Hard GCSE Arcs and sectors Questions

Challenging, exam-style GCSE Arcs and sectors questions with worked solutions. Stretch yourself on the hardest perimeter of a sector, arc plus two radii, exact answers in terms of pi, arc length problems.

perimeter of a sectorarc plus two radiiexact answers in terms of piarc lengthfraction of a circleworking backwards
GCSE Foundation34 questionsStep-by-step solutions
Question 1
5 markschallenging
The diagram shows a running track made from a rectangle of length 8080 m and width 5050 m, with a semicircle on each of the two shorter sides. Work out the perimeter of the track. Give your answer in terms of π\pi.
Show worked solution

Worked solution

  1. Say what the perimeter of the track is made of.

    perimeter=2×length+two semicircular ends\text{perimeter} = 2 \times \text{length} + \text{two semicircular ends}

    Going once round the outside you travel along the two long straight sides and round the two curved ends.

  2. Put the two semicircular ends together.

    12 circle+12 circle=1 whole circle\tfrac{1}{2}\text{ circle} + \tfrac{1}{2}\text{ circle} = 1 \text{ whole circle}

    The two ends are identical semicircles, so together they make one complete circle. That is much quicker than working each one out.

  3. Work out the radius of each semicircular end.

    r=502=25r = \frac{50}{2} = 25

    The width of the rectangle is the DIAMETER of each end, so the radius is half of it, 2525 m.

  4. Work out the circumference of that circle.

    C=2×π×25=50πC = 2 \times \pi \times 25 = 50\pi

    The two curved ends together are 50π50\pi m.

  5. Work out one semicircular end on its own as a check.

    12×50π=25π\tfrac{1}{2} \times 50\pi = 25\pi

    Each end is 25π25\pi m, and two of them give 50π50\pi m, as expected.

  6. Work out the total length of the straight sides.

    2×80=1602 \times 80 = 160

    There are two straight sides, each 8080 m long.

  7. Add the curved part and the straight part.

    P=50π+160=50π+160P = 50\pi + 160 = 50\pi + 160

    The π\pi term and the number term cannot be combined, so the perimeter is 50π+16050\pi + 160 m.

  8. Check which lengths are actually on the outside.

    50 is not part of the perimeter50 \text{ is not part of the perimeter}

    The two short sides of the rectangle are inside the track, not on its edge, so the width is never added on.

  9. Note the most common error here.

    2×π×50≠50π2 \times \pi \times 50 \neq 50\pi

    Using the width 5050 m as the radius instead of the diameter would double the curved part. The radius is half the width.

  10. Compare the curved part with the straight part.

    50π≈157<16050\pi \approx 157 < 160

    The two curved ends come to about 157157 m, and the two straights come to 160160 m, so the straights are the longer part of one lap.

  11. Check the size of the answer using an approximate value for pi.

    50π+160≈50×3.14+160=31750\pi + 160 \approx 50 \times 3.14 + 160 = 317

    Taking π≈3.14\pi \approx 3.14 gives the perimeter as roughly 317317 m, which is a sensible size.

  12. Check what happens if the track is made longer.

    extra length⇒extra 2×length\text{extra length} \Rightarrow \text{extra } 2 \times \text{length}

    Stretching the rectangle changes only the straight part; the curved ends still make one circle of radius 2525 m, so only the number term would change.

  13. Check the width is used only through the radius.

    r=502=25⇒C=50πr = \frac{50}{2} = 25 \Rightarrow C = 50\pi

    The width enters the answer only by fixing the radius of the ends; it is never added on as a straight edge.

  14. Check the units of the answer.

    units:m\text{units}: \text{m}

    A perimeter is a length, so the answer is in m.

  15. State the perimeter of the track.

    P=(50π+160) mP = (50\pi + 160)\text{ m}

    The perimeter of the track is 50π+16050\pi + 160 m.

Answer
P=(50π+160) mP = (50\pi + 160)\text{ m}
Question 2
6 markschallenging
The diagram shows a running track made from a rectangle of length 5050 m and width 3030 m, with a semicircle on each of the two shorter sides. Work out the area of the track. Give your answer in terms of π\pi.
Show worked solution

Worked solution

  1. Split the track into shapes you know.

    area=rectangle+two semicircles\text{area} = \text{rectangle} + \text{two semicircles}

    The track is a rectangle with a semicircle stuck on each of the two shorter sides.

  2. Put the two semicircles together.

    12 circle+12 circle=1 whole circle\tfrac{1}{2}\text{ circle} + \tfrac{1}{2}\text{ circle} = 1 \text{ whole circle}

    The two semicircles are identical, so together they make one complete circle.

