# 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 $$\frac{\pi }{3}$$Â radians.

Find the length of the arc ACB .

[2]
a.

Find the area of the shaded region.

[4]
b.

## Markscheme

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

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

arc length $$= \frac{{20\pi }}{6}$$ $$\left( { = \frac{{10\pi }}{3}} \right)$$Â Â Â Â  A1Â Â Â Â Â  N2

[2 marks]

a.

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

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

evidence of valid approach (seen anywhere)Â Â Â Â  M1

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

area shaded $$= 6\pi$$Â (accept $$\frac{{36\pi }}{6}$$ , etc.)Â Â Â  Â A1Â Â Â Â  N3

[4 marks]

b.

## Question

The following diagram shows a circle with centreÂ $$O$$ and a radius ofÂ $$10$$ cm. Points $$A$$,Â $$B$$ andÂ $$C$$ lie on the circle.

AngleÂ $$AOB$$ isÂ $$1.2$$ radians.

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

[2]
a.

Find the perimeter of the shaded region.

[3]
b.

## Markscheme

correct substitution Â  Â  (A1)

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

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

[2 marks]

a.

valid approach to find major arc Â  Â  (M1)

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

correct working for arc length Â  Â  (A1)

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

perimeter is $$20\piÂ + 8\;\;\;( = 70.8){\text{ (cm)}}$$ Â  Â  A1 Â  Â  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.

$${\text{AB}} = 2\sqrt 3 {\text{ cm, BC}} = 6{\text{ cm, area of triangle ABC}} = 3\sqrt 3{\text{ c}}{{\text{m}}^{\text{2}}}{\rm{, A\hat BC}}$$Â is obtuse.

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

[5]
a.

Find the exact area of the sector BDC.

[3]
b.

## Markscheme

METHOD 1

correct substitution into formula for area of triangle Â  Â  (A1)

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

correct working Â  Â  (A1)

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

$$\sin B = \frac{1}{2}$$ Â  Â (A1)

$$\frac{\pi }{6}(30^\circ )$$ Â  Â (A1)

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

METHOD 2

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

correct substitution into formula for area of triangle Â  Â  (A1)

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

correct working Â  Â  (A1)

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

height of triangle is 3 Â  Â  A1

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

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

[5 marks]

a.

recognizing supplementary angle Â  Â  (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 Â  Â  (A1)

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

$${\text{area}} = 3\pi {\text{ }}({\text{c}}{{\text{m}}^2})$$Â Â  Â  A1 Â  Â  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 $${\text{AC}} = 7{\text{ cm}}$$.

[4]
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.

[3]
b.

## Markscheme

evidence of choosing the cosine rule Â  Â  (M1)

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

correct substitution into RHS of cosine rule Â  Â  (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) Â  Â  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$$\,\,\,\,\,$$$${\text{A}}{{\text{C}}^2} = 49,{\text{ }}b = \sqrt {49}$$

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

Note: Â  Â  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 Â  Â  (A1)

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

valid approach (seen anywhere) Â  Â  (M1)

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

$$11 + \frac{7}{2}\pi {\text{ }}( = 3.5\piÂ + 11){\text{ (cm)}}$$ Â  Â  A1 Â  Â  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 $${\text{A}}\mathop {\text{O}}\limits^ \wedgeÂ {\text{B}}$$Â = Î¸. The area of the shaded sector AOB is 12 cm2 and the length of arc AB is 6â€‰cm.

Find the value of r.

## Markscheme

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

attempt to eliminate one variableÂ  Â  Â  (M1)
egÂ  $$r = \frac{6}{\theta },\,\,\thetaÂ = \frac{1}{r},\,\,\frac{{\frac{1}{2}\theta {r^2}}}{{r\theta }} = \frac{{12}}{6}$$

correct eliminationÂ  Â  Â  (A1)
egÂ Â $$\frac{1}{2} \times \frac{6}{r} \times {r^2} = 12,\,\,\frac{1}{2}\thetaÂ \times {\left( {\frac{6}{\theta }} \right)^2} = 12,\,\,A = \frac{1}{2} \times {r^2} \times \frac{l}{r},\,\,\frac{{{r^2}}}{{2r}} = 2$$

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

correct workingÂ  Â  Â Â (A1)
egÂ Â $$3r = 12,\,\,\frac{{18}}{\theta } = 12,\,\,\frac{r}{2} = 2,\,\,24 = 6r$$

r = 4 (cm)Â  Â  Â  A1 N2

[7 marks]

MAA SL 3.4 ARCS AND SECTORS [concise]-manav

### Question

[Maximum mark: 6] [without GDC]
(i) 20Â° (ii) 18Â° (iii) 540Â°Â  [3]
(b) Express the following angles in degrees
(i) $$\frac{\pi}{18} \mathrm{rad}$$Â  (ii) $$\frac{\pi}{5} \mathrm{rad}$$Â  (iii) $$2.5 \pi$$Â  [3]

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

### Question

[Maximum mark: 8] [without GDC]
In the diagram below, the point A represents the angle of 30Â°, or otherwise $$\frac{\pi}{6}$$Â 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 $$0 \leq \theta<2 \pi$$ [4]

(b) Suppose now that C represents the angle of 220Â°, or otherwise $$\frac{11 \pi}{9}$$ rad,
Complete in a similar way as in (a) the following table: [4]

Ans:
(a)Â
(b)

### Question

[Maximum mark: 8] [without GDC]
In the diagram below, the point A represents the angle of 30Â°, 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 AÃ”B is 1.5 rad.

