeter of the shaded region}\\ =7+7+10.997+16.496+12.934\\ =54.427\text{ cm}\end{array}$

(c)
$\begin{array}{l}\text{Area of shaded region}\\ =\left(\frac{1}{2}\times 7^2\times 1.571\right)+\left(\frac{1}{2}\times 7^2\times 2.3565\right)\\ -\left(\frac{1}{2}\times 7\times 7\times \text{sin}{135}^o\right)\leftarrow \boxed{\begin{array}{l}\text{refer form 4 chapter 10}\\ \left(\text{solution of triangle}\right)\text{for area rule}\end{array}}\\ =38.4895+57.7343-17.3241\\ =78.8997{\text{ cm}}^2\end{array}$

Question 5:
Diagram below shows a semicircle PTS, centre O and radius 8 cm. PTR is a sector of a circle with centre P and Q is the midpoint of OS.

[Use π = 3.142]
Calculate
(a) ∠TOQ, in radians,
(b) the length , in cm , of the arc TR ,
(c) the area, in cm² ,of the shaded region.

Solution:
(a)
$\begin{array}{l}\cos \angle TOQ=\frac{4}{8}=\frac{1}{2}\\ \angle TOQ={60}^{\text{o}}\\ =60\times \frac{\pi }{180}\\ =1.047\text{ radians}\end{array}$

(b)


$\begin{array}{l}\angle TPO={30}^{\text{o}}\\ =30\times \frac{\pi }{180}\\ =0.5237\\ P{T}^2=8^2+8^2-2\left(8\right)\left(8\right)\cos 120\\ P{T}^2=192\\ PT=\sqrt{192}\\ PT=13.86\text{ cm}\\ \\ \text{Length of arc }TR=13.86\times 0.5237\\ =7.258\text{ cm}\end{array}$

(c)
$\begin{array}{l}\text{Area of sector }PTR\\ =\frac{1}{2}\times {13.86}^2\times 0.5237\\ =50.30{\text{ cm}}^2\\ \\ \text{Length }TQ\\ =\sqrt{P{T}^2-P{Q}^2}\\ =\sqrt{{13.86}^2-{12}^2}\\ =6.935\text{ cm}\\ \\ \text{Area of }△PTQ\\ =\frac{1}{2}\times 12\times 6.935\\ =41.61{\text{ cm}}^2\\ \\ \text{Area of shaded region}\\ =50.30-41.61\\ =8.69{\text{ cm}}^2\end{array}$

Question 4:



Solution:
(a)

Question 5:
Solution:

Question 6:
Solution: