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Worked examples · Circular Measure

Circular Measure, Worked Examples (easy)

These easy Circular Measure examples rehearse the four everyday moves, converting between degrees and radians, finding arc length with s=rθs=r\theta, finding sector area with A=12r2θA=\tfrac{1}{2}r^{2}\theta, and building the perimeter of a sector. Keep the angle in radians throughout, try each on paper first, then check every line against our full solution.

What these examples cover

These easy Circular Measure examples build the everyday moves that open almost every question in the chapter: converting an angle between degrees and radians, finding an arc length from s=rθs=r\theta, finding a sector area from A=12r2θA=\tfrac{1}{2}r^{2}\theta, and adding two radii to an arc to get the perimeter of a sector. Each one uses small, clean numbers so you can follow every line while your calculator does only the arithmetic.

The single most important habit is to keep the angle in radians whenever you use s=rθs=r\theta or A=12r2θA=\tfrac{1}{2}r^{2}\theta, these formulas are simply wrong in degrees. 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 four. Attempt each fully before you read the matching solution, and notice how the same discipline, write the formula, substitute one value at a time, then simplify, runs through conversion, arc length, sector area and perimeter alike.

Q1[3 marks]

(a) Convert 135135^{\circ} to radians, giving your answer in terms of π\pi. (b) Convert 5π6\dfrac{5\pi}{6} radians to degrees.

Show worked solution

The whole conversion rests on one fact: a straight angle is 180180^{\circ} and also π\pi radians. So multiply by π180\dfrac{\pi}{180^{\circ}} to turn degrees into radians, and by 180π\dfrac{180^{\circ}}{\pi} to go the other way.

(a) Multiply 135135^{\circ} by π180\dfrac{\pi}{180^{\circ}}, then cancel down the fraction:

135×π180=135180π=34π135^{\circ}\times\dfrac{\pi}{180^{\circ}}=\dfrac{135}{180}\,\pi=\dfrac{3}{4}\,\pi

(b) Multiply 5π6\dfrac{5\pi}{6} by 180π\dfrac{180^{\circ}}{\pi}; the π\pi cancels, leaving only numbers:

5π6×180π=5×1806=9006=150\dfrac{5\pi}{6}\times\dfrac{180^{\circ}}{\pi}=\dfrac{5\times180^{\circ}}{6}=\dfrac{900^{\circ}}{6}=150^{\circ}

Answer

(a) 135=3π4135^{\circ}=\dfrac{3\pi}{4} rad. (b) 5π6\dfrac{5\pi}{6} rad =150=150^{\circ}.

A quick check: 3π4\dfrac{3\pi}{4} is a little more than π2=90\dfrac{\pi}{2}=90^{\circ}, which fits 135135^{\circ}; and 150150^{\circ} is a little less than 180=π180^{\circ}=\pi, which fits 5π6\dfrac{5\pi}{6}.

Q2[2 marks]

An arc of a circle of radius 88 cm subtends an angle of 1.51.5 radians at the centre. Find the length of the arc.

Show worked solution

Arc length uses the formula s=rθs=r\theta, and the angle is already in radians, so no conversion is needed:

Must memorise
s=rθs=r\theta

Substitute r=8r=8 and θ=1.5\theta=1.5:

s=8×1.5=12s=8\times1.5=12

Answer

The arc length is 1212 cm. Because θ=1.5\theta=1.5 rad is a little under a quarter of the full turn 2π6.282\pi\approx6.28, the arc is a little under a quarter of the circumference 2π(8)50.32\pi(8)\approx50.3 cm, and 1212 sits sensibly in that range.

Q3[2 marks]

A sector of a circle has radius 66 cm and the angle at the centre is 22 radians. Find the area of the sector.

