Perimeter & area (incl. compound shapes, circles, arcs & sectors) – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationCalculator5 marks

The diagram shows a shape made from rectangles. All corners are right angles.

(a) Work out the perimeter of the shape. [2]

(b) Work out the area of the shape. [3]

(a) finds both missing lengths, $10-6=4$ and $8-3=5$, or uses perimeter $=2(10+8)$M1
36 cmA1
(b) splits into two rectangles with correct dimensions, e.g. $6\times 8$ and $4\times 5$, or $6\times 3$ and $10\times 5$, or uses $10\times 8$ and $4\times 3$M1
adds (or subtracts) the two areas, e.g. $48+20$ or $80-12$M1
68 cm$^2$A1
Answer: (a) 36 cm (b) 68 cm$^2$
Examiner tip: Label every missing length on the diagram before you add – most lost marks come from leaving a side out of the perimeter.

2Grade 4–5Calculator4 marks

A circle has radius 6.5 cm.

(a) Work out the circumference of the circle. Give your answer correct to 1 decimal place. [2]

(b) Work out the area of the circle. Give your answer correct to 1 decimal place. [2]

(a) $2\times\pi\times 6.5$ or $\pi\times 13$M1
40.8 cm (accept 40.82 to 40.84 seen before rounding)A1
(b) $\pi\times 6.5^2$M1
132.7 cm$^2$ (accept 132.6 to 132.8)A1
Answer: (a) 40.8 cm (b) 132.7 cm$^2$
Examiner tip: Using the diameter in $\pi r^2$ (i.e. $\pi\times13^2$) is the most common error – check whether you've been given the radius or diameter.

3Grade 4–5Non-calculator3 marks

The shape is made from a rectangle and a semicircle. The rectangle measures 12 cm by 8 cm. The diameter of the semicircle is the 8 cm side of the rectangle.

Work out the area of the shape. Give your answer in terms of $\pi$. [3]

area of rectangle $12\times 8\;(=96)$M1
area of semicircle $\tfrac12\times\pi\times 4^2\;(=8\pi)$M1
$96+8\pi$ cm$^2$A1
Answer: $96+8\pi$ cm$^2$
Examiner tip: The radius is 4 cm, not 8 cm – halve the diameter before squaring, then halve the circle's area for the semicircle.

4Grade 4–5Calculator5 marks

The diagram shows a sector of a circle, centre O. The radius is 9 cm and the angle at the centre is 140°.

(a) Work out the arc length of the sector. Give your answer correct to 3 significant figures. [2]

(b) Work out the perimeter of the sector. [1]

(a) $\dfrac{140}{360}\times 2\times\pi\times 9$M1
22.0 cm (accept 21.99)A1
(b) 40.0 cm (accept 39.99 or 40; ft their arc length + 18)B1
(c) $\dfrac{140}{360}\times\pi\times 9^2$M1
99.0 cm$^2$ (accept 98.9 to 99.0)A1
Answer: (a) 22.0 cm (b) 40.0 cm (c) 99.0 cm$^2$
Examiner tip: The perimeter of a sector includes the two radii – adding only the arc loses the mark.

5Grade 6–7HigherCalculator4 marks

A sector of a circle has radius 8 cm and area 50 cm$^2$.

Work out the perimeter of the sector. [4]

$\dfrac{\theta}{360}\times\pi\times 8^2=50$, or fraction of circle $=\dfrac{50}{\pi\times 8^2}$, or $\tfrac12\times 8\times \text{arc}=50$M1
$\theta=89.5^\circ$ (89.52…), or fraction $0.2487\ldots$, or arc $=12.5$A1
arc length $\dfrac{\theta}{360}\times 2\times\pi\times 8$ $(=12.5)$ and adds $2\times 8$ (or their arc $+16$)M1
28.5 cm (accept 28.49 to 28.51)A1
Answer: 28.5 cm
Examiner tip: Keep the unrounded angle in your calculator – rounding θ to 90° early gives 28.6 cm and loses the accuracy mark.

6Grade 6–7HigherNon-calculator4 marks

ABCD is a square of side 10 cm. An arc of a circle, centre A, radius 10 cm, joins B to D. The region outside the arc but inside the square is shaded.

