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Date	November 2017	Marks available	3	Reference code	17N.1.SL.TZ0.S_4
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 0
Command term	Find	Question number	S_4	Adapted from	N/A

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]

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]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

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]

Examiners report

[N/A]

Syllabus sections

Topic 3—Geometry and trigonometry » AHL 3.7—Radians

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Boat A is situated 10km away from boat B, and each boat has a marine radio transmitter on board. The range of the transmitter on boat A is 7km, and the range of the transmitter on boat B is 5km. The region in which both transmitters can be detected is represented by the shaded region in the following diagram. Find the area of this region.
16N.2.SL.TZ0.S_3a:
Write down the exact value of $\theta$ in radians.
20N.2.SL.TZ0.S_7b:
Find the value of $θ$ when the area of the shaded region is equal to the area of sector $OADB$ .
17M.2.AHL.TZ1.H_8b:
Calculate $\frac{{{\text{d}}\theta }}{{{\text{d}}t}}$ when $\theta = \frac{\pi }{3}$ .
21N.1.AHL.TZ0.15b.i:
Find an expression for $V$ in terms of $θ$ .
21N.1.AHL.TZ0.15b.ii:
Find the expression $\frac{d V}{d θ}$ .
21N.1.AHL.TZ0.15b.iii:
Solve algebraically $\frac{d V}{d θ} = 0$ to find the value of $θ$ that will maximize the volume, $V$ .
19M.1.AHL.TZ1.H_3:
A sector of a circle with radius $r$ cm , where $r$ > 0, is shown on the following diagram.
The sector has an angle of 1 radian at the centre.

Let the area of the sector be $A$ cm² and the perimeter be $P$ cm. Given that $A = P$ , find the value of $r$ .
16N.2.SL.TZ0.S_3c:
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19M.2.SL.TZ2.S_4b:
Find the area of triangle OBC in terms of $r$ and θ.
SPM.1.AHL.TZ0.5a:
Find 50° in radians.
17M.2.SL.TZ2.S_1a:
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18M.2.AHL.TZ2.H_4a.i:
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17M.2.SL.TZ2.S_1b:
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17M.2.SL.TZ2.S_1c:
Find the area of sector OABC.
16N.2.AHL.TZ0.H_9a:
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16N.2.AHL.TZ0.H_9c:
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16N.2.AHL.TZ0.H_9b:
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SPM.1.AHL.TZ0.5b:
Find the volume of this log.
18M.2.AHL.TZ2.H_4b:
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20N.2.SL.TZ0.S_7a:
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17M.2.AHL.TZ1.H_8a:
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16N.2.AHL.TZ0.H_7a:
Use the cosine rule to find the two possible values for AC.
16N.2.AHL.TZ0.H_7b:
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16N.2.SL.TZ0.S_3b:
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17N.1.SL.TZ0.S_4a:
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17M.2.SL.TZ1.S_5:
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Find the area of the shaded segment.
21N.1.AHL.TZ0.15a:
Show that $r = \frac{6}{2 + θ}$ .

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Topic 3—Geometry and trigonometry

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