Circular Measure Worked Examples (Medium)

Q: Why does dividing the area equation by the arc equation work so cleanly?

Both equations share the factor rθr\thetarθ. When you compute As=12r2θrθ\dfrac{A}{s}=\dfrac{\tfrac{1}{2}r^{2}\theta}{r\theta}sA=rθ21r2θ, that shared factor cancels and you are left with 12r\tfrac{1}{2}r21r. It isolates the radius in one step, which you can then feed back to find the angle.

Q: How do I find the chord of a segment?

Use AB=2rsin⁡ ⁣(θ2)AB=2r\sin\!\left(\dfrac{\theta}{2}\right)AB=2rsin(2θ), which comes from splitting the isosceles triangle down its axis of symmetry into two right-angled triangles. As a check, the cosine rule AB2=2r2(1−cos⁡θ)AB^{2}=2r^{2}(1-\cos\theta)AB2=2r2(1−cosθ) gives the same length.

What these examples cover

These medium Circular Measure examples build the moves that mark the step up from routine questions: working backwards from an arc length and a sector area to recover the radius and the angle, finding the area of a segment by taking the triangle away from the sector, and finding the perimeter of a segment by adding its arc to its chord. Each still uses clean numbers, but now two ideas meet in one question, so the order of your steps matters.

The key habit is to write down every formula you will use before you substitute, then decide which unknown each equation can release first. Use the set the honest way, cover the solution, attempt the question in full on paper, and only then check line by line against our working.

Where your answer differs, find the exact step where the two solutions part company; that single line is usually where the real learning is.

Worked examples

Work through all three. Attempt each fully before you read the matching solution, and watch how dividing one equation by another, or splitting a region into a sector and a triangle, turns a two-unknown problem into two single steps.

Q1[4 marks]

A sector of a circle has an arc length of $12$ cm and an area of $48$ cm $^{2}$ . Find the radius of the circle and the angle of the sector in radians.

Show worked solution

Write both formulas with the values in place. The arc gives one equation and the area gives another, both in the two unknowns $r$ and $\theta$ :

s=r\theta=12\qquad A=\tfrac{1}{2}r^{2}\theta=48

Dividing the area equation by the arc equation is the neat move: the common factor $r\theta$ cancels, leaving only $r$ .

\dfrac{A}{s}=\dfrac{\tfrac{1}{2}r^{2}\theta}{r\theta}=\tfrac{1}{2}r=\dfrac{48}{12}=4

So $\tfrac{1}{2}r=4$ , which gives $r=8$ . Now substitute back into the arc equation to find $\theta$ :

r\theta=12\;\Rightarrow\;8\theta=12\;\Rightarrow\;\theta=\dfrac{12}{8}=1.5

Answer

The radius is $r=8$ cm and the angle is $\theta=1.5$ radians. Check both: the arc is $8\times1.5=12$ cm, and the area is $\tfrac{1}{2}(8^{2})(1.5)=32\times1.5=48$ cm $^{2}$ .

Both match the question.

Q2[3 marks]

A chord divides a circle of radius $9$ cm so that the minor sector has an angle of $2$ radians at the centre. Find the area of the minor segment, correct to two decimal places.

Show worked solution

A segment is what remains when the triangle formed by the two radii is cut away from the sector. So its area is the sector area minus the triangle area:

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 $r=9$ and $\theta=2$ . Keep $\theta$ in radians, so set your calculator to radian mode before finding $\sin 2$ :

A_{\text{segment}}=\tfrac{1}{2}(9^{2})\,(2-\sin 2)=\tfrac{1}{2}(81)\,(2-\sin 2)

In radian mode $\sin 2 = 0.9093$ , so the bracket is $2-0.9093=1.0907$ :

A_{\text{segment}}=40.5\times1.0907=44.17

Answer

The area of the minor segment is $44.17$ cm $^{2}$ (2 d.p.). The commonest error is leaving the calculator in degree mode, which would give $\sin 2^{\circ}=0.0349$ and a badly wrong bracket, always confirm radian mode when the angle is in radians.

Q3[4 marks]

In a circle of centre $O$ and radius $10$ cm, two radii $OA$ and $OB$ make an angle of $1.2$ radians. Find the perimeter of the minor segment bounded by the chord $AB$ and the minor arc $AB$ , correct to two decimal places.