  3. Work out the radius of each semicircle.

    r=302=15r = \frac{30}{2} = 15

    The width of the rectangle is the diameter of each semicircle, so the radius is 1515 m.

  4. Work out the area of the rectangle.

    50×30=150050 \times 30 = 1500

    The rectangle is 5050 m by 3030 m, so its area is 15001500 square m.

  5. Work out the area of the full circle.

    π×152=225π\pi \times 15^2 = 225\pi

    The two semicircular ends together have the area of one circle of radius 1515 m, which is 225π225\pi square m.

  6. Work out the area of one semicircle as a check.

    12×225π=112.5π\tfrac{1}{2} \times 225\pi = 112.5\pi

    Each end has area 112.5π112.5\pi square m, and two of them give 225π225\pi square m.

  7. Add the rectangle and the circle.

    A=1500+225π=225π+1500A = 1500 + 225\pi = 225\pi + 1500

    The total area is 225π+1500225\pi + 1500 square m, left exactly in terms of π\pi.

  8. Check the answer is bigger than the rectangle alone.

    225π+1500>1500225\pi + 1500 > 1500

    The two ends add area, so the total must be more than the 15001500 square m of the rectangle.

  9. Note the most common error here.

    π×302≠π×152\pi \times 30^2 \neq \pi \times 15^2

    The area formula uses the RADIUS. Using the width 3030 m would make the curved part four times too big.

  10. Compare the curved part with the rectangle.

    225π≈706.5<1500225\pi \approx 706.5 < 1500

    The two ends come to about 706.5706.5 square m, which is less than the 15001500 square m of the rectangle.

  11. Check the width is used only through the radius.

    r=302=15⇒πr2=225πr = \frac{30}{2} = 15 \Rightarrow \pi r^2 = 225\pi

    The width fixes the radius of the two ends, and nothing else in the area depends on it apart from the rectangle.

  12. Check the size of the answer using an approximate value for pi.

    225π+1500≈225×3.14+1500=2206.5225\pi + 1500 \approx 225 \times 3.14 + 1500 = 2206.5

    Taking π≈3.14\pi \approx 3.14 gives the area as roughly 2206.52206.5 square m, which is a sensible size.

  13. Note that the perimeter is a different calculation.

    P=30π+100P = 30\pi + 100

    The perimeter uses the CIRCUMFERENCE of the end circle, not its area, so the two questions must not be mixed up.

  14. Check the units of the answer.

    units:m2\text{units}: \text{m}^2

    An area is measured in square units, so the answer is in m2\text{m}^2.

  15. State the area of the track.

    A=(225π+1500) m2A = (225\pi + 1500)\text{ m}^2

    The area of the track is 225π+1500225\pi + 1500 square m.

Answer
A=(225π+1500) m2A = (225\pi + 1500)\text{ m}^2
Question 3
5 markschallenging
The diagram shows a running track made from a rectangle of length 6060 m and width 4040 m, with a semicircle on each of the two shorter sides. Work out the perimeter of the track. Give your answer in terms of π\pi.
Show worked solution

Worked solution

  1. Say what the perimeter of the track is made of.

    perimeter=2×length+two semicircular ends\text{perimeter} = 2 \times \text{length} + \text{two semicircular ends}

    Going once round the outside you travel along the two long straight sides and round the two curved ends.

  2. Put the two semicircular ends together.

    12 circle+12 circle=1 whole circle\tfrac{1}{2}\text{ circle} + \tfrac{1}{2}\text{ circle} = 1 \text{ whole circle}

    The two ends are identical semicircles, so together they make one complete circle. That is much quicker than working each one out.

  3. Work out the radius of each semicircular end.

    r=402=20r = \frac{40}{2} = 20

    The width of the rectangle is the DIAMETER of each end, so the radius is half of it, 2020 m.

  4. Work out the circumference of that circle.

    C=2×π×20=40πC = 2 \times \pi \times 20 = 40\pi

    The two curved ends together are 40π40\pi m.

  5. Work out one semicircular end on its own as a check.

    12×40π=20π\tfrac{1}{2} \times 40\pi = 20\pi

    Each end is 20π20\pi m, and two of them give 40π40\pi m, as expected.

  6. Work out the total length of the straight sides.

    2×60=1202 \times 60 = 120

    There are two straight sides, each 6060 m long.

  7. Add the curved part and the straight part.

    P=40π+120=40π+120P = 40\pi + 120 = 40\pi + 120

    The π\pi term and the number term cannot be combined, so the perimeter is 40π+12040\pi + 120 m.

  8. Check which lengths are actually on the outside.

    40 is not part of the perimeter40 \text{ is not part of the perimeter}

    The two short sides of the rectangle are inside the track, not on its edge, so the width is never added on.

