Arc Length and Sector Area
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Arc Length and Sector Area — Free MYP4 Mathematics (Extended) Practice Questions

1QuestionLength of arc and chord, perimeter and area of sector and segmentAssessment Practice
5 marks~8 minCriterion A
The diagram shows a sector with radius r=10r = 10 cm and central angle θ=72°\theta = 72°.
a
Write down the arc length of this sector, giving your answer in the form kπk\pi cm, where kk is an integer. [1]
b
A second sector has the same central angle of 72°72° and an arc length of 10π10\pi cm. Find the radius RR of the second sector. Give your answer in centimetres. [2]
c
A student claims: "If the central angle stays the same, doubling the arc length always doubles the radius." Justify whether this claim is correct, using the arc length formula and your results from parts (a) and (b). [2]
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Solutions

2QuestionLength of arc and chord, perimeter and area of sector and segmentAssessment Practice
6 marks~9 minCriterion D
A pizza restaurant advertises that its Large pizza has a diameter of 40 cm and is cut into 8 equal slices. The restaurant claims each slice has a crust length (arc length) of 16 cm.

The diagram shows one slice of the pizza, with centre OO, radius rr, and the arc length of one slice labelled.
a
Calculate the actual arc length of one slice. Give your answer in exact form and to 3 significant figures. [2]
b
The restaurant's manager states: "The percentage error in our advertised crust length is less than 10%."

The percentage error is calculated using:

percentage error=∣claimed−actual∣actual×100%\text{percentage error} = \frac{|\text{claimed} - \text{actual}|}{\text{actual}} \times 100\%

Determine whether the manager's statement is correct. Show all working and give your percentage error to 1 decimal place. [2]
c
The restaurant now considers cutting the same Large pizza into nn equal slices so that each slice has an arc length of exactly 5π2\dfrac{5\pi}{2} cm.

Find the value of nn and justify whether this cutting is physically possible for the restaurant. [2]
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Solutions

3QuestionLength of arc and chord, perimeter and area of sector and segmentAssessment Practice
8 marks~12 minCriterion B
The table shows arc length and sector area values for four sectors, each from a circle of radius 10 cm.

Central angle (degrees)306090120
Arc length (cm)5.2410.4715.7120.94
Sector area (cm²)26.1852.3678.54104.72
a
Describe the relationship between the central angle θ\theta and the arc length LL, referring to at least two values in the table. [2]
b
Using the pattern in the table, write down a formula for the arc length LL of a sector with central angle θ\theta degrees and radius rr cm. Hence find the arc length of a sector with central angle 150∘150^\circ in the same circle (radius 10 cm). Give your answer in exact form and correct to 1 decimal place. [3]
c
A different circle has radius rr cm. Two sectors of this circle have central angles α\alpha degrees and β\beta degrees respectively, where α+β=180\alpha + \beta = 180. Show that the sum of their arc lengths equals πr\pi r, and justify why this result holds regardless of the individual values of α\alpha and β\beta. [3]
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Solutions

4QuestionLength of arc and chord, perimeter and area of sector and segmentAssessment Practice
8 marks~12 minCriterion C
A circle has centre OO and radius r=10r = 10 cm. Points AA and BB lie on the circumference. The shaded region is the minor segment cut off by chord ABAB, where the central angle ∠AOB=θ\angle AOB = \theta degrees. The area of the minor segment is given by

A(θ)=θ360×πr2−12r2sin⁡θA(\theta) = \frac{\theta}{360} \times \pi r^2 - \frac{1}{2}r^2 \sin\theta

where r=10r = 10 cm.

θ\theta (degrees): 60, 90, 120
a
Calculate the area of the minor segment for each value of θ\theta. Give each answer correct to 1 decimal place. [3]
b
A student claims: "The segment area increases by the same amount each time θ\theta increases by 30°30°." Using your values from part (a), determine whether this claim is correct. Show all working. [2]
c
The formula contains a sector-area term θ360πr2\dfrac{\theta}{360}\pi r^2 and a triangle-area term 12r2sin⁡θ\dfrac{1}{2}r^2\sin\theta. Justify why the segment area increases as θ\theta increases from 60°60° to 120°120°, by analysing the behaviour of each term separately over this interval. [3]
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Solutions

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