# IB Math Analysis & Approaches Questionbank-Topic: SL 3.4-The circle SL Paper 1

## Question

The diagram shows two concentric circles with centre O. The radius of the smaller circle is 8 cm and the radius of the larger circle is 10 cm.

Points A, B and C are on the circumference of the larger circle such that $${\rm{A}}\widehat {\rm{O}}{\rm{B}}$$ is $$\frac{\pi }{3}$$ radians.

Find the length of the arc ACB .


a.

Find the area of the shaded region.


b.

## Markscheme

correct substitution in $$l = r\theta$$     (A1)

e.g. $$10 \times \frac{\pi }{3}$$ , $$\frac{1}{6} \times 2\pi \times 10$$

arc length $$= \frac{{20\pi }}{6}$$ $$\left( { = \frac{{10\pi }}{3}} \right)$$     A1      N2

[2 marks]

a.

area of large sector $$= \frac{1}{2} \times {10^2} \times \frac{\pi }{3}$$ $$\left( { = \frac{{100\pi }}{6}} \right)$$     (A1)

area of small sector $$= \frac{1}{2} \times {8^2} \times \frac{\pi }{3}$$ $$\left( { = \frac{{64\pi }}{6}} \right)$$     (A1)

evidence of valid approach (seen anywhere)     M1

e.g. subtracting areas of two sectors, $$\frac{1}{2} \times \frac{\pi }{3}({10^2} – {8^2})$$

area shaded $$= 6\pi$$ (accept $$\frac{{36\pi }}{6}$$ , etc.)     A1     N3

[4 marks]

b.

## Question

The following diagram shows a circle with centre $$O$$ and a radius of $$10$$ cm. Points $$A$$, $$B$$ and $$C$$ lie on the circle. Angle $$AOB$$ is $$1.2$$ radians.

Find the length of $${\text{arc ACB}}$$.


a.

Find the perimeter of the shaded region.


b.

## Markscheme

correct substitution     (A1)

eg$$\;\;\;10(1.2)$$

$$ACB$$ is $$12{\text{ (cm)}}$$     A1     N2

[2 marks]

a.

valid approach to find major arc     (M1)

eg$$\;\;\;$$circumference $$– {\text{AB}}$$, major angle $${\text{AOB}} \times {\text{radius}}$$

correct working for arc length     (A1)

eg$$\;\;\;2\pi (10) – 12,{\text{ }}10(2 \times 3.142 – 1.2),{\text{ }}2\pi (10) – 12 + 20$$

perimeter is $$20\pi + 8\;\;\;( = 70.8){\text{ (cm)}}$$     A1     N2

[3 marks]

Total [5 marks]

b.

## Question

The following diagram shows a triangle ABC and a sector BDC of a circle with centre B and radius 6 cm. The points A , B and D are on the same line. $${\text{AB}} = 2\sqrt 3 {\text{ cm, BC}} = 6{\text{ cm, area of triangle ABC}} = 3\sqrt 3{\text{ c}}{{\text{m}}^{\text{2}}}{\rm{, A\hat BC}}$$ is obtuse.

Find $${\rm{A\hat BC}}$$.


a.

Find the exact area of the sector BDC.


b.

## Markscheme

METHOD 1

correct substitution into formula for area of triangle     (A1)

eg$$\,\,\,\,\,$$$$\frac{1}{2}(6)\left( {2\sqrt 3 } \right)\sin B,{\text{ }}6\sqrt 3 \sin B,{\text{ }}\frac{1}{2}(6)\left( {2\sqrt 3 } \right)\sin B = 3\sqrt 3$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$6\sqrt 3 \sin B = 3\sqrt 3 ,{\text{ }}\sin B = \frac{{3\sqrt 3 }}{{\frac{1}{2}(6)2\sqrt 3 }}$$

$$\sin B = \frac{1}{2}$$    (A1)

$$\frac{\pi }{6}(30^\circ )$$    (A1)

$${\rm{A\hat BC}} = \frac{{5\pi }}{6}(150^\circ )$$     A1     N3

METHOD 2

(using height of triangle ABC by drawing perpendicular segment from C to AD)

correct substitution into formula for area of triangle     (A1)

eg$$\,\,\,\,\,$$$$\frac{1}{2}\left( {2\sqrt 3 } \right)(h) = 3\sqrt 3 ,{\text{ }}h\sqrt 3$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$h\sqrt 3 = 3\sqrt 3$$

height of triangle is 3     A1

$${\rm{C\hat BD}} = \frac{\pi }{6}(30^\circ )$$    (A1)

$${\rm{A\hat BC}} = \frac{{5\pi }}{6}(150^\circ )$$     A1     N3

[5 marks]

a.