Show worked solution

The perimeter of the segment has two parts: the curved arc $AB$ and the straight chord $AB$ . Find the arc first with $s=r\theta$ :

\text{arc } AB=r\theta=10\times1.2=12

For the chord, drop the radius that bisects the angle to make two right-angled triangles; each has half-angle $\tfrac{\theta}{2}$ and hypotenuse $r$ . The chord is twice the opposite side:

AB=2r\sin\!\left(\dfrac{\theta}{2}\right)=2(10)\sin\!\left(\dfrac{1.2}{2}\right)=20\sin 0.6

In radian mode $\sin 0.6 = 0.5646$ , so the chord is:

AB=20\times0.5646=11.29

Add the arc and the chord to close the segment:

P=\text{arc}+\text{chord}=12+11.29=23.29

Answer

The perimeter of the minor segment is $23.29$ cm (2 d.p.). A quick check on the chord: the cosine rule gives $AB^{2}=10^{2}+10^{2}-2(10)(10)\cos 1.2=200(1-0.3624)=127.52$ , so $AB=\sqrt{127.52}=11.29$ cm, the same value by a second route.

Q4[3 marks]

A sector of a circle has radius $5$ cm and area $20$ cm $^{2}$ . Find the perimeter of the sector.

Show worked solution

The area formula has only one unknown once $r$ is known, so find $\theta$ first:

A=\tfrac{1}{2}r^{2}\theta\;\Rightarrow\;20=\tfrac{1}{2}(5^{2})\theta=12.5\theta\;\Rightarrow\;\theta=\dfrac{20}{12.5}=1.6

The perimeter of a sector is the two straight radii plus the curved arc, $P=2r+r\theta$ :

P=2(5)+5(1.6)=10+8=18

Answer

The perimeter of the sector is $18$ cm. Check: the arc length is $s=r\theta=5\times1.6=8$ cm, so $P=2r+s=10+8=18$ cm, both routes agree.

Q5[4 marks]

Two concentric circles have a common centre $O$ , with radii $10$ cm and $6$ cm. The same two radii of $O$ subtend an angle of $60^{\circ}$ at the centre, cutting an arc from each circle.

Find the area of the ring-shaped region between the two arcs, correct to two decimal places.

Show worked solution

The sector formulas need the angle in radians, so convert first:

\theta=60^{\circ}\times\dfrac{\pi}{180}=\dfrac{\pi}{3}\approx1.0472\text{ rad}

The ring-shaped region is the large sector with the small sector removed. Factor out $\theta$ rather than working the two areas separately:

A=\tfrac{1}{2}\theta\,(10^{2}-6^{2})=\tfrac{1}{2}(1.0472)(64)=32\times1.0472=33.51

Answer

The ring-shaped region has area $33.51$ cm $^{2}$ (2 d.p.). Check using degrees directly: $60^{\circ}$ is $\tfrac{1}{6}$ of a full turn, and $\tfrac{1}{6}$ of the full ring's area $\pi(10^{2}-6^{2})=201.06$ cm $^{2}$ is $33.51$ cm $^{2}$ , the same answer.

Q6[3 marks]

The minute hand of a clock is $6$ cm long. As the minute hand moves from the $12$ to the $5$ on the clock face, it sweeps through $5$ of the $12$ equal divisions of a full turn.

Find the length of the arc traced by the tip of the minute hand, correct to two decimal places.

Show worked solution

Moving from the $12$ to the $5$ sweeps $\tfrac{5}{12}$ of one full revolution, which is $2\pi$ radians:

\theta=\dfrac{5}{12}\times2\pi=\dfrac{5\pi}{6}\approx2.6180\text{ rad}

Now apply the arc length formula $s=r\theta$ with $r=6$ :

s=r\theta=6\times\dfrac{5\pi}{6}=5\pi=15.71\text{ cm (2 d.p.)}

Answer

The tip traces an arc of $15.71$ cm. Check by taking the fraction of the whole circumference directly: $\tfrac{5}{12}\times2\pi(6)=\tfrac{5}{12}\times37.70=15.71$ cm, the same answer.

Q7[4 marks]

A chord $AB$ of length $12$ cm lies in a circle of radius $10$ cm, centre $O$ . Find the angle $AOB$ in radians, correct to four significant figures, and hence find the length of the minor arc $AB$ , correct to two decimal places.

Show worked solution

Drop a perpendicular from $O$ to the chord; it bisects both the chord and the angle, giving a right-angled triangle with hypotenuse $r$ and opposite side half the chord:

\sin\!\left(\dfrac{\theta}{2}\right)=\dfrac{6}{10}=0.6\;\Rightarrow\;\dfrac{\theta}{2}=\sin^{-1}(0.6)=0.6435\;\Rightarrow\;\theta=1.287\text{ rad (4 s.f.)}

Now find the minor arc with $s=r\theta$ :

s=r\theta=10\times1.287=12.87\text{ cm (2 d.p.)}

Answer

The angle $AOB\approx1.287$ rad and the minor arc $AB\approx12.87$ cm. Sense check: half the chord is $6$ cm and the perpendicular distance from $O$ to $AB$ is $\sqrt{10^{2}-6^{2}}=8$ cm, a clean $6\text{-}8\text{-}10$ right triangle, which is why $\sin(\theta/2)=0.6$ came out so tidily.

