Circular Measure Practice Questions

Q: Why must the angle be in radians?

The formulas s=rθs=r\thetas=rθ and A=12r2θA=\tfrac{1}{2}r^{2}\thetaA=21r2θ only hold when θ\thetaθ is measured in radians. If a question gives degrees, convert first with π rad=180∘\pi\ \text{rad}=180^{\circ}π rad=180∘, and set your calculator to radian mode before taking any sine or cosine.

Q: What is the difference between a sector and a segment?

A sector is bounded by two radii and an arc, like a slice of pizza. A segment is bounded by a chord and an arc. You find a segment's area as the sector area minus the triangle area: 12r2(θ−sin⁡θ)\tfrac{1}{2}r^{2}(\theta-\sin\theta)21r2(θ−sinθ).

Q: My segment answer is negative or larger than the sector, what went wrong?

Almost always the calculator was in degree mode, so sin⁡θ\sin\thetasinθ came out with the wrong value. In radians sin⁡θ\sin\thetasinθ is smaller than θ\thetaθ for these angles, so θ−sin⁡θ\theta-\sin\thetaθ−sinθ is a small positive number and the segment is a small slice of the sector.

How to use these practice questions

The six questions below rise in difficulty across the whole chapter, from a simple radian–degree conversion to the area of a segment that mixes a sector and a triangle. Give yourself roughly four to seven minutes per question and work on paper first, writing every line the way you would in the real exam, quote the formula, substitute the numbers, then state the answer with its unit.

Set your calculator to radian mode before you start, because every formula here needs $\theta$ in radians.

Only once you have committed to a full answer should you open the solution and mark yourself line by line. When your working differs from ours, stop at the exact line where the two part company; that single step is almost always where the mark was lost.

Because Add Math is marked analytically, quoting $s=r\theta$ or $A=\tfrac{1}{2}r^{2}\theta$ and substituting correctly still earns method marks even when the final arithmetic slips, so always write the formula before you compute.

Two ideas unlock the whole chapter. First, a radian is just another way to measure an angle, tied to the radius, so a full turn is $2\pi$ radians, the same as $360^{\circ}$ .

Second, once $\theta$ is in radians the arc length and sector area follow the tidy formulas $s=r\theta$ and $A=\tfrac{1}{2}r^{2}\theta$ , while a segment is always a sector with its triangle removed. Fix these pictures in your mind, and most questions become a choice of which formula to reach for.

Six practice questions

Q1[2 marks]

(a) Convert $\dfrac{\pi}{5}$ radians to degrees. (b) Convert $150^{\circ}$ to radians, leaving your answer as a multiple of $\pi$ .

Show worked solution

Every conversion rests on the single fact $\pi\ \text{rad}=180^{\circ}$ . To go from radians to degrees, replace $\pi$ with $180^{\circ}$ :

\frac{\pi}{5}=\frac{180^{\circ}}{5}=36^{\circ}

To go the other way, multiply the degrees by $\dfrac{\pi}{180}$ and cancel:

150^{\circ}=150\times\frac{\pi}{180}=\frac{5\pi}{6}\ \text{rad}

Answer

(a) $\dfrac{\pi}{5}\ \text{rad}=36^{\circ}$ . (b) $150^{\circ}=\dfrac{5\pi}{6}\ \text{rad}\approx2.618\ \text{rad}$ .

Q2[3 marks]

A sector of a circle has radius $8\ \text{cm}$ and subtends an angle of $1.5$ radians at the centre. Find (a) the arc length, (b) the area of the sector.

Show worked solution

(a) The arc length uses $s=r\theta$ with $\theta$ already in radians, so no conversion is needed. Substitute $r=8$ and $\theta=1.5$ :

s=r\theta=8\times1.5=12\ \text{cm}

(b) The sector area uses $A=\tfrac{1}{2}r^{2}\theta$ . Square the radius first, then multiply by the half and the angle:

A=\tfrac{1}{2}r^{2}\theta=\tfrac{1}{2}(8)^{2}(1.5)=\tfrac{1}{2}(64)(1.5)=48\ \text{cm}^{2}

Answer

(a) Arc length $=12\ \text{cm}$ . (b) Sector area $=48\ \text{cm}^{2}$ .