recognizing supplementary angle     (M1)

eg$$\,\,\,\,\,$$$${\rm{C\hat BD}} = \frac{\pi }{6},{\text{ sector}} = \frac{1}{2}(180 – {\rm{A\hat BC)(}}{{\text{6}}^2})$$

correct substitution into formula for area of sector     (A1)

eg$$\,\,\,\,\,$$$$\frac{1}{2} \times \frac{\pi }{6} \times {6^2},{\text{ }}\pi ({6^2})\left( {\frac{{30}}{{360}}} \right)$$

$${\text{area}} = 3\pi {\text{ }}({\text{c}}{{\text{m}}^2})$$     A1     N2

[3 marks]

b.

## Question

The following diagram shows triangle ABC, with $${\text{AB}} = 3{\text{ cm}}$$, $${\text{BC}} = 8{\text{ cm}}$$, and $${\rm{A\hat BC = }}\frac{\pi }{3}$$. Show that $${\text{AC}} = 7{\text{ cm}}$$.


a.

The shape in the following diagram is formed by adding a semicircle with diameter [AC] to the triangle. Find the exact perimeter of this shape.


b.

## Markscheme

evidence of choosing the cosine rule     (M1)

eg$$\,\,\,\,\,$$$${c^2} = {a^2} + {b^2} – ab\cos C$$

correct substitution into RHS of cosine rule     (A1)

eg$$\,\,\,\,\,$$$${3^2} + {8^2} – 2 \times 3 \times 8 \times \cos \frac{\pi }{3}$$

evidence of correct value for $$\cos \frac{\pi }{3}$$ (may be seen anywhere, including in cosine rule)     A1

eg$$\,\,\,\,\,$$$$\cos \frac{\pi }{3} = \frac{1}{2},{\text{ A}}{{\text{C}}^2} = 9 + 64 – \left( {48 \times \frac{1}{2}} \right),{\text{ }}9 + 64 – 24$$

eg$$\,\,\,\,\,$$$${\text{A}}{{\text{C}}^2} = 49,{\text{ }}b = \sqrt {49}$$

$${\text{AC}} = 7{\text{ (cm)}}$$     AG     N0

Note:     Award no marks if the only working seen is $${\text{A}}{{\text{C}}^2} = 49$$ or $${\text{AC}} = \sqrt {49}$$ (or similar).

[4 marks]

a.

correct substitution for semicircle     (A1)

eg$$\,\,\,\,\,$$$${\text{semicircle}} = \frac{1}{2}(2\pi \times 3.5),{\text{ }}\frac{1}{2} \times \pi \times 7,{\text{ }}3.5\pi$$

valid approach (seen anywhere)     (M1)

eg$$\,\,\,\,\,$$$${\text{perimeter}} = {\text{AB}} + {\text{BC}} + {\text{semicircle, }}3 + 8 + \left( {\frac{1}{2} \times 2 \times \pi \times \frac{7}{2}} \right),{\text{ }}8 + 3 + 3.5\pi$$

$$11 + \frac{7}{2}\pi {\text{ }}( = 3.5\pi + 11){\text{ (cm)}}$$     A1     N2

[3 marks]

b.

## Question

The following diagram shows a circle with centre O and radius r cm. The points A and B lie on the circumference of the circle, and $${\text{A}}\mathop {\text{O}}\limits^ \wedge {\text{B}}$$ = θ. The area of the shaded sector AOB is 12 cm2 and the length of arc AB is 6 cm.

Find the value of r.

