Edexcel International GCSE Maths Area, Arcs and Sectors Past Paper Questions and Solutions (4MA1 Higher)

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Edexcel International GCSE Maths A (4MA1) · Higher · Geometry and Trigonometry

Area, Perimeter, Arcs and Sectors: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every area, perimeter, arcs and sectors question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 20 questions worth 88 marks. Questions run from easier to harder. Try each one first, then click “Show solution” for a step-by-step method with the reason for every step. All answers are checked against the official Pearson mark schemes.

14 marksHigherJanuary 2020 · Paper 1H (4MA1/1H) · Q3

The diagram shows a shape made of rectangles. The top edge is 12 cm and the left edge is 9 cm. On the right, the shape steps down 6 cm and then out x cm before the bottom right corner. The shape has area 129 cm2. Work out the value of x.

12 cm9 cm6 cmx cm
Show solution
  1. Split the shape into the 12 × 9 rectangle and the strip on the right.
  2. The strip is x long. Its height is 9 − 6 = 3 cm (the step is 6 cm down from the top of the 9 cm side).
  3. Area: 12 × 9 + 3x = 129, so 108 + 3x = 129
  4. 3x = 21

Answer: x = 7

24 marksHigherJune 2021 · Paper 1H (4MA1/1H) · Q3

The diagram shows the shape ABCDE. AB = x cm, BC = 8 cm, AE = 14 cm and ED = 13 cm. Angles A, B and E are right angles. The area of the shape is 91.8 cm2. Work out the value of x.

ABCDEx cm8 cm14 cm13 cm
Show solution
  1. Split the shape with a vertical line down from B to ED. This makes a rectangle on the left and a right-angled triangle below C.
  2. Rectangle: x wide and 14 tall, area = 14x
  3. Triangle: its base is 13 − x and its height is 14 − 8 = 6, so its area = ½ × (13 − x) × 6 = 39 − 3x
  4. Total area: 14x + 39 − 3x = 11x + 39 = 91.8
  5. 11x = 52.8

Answer: x = 4.8

35 marksHigherNovember 2021 · Paper 1H (4MA1/1H) · Q4

The diagram shows the front of a wooden door, 2 m tall and 0.75 m wide, with a semicircular glass window of diameter 0.5 m. Julie wants to apply 2 coats of wood varnish to the front of the door (the wooden part, not the window). 250 millilitres of wood varnish covers 4 m2 of the wood. Work out how many millilitres of wood varnish Julie will need. Give your answer correct to the nearest millilitre.

0.5 m2 m0.75 m
Show solution
  1. Area of the door = 2 × 0.75 = 1.5 m2
  2. The window is a semicircle with radius 0.25 m: area = ½ × π × 0.252 = 0.098 17… m2
  3. Area of wood = 1.5 − 0.098 17… = 1.401 82… m2
  4. Two coats means covering this area twice: 2 × 1.401 82… = 2.803 65… m2
  5. 250 ml covers 4 m2, so 1 m2 needs 250 ÷ 4 = 62.5 ml
  6. Varnish needed = 2.803 65… × 62.5 = 175.228…

Answer: 175 millilitres

45 marksHigherJune 2023 · Paper 1HR (4MA1/1HR) · Q4

The diagram shows a shape made up of three semicircles, enclosing a right-angled triangle ABC. AB, BC and CA are each the diameter of a semicircle. BC = CA = 6 cm. Work out the perimeter of the shape. Give your answer correct to one decimal place.

ACB6 cm6 cm
Show solution
  1. The perimeter is made of the three curved edges only (the triangle’s sides are inside the shape).
  2. A semicircle’s arc = ½ × π × diameter.
  3. Find AB with Pythagoras: AB2 = 62 + 62 = 72, so AB = √72 = 8.485… cm
  4. Arcs on CA and BC: ½ × π × 6 = 3π each, so 6π together.
  5. Arc on AB: ½ × π × 8.485… = 13.328… cm
  6. Perimeter = 6π + 13.328… = 18.849… + 13.328… = 32.178…

Answer: 32.2 cm

53 marksHigherJanuary 2020 · Paper 1HR (4MA1/1HR) · Q5

The diagram shows Yuen’s garden. The garden is in the shape of a semicircle of radius 7.2 m. Yuen is going to cover his garden with grass seed. Yuen has 12 boxes of grass seed. Each box contains enough seed to cover 6 m2 of the garden. Has Yuen enough grass seed for his garden? Show your working clearly.