Show worked solution

Sector area uses A=12r2θA=\tfrac{1}{2}r^{2}\theta, again with θ\theta in radians:

A=12r2θA=\tfrac{1}{2}r^{2}\theta

Substitute r=6r=6 and θ=2\theta=2; square the radius first:

A=12×62×2=12×36×2=36A=\tfrac{1}{2}\times6^{2}\times2=\tfrac{1}{2}\times36\times2=36

Answer

The area of the sector is 3636 cm2^{2}. Notice the 12\tfrac{1}{2} and the ×2\times2 cancel here, so the area equals r2=36r^{2}=36, a neat coincidence only because θ=2\theta=2.

Q4[3 marks]

A sector of a circle has radius 55 cm and the angle at the centre is 1.21.2 radians. Find the perimeter of the sector.

Show worked solution

The perimeter of a sector is the curved arc plus the two straight radii that close it off. First find the arc with s=rθs=r\theta:

s=rθ=5×1.2=6s=r\theta=5\times1.2=6

Now add the two radii. The perimeter is the arc plus 2r2r:

P=s+2r=6+2(5)=6+10=16P=s+2r=6+2(5)=6+10=16

Answer

The perimeter is 1616 cm. The most common slip here is to forget the two radii and report only the arc, a sector is bounded by three edges, not one, so always add 2r2r.

Q5[2 marks]

An arc of a circle of radius 66 cm has length 99 cm. Find the angle subtended at the centre, in radians.

Show worked solution

Arc length uses s=rθs=r\theta; here ss and rr are known and θ\theta is what we need, so rearrange to make θ\theta the subject:

θ=sr\theta=\dfrac{s}{r}

Substitute s=9s=9 and r=6r=6:

θ=96=1.5\theta=\dfrac{9}{6}=1.5

Answer

The angle is θ=1.5\theta=1.5 rad. Since 1.51.5 rad is a little under a quarter turn (π21.57\tfrac{\pi}{2}\approx1.57), the arc should be a little under a quarter of the circumference 2π(6)37.72\pi(6)\approx37.7 cm, a quarter of that is about 9.49.4 cm, close to the given 99 cm.

Q6[3 marks]

A sector of a circle has an area of 88 cm2^{2} and the angle at the centre is 11 radian. Find the radius of the circle.

Show worked solution

Sector area uses A=12r2θA=\tfrac{1}{2}r^{2}\theta. Here AA and θ\theta are known, so rearrange to make r2r^{2} the subject first:

r2=2Aθr^{2}=\dfrac{2A}{\theta}

Substitute A=8A=8 and θ=1\theta=1, then take the square root:

r2=2×81=16r=16=4r^{2}=\dfrac{2\times8}{1}=16 \quad\Rightarrow\quad r=\sqrt{16}=4

Answer

The radius is r=4r=4 cm. Check by substituting back: A=12×42×1=12×16=8A=\tfrac{1}{2}\times4^{2}\times1=\tfrac{1}{2}\times16=8 cm2^{2}, which matches.

Q7[2 marks]

A sector of a circle has radius 55 cm and area 1010 cm2^{2}. Find the angle at the centre, in radians.

Show worked solution

Sector area uses A=12r2θA=\tfrac{1}{2}r^{2}\theta. Here AA and rr are known, so rearrange to make θ\theta the subject:

θ=2Ar2\theta=\dfrac{2A}{r^{2}}

Substitute A=10A=10 and r=5r=5; square the radius first:

θ=2×1052=2025=0.8\theta=\dfrac{2\times10}{5^{2}}=\dfrac{20}{25}=0.8

Answer

The angle is θ=0.8\theta=0.8 rad. As a check, a full circle of this radius has area π(5)278.5\pi(5)^{2}\approx78.5 cm2^{2}; the sector's 1010 cm2^{2} is about an eighth of that, and 0.80.8 rad is indeed close to an eighth of the full turn 2π6.282\pi\approx6.28.

Q8[3 marks]

A sector of a circle has radius 99 cm and the angle at the centre is 6060^{\circ}. Find the length of the arc, in terms of π\pi.