  9. Note the most common error here.

    2×π×40≠40π2 \times \pi \times 40 \neq 40\pi

    Using the width 4040 m as the radius instead of the diameter would double the curved part. The radius is half the width.

  10. Compare the curved part with the straight part.

    40π≈125.6>12040\pi \approx 125.6 > 120

    The two curved ends come to about 125.6125.6 m, and the two straights come to 120120 m, so the curved ends are the longer part of one lap.

  11. Check the size of the answer using an approximate value for pi.

    40π+120≈40×3.14+120=245.640\pi + 120 \approx 40 \times 3.14 + 120 = 245.6

    Taking π≈3.14\pi \approx 3.14 gives the perimeter as roughly 245.6245.6 m, which is a sensible size.

  12. Check what happens if the track is made longer.

    extra length⇒extra 2×length\text{extra length} \Rightarrow \text{extra } 2 \times \text{length}

    Stretching the rectangle changes only the straight part; the curved ends still make one circle of radius 2020 m, so only the number term would change.

  13. Check the width is used only through the radius.

    r=402=20⇒C=40πr = \frac{40}{2} = 20 \Rightarrow C = 40\pi

    The width enters the answer only by fixing the radius of the ends; it is never added on as a straight edge.

  14. Check the units of the answer.

    units:m\text{units}: \text{m}

    A perimeter is a length, so the answer is in m.

  15. State the perimeter of the track.

    P=(40π+120) mP = (40\pi + 120)\text{ m}

    The perimeter of the track is 40π+12040\pi + 120 m.

Answer
P=(40π+120) mP = (40\pi + 120)\text{ m}
Question 4
6 markschallenging
A sector of a circle has arc length 12π12\pi cm and area 90π90\pi cm2^2. Work out the radius of the circle.
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Worked solution

  1. Write down the formula for the arc length of a sector.

    arc length=θ360×2πr\text{arc length} = \frac{\theta}{360} \times 2\pi r

    An arc is part of the circumference. A sector with an angle of θ\theta at the centre takes up θ360\frac{\theta}{360} of the full turn, so its arc takes up the same fraction of the circumference 2πr2\pi r.

  2. Write down the formula for the area of a sector.

    area=θ360×πr2\text{area} = \frac{\theta}{360} \times \pi r^2

    A sector is part of the whole circle. It takes up θ360\frac{\theta}{360} of the full turn, so it takes up the same fraction of the area πr2\pi r^2.

  3. Divide the area of the sector by its arc length.

    areaarc=θ360×πr2θ360×2πr\frac{\text{area}}{\text{arc}} = \frac{\frac{\theta}{360} \times \pi r^2}{\frac{\theta}{360} \times 2\pi r}

    Both formulas contain the same factor θ360×π\frac{\theta}{360} \times \pi, so dividing one by the other cancels the unknown angle completely.

  4. Simplify the division.

    areaarc=r2\frac{\text{area}}{\text{arc}} = \frac{r}{2}

    After cancelling, r22r=r2\frac{r^2}{2r} = \frac{r}{2}, so the area divided by the arc length is always half the radius.

  5. Rearrange to make the radius the subject.

    r=2×areaarcr = \frac{2 \times \text{area}}{\text{arc}}

    Multiplying both sides by 22 gives the radius directly from the two lengths in the question.

  6. Substitute the values from the question.

    r=2×90π12πr = \frac{2 \times 90\pi}{12\pi}

    The area is 90π90\pi square cm and the arc is 12π12\pi cm.

  7. Work out the radius.

    r=18012=15r = \frac{180}{12} = 15

    The π\pi cancels, leaving r=15r = 15 cm.

  8. Work out the circumference of the whole circle.

    C=2πr=2×π×15=30πC = 2\pi r = 2 \times \pi \times 15 = 30\pi

    The circumference of a circle of radius 1515 cm is 2πr2\pi r, which is 30π30\pi cm.

  9. Write the arc as a fraction of the circumference.

    θ360=12π30π=25\frac{\theta}{360} = \frac{12\pi}{30\pi} = \frac{2}{5}

    Now the radius is known, the arc can be compared with the circumference; the π\pi cancels again.

  10. Multiply by 360 to get the angle.

    θ=25×360=144\theta = \frac{2}{5} \times 360 = 144

    The sector is 25\frac{2}{5} of the circle, so its angle is 144∘144^\circ.

  11. Check the area with this angle and radius.

    25×π×152=90π\frac{2}{5} \times \pi \times 15^2 = 90\pi

    The area comes out as 90π90\pi square cm, exactly as the question says.