## Markscheme

evidence of correctly substituting into circle formula (may be seen later)      A1A1
eg  $$\frac{1}{2}\theta {r^2} = 12,\,\,r\theta = 6$$

attempt to eliminate one variable      (M1)
eg  $$r = \frac{6}{\theta },\,\,\theta = \frac{1}{r},\,\,\frac{{\frac{1}{2}\theta {r^2}}}{{r\theta }} = \frac{{12}}{6}$$

correct elimination      (A1)
eg  $$\frac{1}{2} \times \frac{6}{r} \times {r^2} = 12,\,\,\frac{1}{2}\theta \times {\left( {\frac{6}{\theta }} \right)^2} = 12,\,\,A = \frac{1}{2} \times {r^2} \times \frac{l}{r},\,\,\frac{{{r^2}}}{{2r}} = 2$$

correct equation     (A1)
eg  $$\frac{1}{2} \times 6r = 12,\,\,\frac{1}{2} \times \frac{{36}}{\theta } = 12,\,\,12 = \frac{1}{2} \times {r^2} \times \frac{6}{r}$$

correct working      (A1)
eg  $$3r = 12,\,\,\frac{{18}}{\theta } = 12,\,\,\frac{r}{2} = 2,\,\,24 = 6r$$

r = 4 (cm)      A1 N2

[7 marks]

MAA SL 3.4 ARCS AND SECTORS [concise]-manav

### Question

[Maximum mark: 6] [without GDC]
(i) 20° (ii) 18° (iii) 540°  
(b) Express the following angles in degrees
(i) $$\frac{\pi}{18} \mathrm{rad}$$  (ii) $$\frac{\pi}{5} \mathrm{rad}$$  (iii) $$2.5 \pi$$  

Ans:
(a) (i) $\frac{\pi}{9}$
(ii) $\frac{\pi}{10}$
(iii) $3 \pi$
(b) (i) 10°   (ii) 36°   (iii) 450°.

### Question

[Maximum mark: 8] [without GDC]
In the diagram below, the point A represents the angle of 30°, or otherwise $$\frac{\pi}{6}$$ rad, on the unit circle. (a) Write down the values corresponding to the points B, C, and D
(i) in degrees in the interval $$0^{\circ} \leq \theta<360^{\circ}$$
(ii) in radians in the interval $$0 \leq \theta<2 \pi$$ (b) Suppose now that C represents the angle of 220°, or otherwise $$\frac{11 \pi}{9}$$ rad,
Complete in a similar way as in (a) the following table: Ans:
(a) (b) ### Question

[Maximum mark: 8] [without GDC]
In the diagram below, the point A represents the angle of 30°, or otherwise $$\frac{\pi}{6}$$ rad, on the unit circle. The general formula for the angles corresponding to point A is
$$\text { in degrees: } \quad 30^{\circ}+360^{\circ} k \quad k \in Z$$
$$\text { in radians: } \quad \frac{\pi}{6}+2 k \pi \quad k \in Z$$
Determine the values of the angle at point A the following intervals: Ans: ### Question

[Maximum mark: 8] [with / without GDC]
O is the centre of the circle which has a radius of 10 cm. The size of AÔB is 1.5 rad. (a) Find the lengths of the minor arc AB and of the major arc AB. 
(b) Find the areas of the minor sector (shaded region) and of the major sector. 
(c) Find the perimeters of the minor sector (shaded region) and of the major sector. 

Ans:
(a) $$l_{\text {MINOR }}=(10)(1.5)=15 \quad l_{\text {MAJOR }}=(10)(2 \pi-1.5)=20 \pi-15$$
(b) $$A_{M I N O R}=\frac{1}{2}(10)^{2}(1.5)=75 \quad A_{M A J O R}=\frac{1}{2}(10)^{2}(2 \pi-1.5)=200 \pi-75$$
(c) $$P_{\text {MINOR }}=l_{\text {MINOR }}+2 r=35 \quad P_{\text {MAJOR }}=l_{\text {MAJOR }}+2 r=20 \pi+5$$

### Question

[Maximum mark: 6] [with GDC]
The following diagram shows a circle of centre O, and radius 15 cm. The arc ACB subtends an angle of 2 radians at the centre O. (a) Find the length of the arc ACB; 
(b) Find the area of the shaded region. 

