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 .

[2]
a.

Find the area of the shaded region.

[4]
b.
Answer/Explanation

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 \({\text{arc ACB}}\).

[2]
a.

Find the perimeter of the shaded region.

[3]
b.
Answer/Explanation

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.

M16/5/MATME/SP1/ENG/TZ2/05

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

Find \({\rm{A\hat BC}}\).

[5]
a.

Find the exact area of the sector BDC.

[3]
b.
Answer/Explanation

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 \({\rm{A\hat BC = }}\frac{\pi }{3}\).

N17/5/MATME/SP1/ENG/TZ0/04

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.

N17/5/MATME/SP1/ENG/TZ0/04.b

Find the exact perimeter of this shape.

[3]
b.
Answer/Explanation

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

correct working clearly leading to answer     A1

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.

Answer/Explanation

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]
(a) Express the following angles in radians; give your answer in terms of π.
(i) 20° (ii) 18° (iii) 540°  [3]
(b) Express the following angles in degrees
(i) \(\frac{\pi}{18} \mathrm{rad}\)  (ii) \(\frac{\pi}{5} \mathrm{rad}\)  (iii) \(2.5 \pi\)  [3]

Answer/Explanation

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 \(0 \leq \theta<2 \pi\) [4]

(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: [4]

Answer/Explanation

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:

Answer/Explanation

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. [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]

Answer/Explanation

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]

Answer/Explanation

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.

Answer/Explanation

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. [2]
    (b) Find the area of the shaded region. [4]
    Answer/Explanation

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

Answer/Explanation

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.

Answer/Explanation

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 . [4]
(b) Calculate the length of the minor arc ACB. [2]

Answer/Explanation

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. [2]
(b) Find the perimeter of sector OABC. [2]
(c) Find the area of sector OABC. [2]

Answer/Explanation

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.

Answer/Explanation

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.

Answer/Explanation

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

Answer/Explanation

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

Answer/Explanation

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.

Answer/Explanation

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

Answer/Explanation

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 . [2]
(b) Find the angle θ  radians. [4]

Answer/Explanation

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]

Answer/Explanation

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).
Angle ADC = 0.3 radians and angle AOC = 0.8 radians.
(a) Find AD. [3]
(b) Find OD. [4]
(c) Find the area of sector OABC. [2]
(d) Find the area of region ABCD. [4]

Answer/Explanation

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 \(12 \sqrt{2\left(1-\cos 75^{\circ}\right)}\). [2]
(b) Find the length of BP. [3]
(c) Hence find
(i) the area of triangle OBP;    (ii) the area of triangle ABP. [4]
(d) Find the area of sector OAB. [2]
(e) Find the area of the shaded region. [2]

Answer/Explanation

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. [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]

Answer/Explanation

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. [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.

(b) Find AÔB, giving your answer in radians. [2]
(c) Given that APB is 1.63 radians, calculate the area of
(i) sector PAEB;
(ii) sector OADB. [5]
(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]

Answer/Explanation

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. [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]

Answer/Explanation

Ans:
(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 = \(\frac{1}{2} \times 64 \times 0.75=24\)
area of shaded region = area of sector – area of triangle = 2.19 cm2
(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) \(\text { EÔF }=\pi-0.75-1.41=0.985\)
       \(\mathrm{EF}=\sqrt{8^{2}+8^{2}-2 \times 8 \times 8 \times \cos 0.985}=7.57 \mathrm{~cm}\)
(other methods are also possible)

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