1.5.2 Circular Measure SPM Practice (Paper 2 Question 6 – 10)

Question 6:


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:
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 8:
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)
$\begin{array}{l}\text{Area of shaded region}\\ =\left(\frac{1}{2}\times 7^2\times 1.571\right)+\left(\frac{1}{2}\times 7^2\times 2.3565\right)\\ -\left(\frac{1}{2}\times 7\times 7\times \text{sin}{135}^o\right)\leftarrow \boxed{\begin{array}{l}\text{refer form 4 chapter 10}\\ \left(\text{solution of triangle}\right)\text{for area rule}\end{array}}\\ =38.4895+57.7343-17.3241\\ =78.8997{\text{ cm}}^2\end{array}$

(c) the area, in cm², of the shaded region BCDF.   [5 marks]

Question 10 (SPM 2017 - 8 marks):
Diagram 1 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)
$\begin{array}{l}OS=OR=4+10=14\text{ cm}\\ \text{Area of shaded region}\\ =\text{area of }\Delta ORS-\text{ area of sector }OPQ\\ =\left(\frac{1}{2}\times {14}^2\times \sin 0.5\text{ rad}\right)-\left(\frac{1}{2}\times 4^2\times 0.5\right)\\ =42.981{\text{ cm}}^2\end{array}$