(a) Work out the area of the shaded region. Give your answer in terms of $\pi$. [2]

(b) Work out the perimeter of the shaded region. Give your answer in the form $a+b\pi$. [2]

(a) $10^2-\tfrac14\times\pi\times10^2$M1
$100-25\pi$ cm$^2$A1
(b) arc $\tfrac14\times 2\times\pi\times 10\;(=5\pi)$M1
$20+5\pi$ cmA1
Answer: (a) $100-25\pi$ cm$^2$ (b) $20+5\pi$ cm
Examiner tip: For the perimeter, include BC and CD (20 cm) as well as the arc – the radii AB and AD are not on the edge of the shaded region.

7Grade 6–7HigherCalculator4 marks

A circle has the same area as a square of side 7 cm.

Work out the circumference of the circle. Give your answer correct to 1 decimal place. [4]

area of square $=49$ and forms $\pi r^2=49$P1
$r=\sqrt{\dfrac{49}{\pi}}\;(=3.949\ldots)$P1
$2\times\pi\times$ their $r$P1
24.8 cm (accept 24.81 to 24.82 seen)A1
Answer: 24.8 cm
Examiner tip: Find the radius first – a common slip is to set the circumference equal to the square's perimeter.

8Grade 1–3FoundationNon-calculator3 marks

The diagram shows a circle, centre O.

The straight line P has both ends on the circle.

The straight line Q touches the circle at exactly one point.

R is the shaded region between line P and the circle.

Write down the mathematical name for

(a) line P [1]

(b) line Q [1]

(a) chordB1
(b) tangentB1
(c) (minor) segment Do not accept sector, arc or semicircleB1
Answer: (a) chord (b) tangent (c) segment
Examiner tip: A segment is cut off by a chord; a sector is cut off by two radii – mixing these up is the most common lost mark.

9Grade 1–3FoundationNon-calculator4 marks

The diagram shows a parallelogram.

(a) Work out the area of the parallelogram. [2]

A triangle has an area of 24 cm$^2$. The base of the triangle is 8 cm.

(b) Work out the perpendicular height of the triangle. [2]

(a) $11\times 6$M1
66 cm$^2$ Do not award M1 for $11\times 7$ (= 77)A1
(b) $\tfrac12\times 8\times h=24$ oe, e.g. $24\times 2\div 8$ or $48\div 8$M1
6 cmA1
Answer: (a) 66 cm$^2$ (b) 6 cm
Examiner tip: Use the perpendicular height, not the slanted side – $11\times 7=77$ scores zero.

10Grade 4–5Calculator4 marks

The wheel on Amira's bicycle has a diameter of 58 cm.

Amira cycles 1.5 km in a straight line.

(a) Work out the number of complete turns the wheel makes. [3]

Ben says, “A wheel with double the diameter would make half as many turns over the same distance.”

(b) Is Ben correct? Give a reason for your answer. [1]

(a) Circumference $\pi\times 58$ (= 182.2…) cmM1
Consistent units and divides: $150\,000\div 182.2\ldots$ oe (e.g. $1500\div 1.822\ldots$) (= 823.2…)M1
823 Do not accept 824 or 823.2A1
(b) Yes, with reason, e.g. doubling the diameter doubles the circumference, so each turn covers twice the distance (circumference is proportional to diameter)C1
Answer: (a) 823 (b) Yes – the circumference doubles, so each turn covers twice as far
Examiner tip: Convert km to cm before dividing, and round down – the 824th turn is not complete.

11Grade 4–5Non-calculator3 marks

The diagram shows a semicircle with diameter 10 cm.

Kai is asked to work out the perimeter of the semicircle. His answer is $5\pi$ cm.

(a) Explain the mistake Kai has made. [1]

(b) Work out the correct perimeter of the semicircle. Give your answer in terms of $\pi$. [2]

(a) He has only found the curved part (the arc) / he has not added the diameter (the straight edge of 10 cm)C1
(b) $\tfrac12\times\pi\times 10$ oe (or $5\pi$ seen) + 10M1
$10+5\pi$ cm oe (e.g. $5(2+\pi)$; accept $5\pi+10$)A1
Answer: (a) He left out the 10 cm diameter (b) $10+5\pi$ cm
Examiner tip: The perimeter of a semicircle is the arc plus the straight diameter – leaving out the diameter is the classic slip.

12Grade 6–7HigherCalculator5 marks

The diagram shows a sector of a circle, centre O.

The radius is 12 cm and the arc length is 15 cm.