Show solution
  1. Area of the semicircle = ½ × π × 7.22 = 81.43… m2
  2. 12 boxes cover 12 × 6 = 72 m2
  3. 72 m2 is less than 81.43 m2.

Answer: No, 12 boxes cover 72 m2 but the garden is 81.4 m2

63 marksHigherJanuary 2021 · Paper 1H (4MA1/1H) · Q8

The region, shown shaded in the diagram, is a path. The boundary of the path is formed by two semicircles, with the same centre O, and two straight lines. The inner semicircle has a radius of 7 metres. The path has a width of 2 metres. Work out the perimeter of the path. Give your answer correct to one decimal place.

O7 m2 m
Show solution
  1. The outer radius is 7 + 2 = 9 m.
  2. Outer arc = ½ × 2π × 9 = 9π. Inner arc = ½ × 2π × 7 = 7π
  3. The two straight ends are each 2 m.
  4. Perimeter = 9π + 7π + 2 + 2 = 16π + 4 = 54.26…

Answer: 54.3 m

75 marksHigherJanuary 2023 · Paper 1H (4MA1/1H) · Q8

The diagram shows a triangle ABC inside a semicircle. A, B and C are points on the semicircle. AB is the diameter of the semicircle. Angle ACB = 90°, angle BAC = 50° and AC = 18 cm. Work out the perimeter of the semicircle. Give your answer correct to 2 significant figures.

ABC18 cm50°
Show solution
  1. In right-angled triangle ABC, AB is the hypotenuse and AC is adjacent to the 50° angle: cos 50° = 18⁄AB
  2. AB = 18⁄cos 50° = 28.003… cm, so the radius = 14.001… cm
  3. The perimeter of a semicircle = half the circumference + the diameter = π × 14.001… + 28.003…
  4. = 43.98… + 28.00… = 71.99…

Answer: 72 cm

86 marksHigherJune 2024 · Paper 1HR (4MA1/1HR) · Q8

The diagram shows a trapezium ABCD. DAB and ADC are right angles. AD = 15 cm and DC = 14 cm. The area of the trapezium is 360 cm2. Work out the perimeter of the trapezium.

ABCD15 cm14 cm
Show solution
  1. Area of a trapezium = ½(a + b)h, where a and b are the parallel sides AB and DC and the height is AD = 15.
  2. ½(AB + 14) × 15 = 360, so AB + 14 = 360 × 2 ÷ 15 = 48 and AB = 34 cm
  3. To find BC, draw a vertical line down from B to make a right-angled triangle. Its horizontal side is 34 − 14 = 20 cm and its vertical side is 15 cm.
  4. Pythagoras: BC2 = 202 + 152 = 400 + 225 = 625, so BC = 25 cm
  5. Perimeter = AB + BC + CD + DA = 34 + 25 + 14 + 15

Answer: 88 cm

95 marksHigherJanuary 2021 · Paper 1HR (4MA1/1HR) · Q10

The diagram shows one face of a wall. This face is in the shape of a pentagon with exactly one line of symmetry: a 10 m by 6 m rectangle with two sloping edges of 7 m on top. Omondi is going to paint this face of the wall once. He has to buy all the paint that he needs to use. The paint in each tin will cover 16 m2 of the face of the wall. Work out the least number of tins of paint Omondi will need to buy. Show your working clearly.

7 m7 m6 m6 m10 m
Show solution
  1. Split the face into a rectangle and an isosceles triangle.
  2. Rectangle = 10 × 6 = 60 m2
  3. Triangle: base 10, sloping sides 7. Its height splits the base into 5 and 5: height = √72 − 52 = √24 = 4.898… m. Area = ½ × 10 × 4.898… = 24.49… m2
  4. Total area = 60 + 24.49… = 84.49… m2
  5. Tins = 84.49… ÷ 16 = 5.28…. He must buy whole tins and 5 is not enough, so round up.

Answer: 6 tins

102 marksHigherJanuary 2022 · Paper 1HR (4MA1/1HR) · Q11

The diagram shows sector OPQ of a circle, centre O. OP = OQ = 7.1 cm and angle POQ = 110°. Calculate the area of sector OPQ. Give your answer correct to one decimal place.

Show solution
  1. Area = 110⁄360 × π × 7.12 = 48.39…

Answer: 48.4 cm2

112 marksHigherJune 2025 · Paper 1H (4MA1/1H) · Q11

OABC is a sector of a circle, centre O. Angle AOC = 140° and OA = OC = 16 cm. Calculate the area of the sector. Give your answer correct to 3 significant figures.