Ans:
(a) $$l=r \theta \quad \text { or } \quad \mathrm{ACB}=2 \times \mathrm{OA}=30 \mathrm{~cm}$$
(b) AÔB (obtuse) = 2π – 2
$$\text { Area }=\frac{1}{2} \theta r^{2}=\frac{1}{2}(2 \pi-2)(15)^{2}=482 \mathrm{~cm}^{2}(3 \mathrm{sf})$$

### Question

[Maximum mark: 4] [with / without GDC]
The diagram shows a circle of radius 5 cm. Find the perimeter of the shaded region. Ans:
|9\text { Perimeter }=5(2 \pi-1)+10=(10 \pi+5) \mathrm{cm}(=36.4 \text {, to } 3 \mathrm{sf})\)

### Question

[Maximum mark: 6] [without GDC]
The diagram shows two concentric circles with centre O. 1. The radius of the smaller circle is 8 cm, the radius of the larger circle is 10 cm. Points A, B and C are on the circumference of the larger circle such that $$\text { AÔB }$$ is $$\frac{\pi}{3}$$ radians
(a) Find the length of the arc ACB. 
(b) Find the area of the shaded region. 

Ans:
(a) $$l=r \theta=10 \times \frac{\pi}{3} \operatorname{arc} length =\frac{20 \pi}{6}\left(=\frac{10 \pi}{3}\right)$$
(b) $$\text { area of large sector }=\frac{1}{2} \times 10^{2} \times \frac{\pi}{3}\left(=\frac{100 \pi}{6}\right)$$
$$\text { area of small sector }=\frac{1}{2} \times 8^{2} \times \frac{\pi}{3}\left(=\frac{64 \pi}{6}\right)$$
$$\text { area shaded }=\frac{36 \pi}{6}=6 \pi$$

### Question

[Maximum mark: 7] [with GDC]
The circle shown has centre O and radius 3.9 cm. Points A and B lie on the circle and angle AOB is 1.8 radians. (a) Find AB. 
(b) Find the area of the shaded region. 

Ans:
(a) METHOD 1
$$\text { cosine rule } \mathrm{AB}=\sqrt{3.9^{2}+3.9^{2}-2(3.9)(3.9) \cos 1.8}=\mathrm{AB}=6.11(\mathrm{~cm})$$
METHOD 2
using right-angled triangles
$$\sin 0.9=\frac{x}{3.9} \Rightarrow x=\frac{1}{2} \mathrm{AB}=3.9 \sin 0.9$$
AB = 6.11 (cm)
METHOD 3
$$\text { sine rule } \frac{\sin 0.670 \ldots}{3.9}=\frac{\sin 1.8}{\mathrm{AB}} \Rightarrow \mathrm{AB}=6.11(\mathrm{~cm})$$
(b) METHOD 1
For major sector: AÔB = 2π – 1.8 (= 4.4832)
$$A=\frac{1}{2}(3.9)^{2}(4.4832 \ldots)=34.1\left(\mathrm{~cm}^{2}\right)$$
METHOD 2
area of circle A = π(3.9)2(= 47.78…)
area of (minor) sector $$A=\frac{1}{2}(3.9)^{2}(1.8)(=13.68 \ldots)$$
area = 47.8 – 13.7 = 34.1 (cm2)

### Question

[Maximum mark: 4] [with GDC]
O is the centre of the circle which has a radius of 5.4 cm. The area of the shaded sector OAB is 21.6 cm2. Find the length of the minor arc AB.

Ans:
$$\frac{1}{2} \times(5.4)^{2} \theta=21.6 \Rightarrow \theta=\frac{4}{2.7}(=1.481 radians )$$
$$A B=r \theta=5.4 \times \frac{4}{2.7}=8 \mathrm{~cm}$$

### Question

[Maximum mark: 6] [with GDC]
The following diagram shows a circle of centre O, and radius r. The shaded sector OACB has an area of 27 cm2. Angle  AÔB = θ = 1.5 radians. (a) Find the radius r . 
(b) Calculate the length of the minor arc ACB. 