Q3[3 marks]

A sector of a circle has radius $10\ \text{cm}$ and an arc length of $25\ \text{cm}$ . Find (a) the angle $\theta$ of the sector in radians, (b) the perimeter of the sector.

Show worked solution

(a) The same formula $s=r\theta$ works backwards. Rearrange to make $\theta$ the subject, then substitute $s=25$ and $r=10$ :

\theta=\frac{s}{r}=\frac{25}{10}=2.5\ \text{rad}

(b) The perimeter of a sector is its two straight radii plus the curved arc, so $P=2r+s$ , the two straight edges are two separate radii, not a diameter:

P=2r+s=2(10)+25=45\ \text{cm}

Answer

(a) $\theta=2.5\ \text{rad}$ . (b) Perimeter $=45\ \text{cm}$ .

Notice the perimeter adds the two radii, not the diameter.

Q4[4 marks]

A sector of a circle has an angle of $0.5$ radians and an area of $36\ \text{cm}^{2}$ . Find (a) the radius of the circle, (b) the arc length of the sector.

Show worked solution

(a) The unknown here is the radius, so start from the sector-area formula and substitute the two known values $A=36$ and $\theta=0.5$ :

A=\tfrac{1}{2}r^{2}\theta \;\Rightarrow\; 36=\tfrac{1}{2}r^{2}(0.5)=0.25\,r^{2}

Solve for $r^{2}$ , then take the positive square root because a radius cannot be negative:

r^{2}=\frac{36}{0.25}=144 \;\Rightarrow\; r=12\ \text{cm}

(b) With the radius now known, the arc length follows at once from $s=r\theta$ :

s=r\theta=12\times0.5=6\ \text{cm}

Answer

(a) Radius $=12\ \text{cm}$ . (b) Arc length $=6\ \text{cm}$ .

Check: $\tfrac{1}{2}(12)^{2}(0.5)=\tfrac{1}{2}(144)(0.5)=36\ \text{cm}^{2}$ , the given area.

Q5[5 marks]

A chord of a circle of radius $10\ \text{cm}$ subtends an angle of $1.2$ radians at the centre. Find the area of the minor segment cut off by the chord.

Give your answer correct to two decimal places. [Use $\sin 1.2=0.9320$ .]

Show worked solution

A segment is what remains when the triangle formed by the two radii is removed from the sector. So its area is the sector area minus the triangle area, which combine into one formula:

A_{\text{segment}}=\tfrac{1}{2}r^{2}\theta-\tfrac{1}{2}r^{2}\sin\theta=\tfrac{1}{2}r^{2}(\theta-\sin\theta)

Substitute $r=10$ and $\theta=1.2$ , keeping the calculator in radian mode so that $\sin1.2=0.9320$ and not the degree value:

A_{\text{segment}}=\tfrac{1}{2}(10)^{2}(1.2-0.9320)=50\times0.2680

A_{\text{segment}}=13.40\ \text{cm}^{2}

Answer

The minor segment has area $13.40\ \text{cm}^{2}$ . Check the split: sector $=\tfrac{1}{2}(100)(1.2)=60\ \text{cm}^{2}$ , triangle $=\tfrac{1}{2}(100)(0.9320)=46.60\ \text{cm}^{2}$ , and $60-46.60=13.40\ \text{cm}^{2}$ .

Q6[6 marks]

In a circle with centre $O$ and radius $12\ \text{cm}$ , the sector $OAB$ has angle $\angle AOB=1$ radian. Find (a) the arc length $AB$ , (b) the area of the sector $OAB$ , (c) the area of the shaded segment between the chord $AB$ and the arc $AB$ .

Give (c) correct to two decimal places. [Use $\sin 1=0.8415$ .]