140°OABC16 cm16 cm
Show solution
  1. Area of a sector = angle⁄360 × πr2
  2. 140⁄360 × π × 162 = 140⁄360 × 256π = 312.76…

Answer: 313 cm2

122 marksHigherJune 2021 · Paper 1H (4MA1/1H) · Q12

The diagram shows a sector of a circle with radius 7 cm and angle 50°. Work out the length of the arc of the sector. Give your answer correct to one decimal place.

Show solution
  1. Arc length = angle⁄360 × 2πr = 50⁄360 × 2 × π × 7 = 6.108…

Answer: 6.1 cm

134 marksHigherJanuary 2022 · Paper 1H (4MA1/1H) · Q16

The diagram shows a circle with centre O. A, B and C are points on the circle so that the length of the arc ABC is 5 cm. Given that angle AOC = 55°, work out the area of the circle. Give your answer correct to one decimal place.

Show solution
  1. Arc length = 55⁄360 × 2πr = 5
  2. 2πr = 5 × 360⁄55 = 32.727…, so r = 5.2087… cm
  3. Area = πr2 = π × 5.2087…2 = 85.23…

Answer: 85.2 cm2

145 marksHigherJune 2022 · Paper 1H (4MA1/1H) · Q19

OAB is a sector S of a circle with centre O and radius (r + 7) metres. Angle AOB = 45°. A circle C has radius (r − 2) metres. The area of sector S is twice the area of circle C. Find the value of r. Show your working clearly.

OAB45°(r + 7) m
Show solution
  1. Area of sector S = 45⁄360 × π(r + 7)2 = 1⁄8π(r + 7)2
  2. Area of circle C = π(r − 2)2
  3. 1⁄8π(r + 7)2 = 2π(r − 2)2. Cancel π and multiply by 8: (r + 7)2 = 16(r − 2)2
  4. Square root both sides (lengths are positive): r + 7 = 4(r − 2)
  5. r + 7 = 4r − 8, so 15 = 3r

Answer: r = 5

Tip: Expanding instead gives 15r2 − 78r + 15 = 0, with roots 5 and 0.2. Reject 0.2 because the radius r − 2 would be negative.
154 marksHigherJune 2024 · Paper 1H (4MA1/1H) · Q20

The diagram shows a sector OABC of a circle centre O. Angle AOC = 60°. The area of the shaded segment ABC is 38 cm2. Work out the perimeter of the shaded segment ABC. Give your answer correct to one decimal place.

O60°ABC
Show solution
  1. Segment area = sector area − triangle area. With radius r: sector = 60⁄360 × πr2 and triangle AOC = ½r2 sin 60°.
  2. 60⁄360πr2 − ½r2 sin 60° = 38
  3. Factorise: r2(0.52359… − 0.43301…) = 38, so 0.090586…r2 = 38
  4. r2 = 419.49…, so r = 20.481… cm
  5. The perimeter of the segment is the arc ABC plus the chord AC.
  6. Arc = 60⁄360 × 2πr = 21.447… cm
  7. Triangle AOC has OA = OC and a 60° angle between them, so it is equilateral and the chord AC = r = 20.481… cm
  8. Perimeter = 21.447… + 20.481… = 41.92…

Answer: 41.9 cm

165 marksHigherJanuary 2023 · Paper 1H (4MA1/1H) · Q21

The diagram shows the cross section of a circular water pipe of radius 4.8 cm. OABC is a sector of the circle, centre O, with angle AOC = 72°. The shaded segment (between chord AC and arc ABC) represents the water flowing in the pipe. The water flows at 14 cm/s in the pipe. Work out the volume of water that has flowed through the pipe in 3 minutes. Give your answer in cm3 correct to 3 significant figures.

OACB4.8 cm72°
Show solution
  1. Area of the segment = sector − triangle = 72⁄360 × π × 4.82 − ½ × 4.82 × sin 72° = 14.476… − 10.956… = 3.5202… cm2
  2. In 1 second, a column of water 14 cm long passes through, so the volume per second = 3.5202… × 14 cm3
  3. 3 minutes = 180 seconds: volume = 3.5202… × 14 × 180 = 8871.1…

Answer: 8870 cm3

176 marksHigherJune 2020 · Paper 1HR (4MA1/1HR) · Q22

The diagram shows a regular octagon ABCDEFGH. Each side of the octagon has length 10 cm. Find the area of the shaded region ACDEH. Give your answer correct to the nearest cm2.