Ans:
(a) $$A=\frac{1}{2} r^{2} \theta \Leftrightarrow 27=\frac{1}{2}(1.5) r^{2} \Leftrightarrow r^{2}=36 \Rightarrow r=6 \mathrm{~cm}$$
(b) $$Arc length =r \theta=1.5 \times 6=9 \mathrm{~cm}$$

### Question

[Maximum mark: 6] [with GDC]
The diagram below shows a circle centre O, with radius r. The length of arc ABC is 3π cm and $$\mathrm{AOC}=\frac{2 \pi}{9}$$
(a) Find the value of r. 
(b) Find the perimeter of sector OABC. 
(c) Find the area of sector OABC. 

Ans:
(a) $$3 \pi=r \frac{2 \pi}{9} \Leftrightarrow r=13.5(\mathrm{~cm})$$
(b) perimeter = 27+3π (cm)(= 36.4)
(c) $$area =\frac{1}{2} \times 13.5^{2} \times \frac{2 \pi}{9}=20.25 \pi\left(\mathrm{cm}^{2}\right)(=63.6)$$

### Question

[Maximum mark: 6] [without GDC]
The following diagram shows a circle with radius r and centre O. The points A, B and C are on the circle and AÔC = θ. The area of sector OABC is $$\frac{4}{3} \pi$$ the length of arc ABC is $$\frac{2}{3} \pi \text {. }$$.
Find the value of r and of θ.

Ans:
$$A=\frac{1}{2} r^{2} \theta \Leftrightarrow \frac{1}{2} r^{2} \theta=\frac{4}{3} \pi$$
$$l=r \theta \Leftrightarrow r \theta=\frac{2}{3} \pi$$
Solving the system: $$r=4, \quad \theta=\frac{\pi}{6}$$

### Question

[Maximum mark: 6] [with GDC] [diagram as above – exercise 12]
The area of the sector OAB is 180 cm2, the length of the arc AB is 24 cm. Find the value of r and of θ.

Ans:
$$A=\frac{1}{2} r^{2} \theta \Leftrightarrow \frac{1}{2} r^{2} \theta=180$$
$$l=r \theta \Leftrightarrow r \theta=24$$
Solving the system: r = 15, θ = 1.6

### Question

[Maximum mark: 4] [with GDC]
The diagram below shows a sector AOB of a circle of radius 15 cm and centre O. The angle θ at the centre of the circle is 2 radians. (a) Calculate the area of the sector AOB. 
(b) Calculate the area of the shaded region. 

Ans:
(a) $$Area =\frac{1}{2} r^{2} \theta=\frac{1}{2}\left(15^{2}\right)(2)=225\left(\mathrm{~cm}^{2}\right)$$
(b) $$Area \Delta \mathrm{OAB}=\frac{1}{2} 15^{2} \sin 2=102.3$$
Area = 225 – 102.3 = 122.7 (cm2) = 123 (3 sf)

### Question

[Maximum mark: 6] [with GDC]
The following diagram shows a sector of a circle of radius r cm, and angle θ at the centre. The perimeter of the sector is 20 cm. (a) Show that $$\theta=\frac{20-2 r}{r}$$ 
(b) The area of the sector is 25 cm2. Find the value of r. 

Ans:
(a) $$perimeter =r+r+\operatorname{arc} length \Leftrightarrow 20=2 r+r \theta \Leftrightarrow \theta=\frac{20-2 r}{r}$$
(b) $$A=\frac{1}{2} r^{2}\left(\frac{20-2 r}{r}\right) \Leftrightarrow 10 r-r^{2}=25 \Leftrightarrow r=5 \mathrm{~cm}$$

### Question

[Maximum mark: 4] [with / without GDC]
In the following diagram, O is the centre of the circle and (AT) is the tangent to the circle at T. If OA = 12 cm, and the circle has a radius of 6 cm, find the area of the shaded region.

Ans:
$$\mathrm{TO} \mathrm{A}=60^{\circ}$$
$$\text { Area of } \Delta=\frac{1}{2} \times 6 \times 12 \times \sin 60=18 \sqrt{3} \quad \text { Area of sector }=\frac{1}{2} \times 6 \times 6 \times \frac{\pi}{3}=6 \pi$$
$$\text { Shaded area }=18 \sqrt{3}-6 \pi=12.3 \mathrm{~cm}^{2}(3 \mathrm{sf})$$
OR
$$\mathrm{OTA}=90^{\circ}$$
$$\mathrm{AT}=\sqrt{12^{2}-6^{2}}=6 \sqrt{3} \quad \mathrm{TOA}=60^{\circ}=\frac{\pi}{3}$$
Area = area of triangle – area of sector = $$\frac{1}{2} \times 6 \times 6 \sqrt{3}-\frac{1}{2} \times 6 \times 6 \times \frac{\pi}{3}=18 \sqrt{3}-6 \pi=12.3 \mathrm{~cm}^{2}$$