Show worked solution

This question stacks the three main formulas, so take them in order. (a) The arc length uses $s=r\theta$ with $r=12$ and $\theta=1$ :

s=r\theta=12\times1=12\ \text{cm}

(b) The sector area uses $A=\tfrac{1}{2}r^{2}\theta$ with the same values:

A=\tfrac{1}{2}(12)^{2}(1)=\tfrac{1}{2}(144)(1)=72\ \text{cm}^{2}

(c) The shaded region is a segment, so subtract the triangle from the sector using $\tfrac{1}{2}r^{2}(\theta-\sin\theta)$ with $\sin1=0.8415$ :

A_{\text{segment}}=\tfrac{1}{2}(144)(1-0.8415)=72\times0.1585=11.41\ \text{cm}^{2}

Answer

(a) Arc $AB=12\ \text{cm}$ . (b) Sector $OAB=72\ \text{cm}^{2}$ .

(c) Shaded segment $=11.41\ \text{cm}^{2}$ . Check: triangle $=\tfrac{1}{2}(144)(0.8415)=60.59\ \text{cm}^{2}$ , and $72-60.59=11.41\ \text{cm}^{2}$ .

How to mark yourself like an examiner

Add Math is marked analytically, which means marks are attached to steps, not only to the final number. When you check your own script, award yourself credit the way a marker would: look for the correct formula, the calculator in radian mode, a clean substitution, and a final value with the right unit.

Run through the checklist below on every part, and pin down the exact line where a mark was earned or lost.

Radian mark: is $\theta$ in radians for every formula, and is the calculator set to radian mode before you take a sine?
Formula mark: did you write $s=r\theta$ , $A=\tfrac{1}{2}r^{2}\theta$ or $\tfrac{1}{2}r^{2}(\theta-\sin\theta)$ before substituting?
Segment mark: did you form the segment as sector minus triangle, rather than guessing a single formula?
Perimeter mark: for the perimeter of a sector, did you add the two radii plus the arc, and not the diameter?
If your final number is wrong but the formula and substitution are right, give yourself the method marks, that is exactly what a real marker does.

How a teacher helps

Marking yourself is powerful, but it is hard to see your own blind spots. In a one-to-one lesson our teacher watches the exact line where a mark slips away, leaving the calculator in degree mode, adding the diameter instead of two radii, or forgetting that a segment is a sector minus a triangle, and corrects the habit on the spot.

Because our teachers are experienced, you work with someone who explains the why behind each step. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

Get 1-to-1 help.

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Frequently asked questions

Why must the angle be in radians?

The formulas $s=r\theta$ and $A=\tfrac{1}{2}r^{2}\theta$ only hold when $\theta$ is measured in radians. If a question gives degrees, convert first with $\pi\ \text{rad}=180^{\circ}$ , and set your calculator to radian mode before taking any sine or cosine.

What is the difference between a sector and a segment?

A sector is bounded by two radii and an arc, like a slice of pizza. A segment is bounded by a chord and an arc.

You find a segment's area as the sector area minus the triangle area: $\tfrac{1}{2}r^{2}(\theta-\sin\theta)$ .

How do I find the perimeter of a sector?

Add the two straight edges (each a radius) to the curved arc: $P=2r+s=2r+r\theta$ . A common slip is to use the diameter $2r$ as one edge, the two edges are two separate radii.

My segment answer is negative or larger than the sector, what went wrong?

Almost always the calculator was in degree mode, so $\sin\theta$ came out with the wrong value. In radians $\sin\theta$ is smaller than $\theta$ for these angles, so $\theta-\sin\theta$ is a small positive number and the segment is a small slice of the sector.

Does a wrong final answer cost me every mark?

No. Because marking is analytic, quoting the correct formula and substituting correctly still earn method marks even if the arithmetic slips.

Always write the formula line first.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Circular Measure, Practice Questions

How to use these practice questions

Six practice questions

How to mark yourself like an examiner

How a teacher helps

Frequently asked questions

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