IB DP Maths Topic 3.1 The circle: radian measure of angles; length of an arc; area of a sector 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]

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