Show worked solution

The formula s=rθs=r\theta needs θ\theta in radians, so convert 6060^{\circ} first using π180\dfrac{\pi}{180^{\circ}}:

60×π180=π360^{\circ}\times\dfrac{\pi}{180^{\circ}}=\dfrac{\pi}{3}

Now substitute r=9r=9 and θ=π3\theta=\dfrac{\pi}{3} into s=rθs=r\theta:

s=9×π3=3πs=9\times\dfrac{\pi}{3}=3\pi

Answer

The arc length is 3π3\pi cm (about 9.429.42 cm). Since 6060^{\circ} is one-sixth of a full turn, the arc should be one-sixth of the circumference 2π(9)=18π2\pi(9)=18\pi; and 16×18π=3π\tfrac{1}{6}\times18\pi=3\pi, which matches exactly.

Notice how little changes from one example to the next. Once the angle is in radians, arc length is rθr\theta, sector area is 12r2θ\tfrac{1}{2}r^{2}\theta, and the perimeter is just the arc with the two radii added back.

Read what the question asks for, pick the matching formula, and the arithmetic stays short and reliable.

Key method points

These four examples rehearse the tools that open almost every Circular Measure question in Add Math. Keep the following points in mind as you practise more.

  • The bridge for conversion is 180=π180^{\circ}=\pi rad: multiply by π180\dfrac{\pi}{180^{\circ}} for degrees-to-radians, and by 180π\dfrac{180^{\circ}}{\pi} for radians-to-degrees.
  • Arc length is s=rθs=r\theta and sector area is A=12r2θA=\tfrac{1}{2}r^{2}\theta, both demand θ\theta in radians, so convert first if the angle is given in degrees.
  • The perimeter of a sector is the arc plus two radii, P=rθ+2rP=r\theta+2r; the arc alone is never the full perimeter.
  • Square the radius before multiplying in the area formula, and substitute one value at a time to keep the working clean.
  • Do a quick sense check against the whole circle: an angle near 2π2\pi should give an arc near the full circumference 2πr2\pi r.
  • Keep every substitution line, with analytic marking a clear method line still earns method marks even if the final digit slips.

How a teacher helps

When a student drops a mark on questions like these, it is nearly always a small, fixable habit, using s=rθs=r\theta with the angle still in degrees, or giving only the arc when the question asks for a whole perimeter. In a one-to-one lesson our teacher watches the exact line where the slip happens and corrects it on the spot, before it settles into a routine.

Because our teachers are experienced, you work with someone who explains why the angle must be in radians, not just which button to press. Lessons are taught in English, while SPM papers are set in both Malay and English, so 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 for s=rθs=r\theta and A=12r2θA=\tfrac{1}{2}r^{2}\theta?

Both formulas are derived using radian measure, where a full turn is 2π2\pi. If you leave the angle in degrees the numbers no longer match the circle, and the answer is wrong.

When a question gives degrees, convert to radians first, then substitute.

How do I convert between degrees and radians quickly?

Use 180=π180^{\circ}=\pi radians. To go from degrees to radians, multiply by π180\dfrac{\pi}{180^{\circ}}; to go from radians to degrees, multiply by 180π\dfrac{180^{\circ}}{\pi}.

The π\pi cancels neatly in one direction, leaving only numbers.

What is the difference between the arc length and the perimeter of a sector?

The arc length is only the curved edge, s=rθs=r\theta. The perimeter of the sector also includes the two straight radii, so P=rθ+2rP=r\theta+2r.

Read the question carefully, "perimeter" means all three edges.

Should I leave π\pi in my answer or use a decimal?

Follow the question. If it asks for an exact answer or an answer in terms of π\pi, leave the π\pi; if it asks for a length or area to a number of decimal places, use the π\pi key on your calculator and round only at the end.

Source:SRC-DSKP-EN

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

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