BCDEFGHA10 cm
Show solution
  1. Split the shaded shape into the rectangle ADEH and the triangle ACD.
  2. The corner pieces of the octagon are right-angled isosceles triangles with hypotenuse 10, so each short side is 10 ÷ √2 = 7.071… cm.
  3. Width of the octagon AD = 7.071… + 10 + 7.071… = 24.142… cm. Rectangle ADEH = 24.142… × 10 = 241.42… cm2
  4. Triangle ACD: base AD = 24.142… and its height is how far C is above AD, which is 7.071… cm. Area = ½ × 24.142… × 7.071… = 85.355… cm2
  5. Total = 241.42… + 85.355… = 326.78…

Answer: 327 cm2

186 marksHigherJune 2023 · Paper 1H (4MA1/1H) · Q22

The diagram shows two circles with centre O and a regular pentagon ABCDE. A, B, C, D and E are points on the larger circle. The pentagon has sides of length 8 cm. The diagram is shaded such that shaded area = unshaded area. (The shaded parts are the small circle and the five segments between the pentagon and the large circle.) Work out the radius of the smaller circle. Give your answer correct to 3 significant figures.

OBCDEA
Show solution
  1. First find the radius R of the large circle. Join O to two neighbouring vertices: this makes an isosceles triangle with angle 360° ÷ 5 = 72° at O and base 8 cm.
  2. Half of that triangle is right-angled with angle 36° and opposite side 4: R = 4⁄sin 36° = 6.8052… cm
  3. Area of the pentagon = 5 triangles = 5 × ½R2 sin 72° = 110.11… cm2
  4. Area of the large circle = πR2 = 145.49… cm2
  5. Let the small radius be r. Shaded = πr2 + (145.49… − 110.11…). Unshaded = 110.11… − πr2.
  6. Set them equal: πr2 + 35.379… = 110.11… − πr2, so 2πr2 = 74.731… and r2 = 11.893…
  7. r = √11.893… = 3.4487…

Answer: 3.45 cm

196 marksHigherJune 2021 · Paper 1H (4MA1/1H) · Q25

The diagram shows two circles such that the region R, shown shaded, is the region common to both circles. One of the circles has centre O and radius 5 cm. The other circle has centre P and radius 4 cm. Angle AOB = 50°, where A and B are the points where the circles cross. Calculate the area of region R. Give your answer correct to 3 significant figures.

OPABR50°5 cm4 cm
Show solution
  1. R is made of two segments either side of the common chord AB: a segment of the circle centre O and a segment of the circle centre P.
  2. Chord AB: triangle AOB is isosceles with apex 50°, so AB = 2 × 5 × sin 25° = 4.2261… cm
  3. Angle APB: in triangle APB, half of AB is opposite half the angle: sin(½ APB) = 2.1130…⁄4 = 0.5282…, so angle APB = 2 × 31.888…° = 63.777…°
  4. Segment of circle O = 50⁄360π × 52 − ½ × 52 × sin 50° = 10.908… − 9.575… = 1.3327… cm2
  5. Segment of circle P = 63.777…⁄360π × 42 − ½ × 42 × sin 63.777…° = 8.904… − 7.176… = 1.7283… cm2
  6. Area of R = 1.3327… + 1.7283… = 3.061…

Answer: 3.06 cm2

206 marksHigherJune 2025 · Paper 1HR (4MA1/1HR) · Q25

BAC is a sector of a circle, centre A, with angle BAC = 40°. BCD is a sector of a circle, centre C, with angle BCD = 130°. The area of the shaded segment (between chord BC and arc BC of the first sector) is 28 cm2. Find the length of the arc BD. Give your answer correct to 3 significant figures.

ABCD
Show solution
  1. Let the radius of sector BAC be R, so AB = AC = R.
  2. Segment area = sector area − triangle area: 40⁄360πR2 − ½R2 sin 40° = 28
  3. R2(0.34906… − 0.32139…) = 28, so 0.027674…R2 = 28, giving R2 = 1011.8… and R = 31.809… cm
  4. The second sector has centre C and radius CB, so we need the chord BC. Triangle ABC is isosceles with apex angle 40°, so BC = 2R sin 20° = 21.759… cm (or use the cosine rule).
  5. Arc BD = 130⁄360 × 2π × BC = 130⁄360 × 2π × 21.759… = 49.369…

Answer: 49.4 cm

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