(a) Find the lengths of the minor arc AB and of the major arc AB. [3]
(b) Find the areas of the minor sector (shaded region) and of the major sector. [3]
(c) Find the perimeters of the minor sector (shaded region) and of the major sector. [2]

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; [2]
(b) Find the area of the shaded region. [4]

Ans:
(a) $$l=r \theta \quad \text { or } \quad \mathrm{ACB}=2 \times \mathrm{OA}=30 \mathrm{~cm}$$
(b) AÃ”B (obtuse) = 2Ï€ â€“ 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 $$\text { AÃ”B }$$ is $$\frac{\pi}{3}$$ radians
(a) Find the length of the arc ACB. [2]
(b) Find the area of the shaded region. [4]

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

### 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. [3]
(b) Find the area of the shaded region. [4]

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: AÃ”B = 2Ï€ â€“ 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 = Ï€(3.9)2(= 47.78…)
area of (minor) sector $$A=\frac{1}{2}(3.9)^{2}(1.8)(=13.68 \ldots)$$
area = 47.8 â€“ 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 Â AÃ”B = Î¸ = 1.5 radians.

(a) Find the radius r . [4]
(b) Calculate the length of the minor arc ACB. [2]

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 3Ï€ cm and $$\mathrm{AOC}=\frac{2 \pi}{9}$$
(a) Find the value of r. [2]
(b) Find the perimeter of sector OABC. [2]
(c) Find the area of sector OABC. [2]

Ans:
(a) $$3 \pi=r \frac{2 \pi}{9} \Leftrightarrow r=13.5(\mathrm{~cm})$$
(b) perimeter = 27+3Ï€ (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 AÃ”C = Î¸.

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 Î¸.

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 Î¸.

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, Î¸ = 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 Î¸ at the centre of the circle is 2 radians.

(a) Calculate the area of the sector AOB. [2]
(b) Calculate the area of the shaded region. [2]

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 â€“ 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 Î¸ at the centre. The perimeter of the sector is 20 cm.

(a) Show that $$\theta=\frac{20-2 r}{r}$$Â [2]
(b) The area of the sector is 25 cm2. Find the value of r. [4]

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 â€“ 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 AÃ”B 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 Î¸ 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 . [2]
(b) Find the angle Î¸ Â radians. [4]

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. [3]
(b) Calculate the area of the shaded region BECD. [3]

Ans:
(a)Â  Â area of sector Î‘Î’DC = $$\frac{1}{4} \pi(2)^{2}=\pi$$
area of segment BDCP = Ï€ â€“ area of $$\Delta$$ABC = Ï€ â€“ 2
(b)Â  $$\mathrm{BP}=\sqrt{2}$$
area of semicircle of radius BP = $$\frac{1}{2} \pi(\sqrt{2})^{2}=\pi$$
area of shaded region = Ï€ â€“ (Ï€ â€“ 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. [4]
(c) Find the area of sector OABC. [2]
(d) Find the area of region ABCD. [4]

Ans:
(a) $$\frac{\mathrm{AD}}{\sin 0.8}=\frac{4}{\sin 0.3} \Rightarrow \mathrm{AD}=9.71(\mathrm{~cm})$$
(b) OAD = Ï€ â€“ 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: $$A=\frac{1}{2} \times 4 \times 12.1 \times \sin 0.8=17.3067$$
$$\text { (OR } A=\frac{1}{2} \times 4 \times 9.71 \times \sin 2.04=17.3067 \quad \text { OR } \quad A=\frac{1}{2} \times 12.1 \times 9.71 \times \sin 0.3=17.3067$$Â  area ABCD = 17.3067 â€“ 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 75Â° 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 $$12 \sqrt{2\left(1-\cos 75^{\circ}\right)}$$. [2]
(b) Find the length of BP. [3]
(c) Hence find
(i) the area of triangle OBP;Â  Â  (ii) the area of triangle ABP. [4]
(d) Find the area of sector OAB. [2]
(e) Find the area of the shaded region. [2]

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 = 37.5Â°, BP = 12 tan 37.5Â° = 9.21 cm
OR
BPA = 105Â°Â  Â BAP= 37.5Â°
$$\frac{\mathrm{AB}}{\sin 105^{\circ}}=\frac{\mathrm{BP}}{\sin 37.5^{\circ}} \Rightarrow \mathrm{BP}=\frac{\mathrm{AB} \sin 37.5^{\circ}}{\sin 105^{\circ}}=9.21(\mathrm{~cm})$$
(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. [7]
(b) Consider the smaller semi-circle, with centre P.
(i) Write down the size of OPQ.
(ii) Calculate the area of the sector OPQ. [3]
(c) Consider the larger semi-circle, with centre O. Calculate the area of the sector QOS. [3]
(d) Hence calculate the area of the shaded region. [4]

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 4.5Ï€ (= 14.137…)
Area = area of semi-circle – area sector OPQ – area sector QOS + area triangle POQ
= 4.5Ï€ – 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 AÃ”P, giving your answer in radians. [3]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 [4]

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)Â  (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)Â  Â (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. [2]
(b) Hence find the area of the shaded region. [6]
The area of sector OCDE is 45 cm2.
(c) Find the size of angle COE. [2]
(d) Find EF. [5]

(a) $$6=8 \theta \Leftrightarrow \text { AÃ”C }=0.75$$
(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) $$\text { EÃ”F }=\pi-0.75-1.41=0.985$$
Â  Â  Â  Â $$\mathrm{EF}=\sqrt{8^{2}+8^{2}-2 \times 8 \times 8 \times \cos 0.985}=7.57 \mathrm{~cm}$$