(a) Work out the value of $x$. Give your answer correct to 1 decimal place. [3]

(b) Show that the area of the sector is 90 cm$^2$, correct to 2 significant figures. [2]

(a) $\dfrac{x}{360}\times 2\times\pi\times 12 = 15$ oeM1
$x = \dfrac{15\times 360}{24\pi}$ oe (= 71.61…)M1
71.6A1
(b) $\dfrac{'71.6'}{360}\times\pi\times 12^2$ oe, or $\tfrac12\times 12\times 15$M1
89.97… or 90.0… or exactly 90, and states 90 cm$^2$ (2 s.f.) (needs a value to more than 2 s.f., or exactly 90 from an unrounded angle or from $\tfrac12 rl$, with the method shown)A1
Answer: (a) $x = 71.6$ (b) area $=89.97\ldots \approx 90$ cm$^2$
Examiner tip: In a ‘show that’, show the full calculation and the value it gives before you round (e.g. 89.97… if you used x = 71.6) – a bare ‘90’ with no working earns nothing.

13Grade 8–9HigherCalculator5 marks

A, B are points on a circle, centre O, radius 10 cm.

AB is a chord of length 12 cm.

The shaded region is the minor segment cut off by AB.

Work out the area of the shaded segment. Give your answer correct to 3 significant figures. [5]

Method to find angle AOB, e.g. $\sin\left(\tfrac12 AOB\right)=\tfrac{6}{10}$ or cosine rule $\cos AOB=\dfrac{10^2+10^2-12^2}{2\times10\times10}$ (AOB = 73.7…°)P1
Sector area: $\dfrac{'73.7…'}{360}\times\pi\times 10^2$ (= 64.3…)P1
Triangle area: $\tfrac12\times 12\times 8$ (Pythagoras: $\sqrt{10^2-6^2}=8$) or $\tfrac12\times10\times10\times\sin '73.7…'$ (= 48)P1
Sector − triangle: '64.3…' − '48'P1
16.3 to 16.4 cm$^2$ (16.35…) Note: using a rounded angle of 74° gives 16.5 or 16.6 – award P1P1P1P1A0A1
Answer: 16.4 cm$^2$
Examiner tip: Keep the angle unrounded in your calculator – rounding it to 74° pushes the answer outside the accepted range.

14Grade 8–9HigherNon-calculator5 marks

A sector of a circle has radius $r$ cm and angle 120° at the centre.

The area of the sector is equal to the area of a circle with radius 6 cm.

(a) Show that $r = 6\sqrt{3}$. [3]

(b) Work out the perimeter of the sector. Give your answer in the form $a\sqrt{3}+b\sqrt{3}\,\pi$, where $a$ and $b$ are integers. [2]

(a) $\dfrac{120}{360}\times\pi r^2 = \pi\times 6^2$ oeM1
$r^2 = 108$ (from $r^2 = 3\times 36$)M1
$\sqrt{108}=\sqrt{36\times 3}=6\sqrt3$ shown with no errorsA1
(b) Arc $=\dfrac{120}{360}\times 2\times\pi\times 6\sqrt3$ (= $4\sqrt3\,\pi$)M1
$12\sqrt3+4\sqrt3\,\pi$ cm (cao; $a=12,\ b=4$)A1
Answer: (b) $12\sqrt{3}+4\sqrt{3}\,\pi$ cm
Examiner tip: Remember both radii: the perimeter needs $2\times 6\sqrt3 = 12\sqrt3$ as well as the arc.

15Grade 4–5Calculator5 marks

The diagram shows the plan of a garden in the shape of a trapezium.

The shaded flower bed is a semicircle with diameter 4 m along one side of the garden. The rest of the garden is to be covered with turf.

Turf is sold in rolls. One roll covers 5 m$^2$ and costs £14.50. Only whole rolls can be bought.

Work out the total cost of the turf needed. [5]

Area of trapezium: $\tfrac12(12+8)\times 6$ (= 60)P1
Area of semicircle: $\tfrac12\times\pi\times 2^2$ (= 6.28…) Do not award for $\tfrac12\pi\times4^2$P1
Lawn area: '60' − '6.28…' (= 53.7…)P1
'53.7…' ÷ 5 (= 10.7…) and rounds up to 11 rolls, or 11 × 14.50P1
£159.50 (cao; £159.5 acceptable)A1
Answer: £159.50
Examiner tip: You can't buy 10.74 rolls – rounding down to 10, or working out 10.74 × £14.50 (= £155.78), loses the last two marks: always round the number of rolls UP to 11.

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