  12. Check the arc with this angle and radius.

    25×2×π×15=12π\frac{2}{5} \times 2 \times \pi \times 15 = 12\pi

    The arc comes out as 12π12\pi cm, so both pieces of information are satisfied.

  13. Check both answers are sensible.

    15>0 and 0<144<36015 > 0 \text{ and } 0 < 144 < 360

    The radius is a positive length and the angle is less than a full turn.

  14. Check the size of the answer using an approximate value for pi.

    90π≈90×3.14=282.690\pi \approx 90 \times 3.14 = 282.6

    Taking π≈3.14\pi \approx 3.14 gives the area as roughly 282.6282.6 square cm, which is a sensible size.

  15. State the radius of the circle.

    r=15 cmr = 15\text{ cm}

    The radius of the circle is 1515 cm.

Answer
r=15 cmr = 15\text{ cm}
Question 5
6 markschallenging
A sector of a circle has arc length 5π5\pi cm and area 25π25\pi cm2^2. Work out the size of the angle of the sector at the centre.
Show worked solution

Worked solution

  1. Write down the formula for the arc length of a sector.

    arc length=θ360×2πr\text{arc length} = \frac{\theta}{360} \times 2\pi r

    An arc is part of the circumference. A sector with an angle of θ\theta at the centre takes up θ360\frac{\theta}{360} of the full turn, so its arc takes up the same fraction of the circumference 2πr2\pi r.

  2. Write down the formula for the area of a sector.

    area=θ360×πr2\text{area} = \frac{\theta}{360} \times \pi r^2

    A sector is part of the whole circle. It takes up θ360\frac{\theta}{360} of the full turn, so it takes up the same fraction of the area πr2\pi r^2.

  3. Divide the area of the sector by its arc length.

    areaarc=θ360×πr2θ360×2πr\frac{\text{area}}{\text{arc}} = \frac{\frac{\theta}{360} \times \pi r^2}{\frac{\theta}{360} \times 2\pi r}

    Both formulas contain the same factor θ360×π\frac{\theta}{360} \times \pi, so dividing one by the other cancels the unknown angle completely.

  4. Simplify the division.

    areaarc=r2\frac{\text{area}}{\text{arc}} = \frac{r}{2}

    After cancelling, r22r=r2\frac{r^2}{2r} = \frac{r}{2}, so the area divided by the arc length is always half the radius.

  5. Rearrange to make the radius the subject.

    r=2×areaarcr = \frac{2 \times \text{area}}{\text{arc}}

    Multiplying both sides by 22 gives the radius directly from the two lengths in the question.

  6. Substitute the values from the question.

    r=2×25π5πr = \frac{2 \times 25\pi}{5\pi}

    The area is 25π25\pi square cm and the arc is 5π5\pi cm.

  7. Work out the radius.

    r=505=10r = \frac{50}{5} = 10

    The π\pi cancels, leaving r=10r = 10 cm.

  8. Work out the circumference of the whole circle.

    C=2πr=2×π×10=20πC = 2\pi r = 2 \times \pi \times 10 = 20\pi

    The circumference of a circle of radius 1010 cm is 2πr2\pi r, which is 20π20\pi cm.

  9. Write the arc as a fraction of the circumference.

    θ360=5π20π=14\frac{\theta}{360} = \frac{5\pi}{20\pi} = \frac{1}{4}

    Now the radius is known, the arc can be compared with the circumference; the π\pi cancels again.

  10. Multiply by 360 to get the angle.

    θ=14×360=90\theta = \frac{1}{4} \times 360 = 90

    The sector is 14\frac{1}{4} of the circle, so its angle is 90∘90^\circ.

  11. Check the area with this angle and radius.

    14×π×102=25π\frac{1}{4} \times \pi \times 10^2 = 25\pi

    The area comes out as 25π25\pi square cm, exactly as the question says.

  12. Check the arc with this angle and radius.

    14×2×π×10=5π\frac{1}{4} \times 2 \times \pi \times 10 = 5\pi

    The arc comes out as 5π5\pi cm, so both pieces of information are satisfied.

  13. Check both answers are sensible.

    10>0 and 0<90<36010 > 0 \text{ and } 0 < 90 < 360

    The radius is a positive length and the angle is less than a full turn.

  14. Check the size of the answer using an approximate value for pi.

    25π≈25×3.14=78.525\pi \approx 25 \times 3.14 = 78.5

    Taking π≈3.14\pi \approx 3.14 gives the area as roughly 78.578.5 square cm, which is a sensible size.

  15. State the angle of the sector.

    θ=90∘\theta = 90^\circ

    The angle at the centre is 90∘90^\circ.

Answer
θ=90∘\theta = 90^\circ

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