### Question

[Maximum mark: 6] [with GDC]
The diagram below shows a circle of radius 5 cm with centre O. Points A and B are on the circle, and AÔB is 0.8 radians. The point N is on [OB] such that [AN] is perpendicular to [OB]. Find the area of the shaded region

Ans:
Area sector OAB =$$\frac{1}{2}(5)^{2}(0.8)=10$$
cos 0.8 = ON/5 $$\Rightarrow \mathrm{ON}=5 \cos 0.8 \quad(=3.483 \ldots)$$
$$\text { Area of } \Delta \mathrm{AON}=\frac{1}{2} \mathrm{ON} \times 5 \times \sin 0.8=6.249 \ldots$$
Shaded area = 10 – 6.249.. = 3.75

### Question

[Maximum mark: 6] [without GDC]
The diagrams show a circular sector of radius 10 cm and angle θ radians which is formed into a cone of slant height 10 cm. The vertical height h of the cone is equal to the radius r of its base. (a) Find the value of r . 
(b) Find the angle θ  radians. 

Ans:
$$h=r \text { so } 2 r^{2}=100 \Rightarrow r^{2}=50 \Rightarrow r=5 \sqrt{2} \quad \text { Hence circumference }=2 \pi r=10 \pi \sqrt{2}$$
$$l=10 \theta=10 \pi \sqrt{2} \Rightarrow \theta=\pi \sqrt{2}$$

### Question

[Maximum mark: 6] [without GDC]
The diagram below shows a triangle and two arcs of circles.
The triangle ABC is a right-angled isosceles triangle, with AB = AC = 2. The point P is the midpoint of [BC].
The arc BDC is part of a circle with centre A.
The arc BEC is part of a circle with centre P. (a) Calculate the area of the segment BDCP. 
(b) Calculate the area of the shaded region BECD. 

Ans:
(a)   area of sector ΑΒDC = $$\frac{1}{4} \pi(2)^{2}=\pi$$
area of segment BDCP = π – area of $$\Delta$$ABC = π – 2
(b)  $$\mathrm{BP}=\sqrt{2}$$
area of semicircle of radius BP = $$\frac{1}{2} \pi(\sqrt{2})^{2}=\pi$$
area of shaded region = π – (π – 2) = 2

### Question

[Maximum mark: 13] [with GDC]
The following diagram shows a circle with centre O and radius 4 cm. The points A, B and C lie on the circle. The point D is outside the circle, on (OC).
(b) Find OD. 
(c) Find the area of sector OABC. 
(d) Find the area of region ABCD. 

Ans:
(a) $$\frac{\mathrm{AD}}{\sin 0.8}=\frac{4}{\sin 0.3} \Rightarrow \mathrm{AD}=9.71(\mathrm{~cm})$$
(b) OAD = π – 1.1 = (2.04)
EITHER $$\mathrm{OD}^{2}=9.71^{2}+4^{2}-2 \times 9.71 \times 4 \times \cos (\pi-1.1) \Rightarrow \mathrm{OD}=12.1$$
OR $$\frac{\mathrm{OD}}{\sin (\pi-1.1)}=\frac{9.71}{\sin 0.8}=\frac{4}{\sin 0.3} \Rightarrow \mathrm{OD}=12.1$$
(c) $$\text { area }=0.5 \times 4^{2} \times 0.8=6.4$$
(d) area of triangle OAD: $$A=\frac{1}{2} \times 4 \times 12.1 \times \sin 0.8=17.3067$$
$$\text { (OR } A=\frac{1}{2} \times 4 \times 9.71 \times \sin 2.04=17.3067 \quad \text { OR } \quad A=\frac{1}{2} \times 12.1 \times 9.71 \times \sin 0.3=17.3067$$  area ABCD = 17.3067 – 6.4 = 10.9 (cm2)

### Question

[Maximum mark: 13] [with GDC]
The diagram below shows a circle, centre O, with a radius 12 cm. The chord AB subtends at an angle of 75° at the centre. The tangents to the circle at A and at B meet at P. (a) Using the cosine rule, show that the length of AB is $$12 \sqrt{2\left(1-\cos 75^{\circ}\right)}$$. 
(b) Find the length of BP. 
(c) Hence find
(i) the area of triangle OBP;    (ii) the area of triangle ABP. 
(d) Find the area of sector OAB. 
(e) Find the area of the shaded region. 