Q8[4 marks]

A circular disc of radius $7$ cm is cut from its centre into three sectors whose angles are in the ratio $2:3:4$ . Find the area of the largest sector, correct to two decimal places.

Show worked solution

The three sectors share a centre, so their angles add up to one full revolution, $2\pi$ radians. Split $2\pi$ into $2+3+4=9$ equal ratio-parts, and take $4$ of them for the largest sector:

\theta_{\text{largest}}=\dfrac{4}{9}\times2\pi=\dfrac{8\pi}{9}\approx2.7925\text{ rad}

Apply the sector area formula with $r=7$ :

A=\tfrac{1}{2}r^{2}\theta=\tfrac{1}{2}(7^{2})(2.7925)=24.5\times2.7925=68.42\text{ cm}^{2}\text{ (2 d.p.)}

Answer

The largest sector has area $68.42$ cm $^{2}$ . Check: the whole disc has area $\pi r^{2}=\pi(7^{2})=153.94$ cm $^{2}$ , and $\tfrac{4}{9}$ of that is $68.42$ cm $^{2}$ , the same answer.

Across all three, the pattern is the same: name every quantity, write the formula that contains it, and release the unknowns in an order that keeps the arithmetic clean. Dividing two equations, or splitting a region into a sector and a triangle, is often what turns a hard-looking question into two short ones.

Key method points

These three examples rehearse the reverse and two-step moves that mark the medium band in Circular Measure. Keep the following points in mind as you practise more.

When you know the arc and the sector area, divide $A$ by $s$ : the factor $r\theta$ cancels and $\tfrac{1}{2}r=\dfrac{A}{s}$ gives the radius at once.
The area of a segment is the sector minus the triangle: $A=\tfrac{1}{2}r^{2}(\theta-\sin\theta)$ , with $\theta$ in radians.
The chord of a segment is $AB=2r\sin\!\left(\dfrac{\theta}{2}\right)$ , or you can find it from the cosine rule as a check.
The perimeter of a segment is its arc plus its chord, never the arc alone.
Set the calculator to radian mode before taking $\sin\theta$ or $\sin\!\left(\dfrac{\theta}{2}\right)$ ; a degree-mode value will be far too small.
Verify a recovered radius and angle by substituting back into both original equations before you move on.

How a teacher helps

The medium band is where the order of steps starts to decide the mark. Students often have the right formulas but reach for them in the wrong sequence, or slip into degree mode halfway through.

In a one-to-one lesson our teacher watches the exact line where the plan goes astray and shows the cleaner route, dividing two equations, or splitting a segment into a sector and a triangle, before the habit sets. Because our teachers are experienced, you work with someone who explains why the segment formula is a subtraction, not just how to key it in.

Lessons are taught in English, while SPM papers are set in both Malay and English.

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Frequently asked questions

Why does dividing the area equation by the arc equation work so cleanly?

Both equations share the factor $r\theta$ . When you compute $\dfrac{A}{s}=\dfrac{\tfrac{1}{2}r^{2}\theta}{r\theta}$ , that shared factor cancels and you are left with $\tfrac{1}{2}r$ .

It isolates the radius in one step, which you can then feed back to find the angle.

Where does the segment formula

\tfrac{1}{2}r^{2}(\theta-\sin\theta)

come from?

A segment is a sector with the triangle removed. The sector area is $\tfrac{1}{2}r^{2}\theta$ and the triangle formed by the two radii has area $\tfrac{1}{2}r^{2}\sin\theta$ .

Subtracting gives $\tfrac{1}{2}r^{2}(\theta-\sin\theta)$ .

How do I find the chord of a segment?

Use $AB=2r\sin\!\left(\dfrac{\theta}{2}\right)$ , which comes from splitting the isosceles triangle down its axis of symmetry into two right-angled triangles. As a check, the cosine rule $AB^{2}=2r^{2}(1-\cos\theta)$ gives the same length.

My segment answer is far too small, what went wrong?

Almost always the calculator is in degree mode. When the angle is in radians you must take $\sin\theta$ in radian mode; otherwise $\sin 2$ is read as $\sin 2^{\circ}\approx0.035$ and the whole bracket collapses.

Switch to radian mode and redo the trigonometric line.

Source:SRC-DSKP-EN

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

Circular Measure, Worked Examples (medium)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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