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Date	May 2022	Marks available	3	Reference code	22M.2.SL.TZ1.3
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Find	Question number	3	Adapted from	N/A

Question

A company is designing a new logo. The logo is created by removing two equal segments from a rectangle, as shown in the following diagram.

The rectangle measures $5 cm$ by $4 cm$ . The points $A$ and $B$ lie on a circle, with centre $O$ and radius $2 cm$ , such that $AÔB = θ$ , where $0 < θ < π$ . This information is shown in the following diagram.

Find the area of one of the shaded segments in terms of $θ$ .

[3]

Given that the area of the logo is $13.4 {cm}^{2}$ , find the value of $θ$ .

[3]

Markscheme

valid approach to find area of segment by finding area of sector – area of triangle (M1)

$\frac{1}{2} r^{2} θ - \frac{1}{2} r^{2} \sin θ$

$\frac{1}{2} {(2)}^{2} θ - \frac{1}{2} {(2)}^{2} \sin θ$ (A1)

area $= 2 θ - 2 \sin θ$ A1

[3 marks]

EITHER

area of logo = area of rectangle – area of segments (M1)

$5 \times 4 - 2 \times (2 θ - 2 \sin θ) = 13.4$ (A1)

area of one segment $= \frac{20 - 13.4}{2} (= 3.3)$ (M1)

$2 θ - 2 \sin θ = 3.3$ (A1)

THEN

$θ = 2.35672 \dots$

$θ = 2.36$ (do not accept an answer in degrees) A1

Note: Award (M1)(A1)A0 if there is more than one solution.
Award (M1)(A1FT)A0 if the candidate works in degrees and obtains a final answer of $135.030 \dots$

[3 marks]

Examiners report

The first part of this question proved particularly challenging for many candidates. Most were able to obtain the area of the sector, but then did not recognise that the segment was formed by subtracting the area of the triangle. Those that did, often struggled to find the appropriate triangle area formula to do so.

The second part of the question was generally answered more successfully, although sometimes only one of the segments was subtracted from the whole rectangle. Many candidates who had a correct equation lacked the requisite calculator skills to obtain a solution. Those that had an incorrect expression for the area from part (a), often obtained $20 - 4 θ = 13.4$ . This was a significantly easier equation to solve, and consequently these candidates were not awarded the final A mark for a final answer of $θ = 1.65$ .

[N/A]

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.4—Circle: radians, arcs, sectors

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