Ans:
(a) $$\mathrm{AB}^{2}=12^{2}+12^{2}-2 \times 12 \times 12 \times \cos 75^{\circ}=12^{2}\left(2-2 \cos 75^{\circ}\right)=12^{2} \times 2\left(1-\cos 75^{\circ}\right)$$
$$\Rightarrow \mathrm{AB}=12 \sqrt{2\left(1-\cos 75^{\circ}\right)}$$
(b) POB = 37.5°, BP = 12 tan 37.5° = 9.21 cm
OR
BPA = 105°   BAP= 37.5°
$$\frac{\mathrm{AB}}{\sin 105^{\circ}}=\frac{\mathrm{BP}}{\sin 37.5^{\circ}} \Rightarrow \mathrm{BP}=\frac{\mathrm{AB} \sin 37.5^{\circ}}{\sin 105^{\circ}}=9.21(\mathrm{~cm})$$
(c) $$\text { (i) Area } \Delta \mathrm{OBP}=\frac{1}{2} \times 12 \times 9.21=55.3\left(\mathrm{~cm}^{2}\right)\left(\operatorname{accept} 55.2 \mathrm{~cm}^{2}\right)$$
$$\text { (ii) Area } \Delta \mathrm{ABP}=\frac{1}{2}(9.21)^{2} \sin 105^{\circ}=41.0\left(\mathrm{~cm}^{2}\right)\left(\text { accept } 40.9 \mathrm{~cm}^{2}\right)$$
(d) $$\text { Area of sector }=\frac{1}{2} \times 12^{2} \times 75 \times \frac{\pi}{180}=94.2\left(\mathrm{~cm}^{2}\right)\left(\text { accept } 30 \pi \text { or } 94.3\left(\mathrm{~cm}^{2}\right)\right)$$
(e) $$\text { Shaded area }=2 \times \text { area } \Delta \mathrm{OPB}-\text { area sector }=16.4\left(\mathrm{~cm}^{2}\right)\left(\text { accept } 16.2 \mathrm{~cm}^{2}, 16.3 \mathrm{~cm}^{2}\right)$$

### Question

[Maximum mark: 17] [with GDC]
The following diagram shows two semi-circles. The larger one has centre O and radius 4 cm. The smaller one has centre P, radius 3 cm, and passes through O. The line (OP) meets the larger semi-circle at S. The semi-circles intersect at Q. (a) (i) Explain why OPQ is an isosceles triangle.
(ii) Use the cosine rule to show that $$cos\mathrm{OPQ}=\frac{1}{9}$$
(iii) Hence show that sin $$\mathrm{OPQ}=\frac{\sqrt{80}}{9}$$
(iv) Find the area of the triangle OPQ. 
(b) Consider the smaller semi-circle, with centre P.
(i) Write down the size of OPQ.
(ii) Calculate the area of the sector OPQ. 
(c) Consider the larger semi-circle, with centre O. Calculate the area of the sector QOS. 
(d) Hence calculate the area of the shaded region. 

Ans:
(a) (i) $$\mathrm{OP}=\mathrm{PQ}(=3 \mathrm{~cm}) \text { So } \Delta \mathrm{OPQ} \text { is isosceles }$$
(ii) $$\cos \mathrm{OP} \mathrm{Q}=\frac{3^{2}+3^{2}-4^{2}}{2 \times 3 \times 3}=\frac{9+9-16}{18}\left(=\frac{2}{18}\right)=\frac{1}{9}$$
(iii) $$\sin ^{2} A+\cos ^{2} A=1 \Rightarrow \sin \mathrm{OPQ}=\sqrt{1-\frac{1}{81}}\left(=\sqrt{\frac{80}{81}}\right)=\frac{\sqrt{80}}{9}$$
(iv) $$\text { Area triangle } \mathrm{OPQ}=\frac{1}{2} \times \mathrm{OP} \times \mathrm{PQ} \sin \mathrm{P}=\frac{\sqrt{80}}{2}(=\sqrt{20})(=4.47)$$
(b) (i) OPQ = 1.4594… = 1.46,
(ii) Area sector OPQ = $$\frac{1}{2} \times 3^{2} \times 1.4594 \ldots=6.57$$
(c) $$\text { QOP }=\frac{\pi-1.4594 \ldots}{2}(=0.841), \text { Area sector QOS }=\frac{1}{2} \times 4^{2} \times 0.841=6.73$$
(d) Area of small semi-circle is 4.5π (= 14.137…)
Area = area of semi-circle – area sector OPQ – area sector QOS + area triangle POQ
= 4.5π – 6.5675… – 6.7285… + 4.472… = 5.31

### Question

[Maximum mark: 14] [with GDC]
The following diagram shows the triangle AOP, where OP = 2cm, AP = 4cm and AO = 3cm. (a) Calculate AÔP, giving your answer in radians. The following diagram shows two circles which intersect at the points A and B. The smaller circle C1 has centre O and radius 3 cm, the larger circle C2 has centre P and radius 4 cm, and OP = 2 cm. The point D lies on the circumference of C1 and E on the circumference of C2. Triangle AOP is the same as triangle AOP in the diagram above. (c) Given that APB is 1.63 radians, calculate the area of
(i) sector PAEB;
(d) The area of the quadrilateral AOBP is 5.81 cm2.
(i) Find the area of AOBE. (ii) Hence find the area of the shaded region AEBD 

Ans:
(a) $$\text { cosine rule; } 4^{2}=3^{2}+2^{2}-2 \times 3 \times 2 \cos \mathrm{AOP} \quad \Rightarrow \mathrm{AOP}=1.82\left(=\frac{26 \pi}{45}\right) \text { (radians) }$$
(b) $$\mathrm{AOB}=2(\pi-1.82)=2 \pi-3.64=2.64\left(=\frac{38 \pi}{45}\right) \text { (radians) }$$
(c)  (i) Area of sector PAEB = $$\frac{1}{2} \times 4^{2} \times 1.63=13.04\left(\mathrm{~cm}^{2}\right)$$
(ii) Area of sector OADB = $$\frac{1}{2} \times 3^{2} \times 2.64=11.9\left(\mathrm{~cm}^{2}\right)$$
(d)   (i) Area AOBE = Area PAEB – Area AOBP (= 13.0 – 5.81) = 7.23
(ii) Area shaded = Area OADB – Area AOBE = 11.9 – 7.23 = 4.67
(accept answers between 4.63 and 4.72)

### Question

[Maximum mark: 15] [with GDC]
The diagram below shows a circle with centre O and radius 8 cm. The points A, B, C, D, E and F are on the circle, and [AF] is a diameter. The length of
arc ABC is 6 cm.
(a) Find the size of angle AOC. 
(b) Hence find the area of the shaded region. 
The area of sector OCDE is 45 cm2.
(c) Find the size of angle COE. 
(d) Find EF. 

(a) $$6=8 \theta \Leftrightarrow \text { AÔC }=0.75$$
(b) area of triangle = $$\frac{1}{2} \times 8 \times 8 \times \sin (0.75)=21.8 \ldots$$
area of sector = $$\frac{1}{2} \times 64 \times 0.75=24$$
(or directly area of segment = = $$\frac{1}{2} \times 8^{2}\left(0.75-\sin 0.75=2.19 \mathrm{~cm}^{2}\right) .$$
(c) $$45=\frac{1}{2} \times 8^{2} \times \theta \Rightarrow \mathrm{COE}=1.40625(1.41 \text { to } 3 \mathrm{sf})$$
(d) $$\text { EÔF }=\pi-0.75-1.41=0.985$$
$$\mathrm{EF}=\sqrt{8^{2}+8^{2}-2 \times 8 \times 8 \times \cos 0.985}=7.57 \mathrm{~cm}$$