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Cambridge IGCSE Mathematics 0580 · Core and Extended · Topic 5

Mensuration

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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →

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+ Add to test✓ In your test1IGMC-045-00-01Perimeter and area of plane shapes Paper CEasy4 marks

Calculator

A rectangular vegetable garden is 12.5 m long and 8.4 m wide.

(a)

Calculate the perimeter of the garden.

(2)

(b)

Calculate the area of the garden.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × (12.5 + 8.4) or 12.5 + 8.4 + 12.5 + 8.4	✓ 1
A1 41.8 m	✓ 1
Part (b)
M1 12.5 × 8.4	✓ 1
A1 105 m²	✓ 1

Common error: Perimeter is the distance around the edge, so all four sides are added. Adding just the length and the width gives only half the perimeter.

+ Add to test✓ In your test2IGMC-045-00-02Perimeter and area of plane shapes Paper CEasy4 marks

No calculator

Triangle A has a base of 9 cm and a perpendicular height of 6.4 cm.

(a)

Work out the area of triangle A.

(2)

(b)

Triangle B has the same area as triangle A. The base of triangle B is 12 cm.
Work out the perpendicular height of triangle B.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{1}{2}$ × 9 × 6.4	✓ 1
A1 28.8 cm²	✓ 1
Part (b)
M1 $\frac{1}{2}$ × 12 × h = 28.8 or h = 28.8 × 2 ÷ 12	✓ 1
A1 4.8 cm	✓ 1

Common error: Forgetting the $\frac{1}{2}$ in the triangle area formula doubles the answer. A triangle is half of a rectangle with the same base and height.

+ Add to test✓ In your test3IGMC-045-00-03Perimeter and area of plane shapes Paper CEasy4 marks

No calculator

A floor tile is in the shape of a trapezium. Its parallel sides are 7 cm and 13 cm long and the perpendicular distance between them is 5 cm.

(a)

Work out the area of one tile.

(2)

(b)

The tiles fit together with no gaps.
Work out how many tiles are needed to cover an area of 1 m².

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{1}{2}$ × (7 + 13) × 5	✓ 1
A1 50 cm²	✓ 1
Part (b)
B1 1 m² = 10 000 cm²	✓ 1
A1 200 tiles	✓ 1

Common error: 1 m² is 100 × 100 = 10 000 cm², not 100 cm². Area conversions use the square of the length conversion.

+ Add to test✓ In your test4IGMC-045-00-04Perimeter and area of plane shapes Paper CEasy4 marks

Calculator

A circular coaster has a radius of 6 cm.

(a)

Calculate the circumference of the coaster.

(2)

(b)

Calculate the area of the coaster.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × π × 6 or π × 12	✓ 1
A1 37.7 cm	✓ 1
Part (b)
M1 π × 6²	✓ 1
A1 113 cm²	✓ 1

Common error: In πr² only the radius is squared. Working out (π × 6)² gives 355, which is far too big.

+ Add to test✓ In your test5IGMC-045-00-05Perimeter and area of plane shapes Paper CEasy3 marks

No calculator

A parallelogram has a base of 15 cm and a perpendicular height of 8 cm. The sloping sides are each 10 cm long.

(a)

Work out the area of the parallelogram.

(2)

(b)

Work out the perimeter of the parallelogram.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 15 × 8	✓ 1
A1 120 cm²	✓ 1
Part (b)
B1 50 cm	✓ 1

Common error: The area of a parallelogram uses the perpendicular height (8 cm), not the sloping side (10 cm). Using 15 × 10 gives an incorrect answer of 150 cm².

+ Add to test✓ In your test6IGMC-045-01-01Circles, arcs and sectors Paper CEasy4 marks

Calculator

The face of a circular wall clock has a radius of 6.5 cm.

(a)

Calculate the circumference of the clock face.

(2)

(b)

Calculate the area of the clock face.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 2 × π × 6.5	✓ 1
A1 40.8 cm (accept 40.84…)	✓ 1
Part (b)
M1 for π × 6.5²	✓ 1
A1 133 cm² (accept 132.7…)	✓ 1

Common error: Squaring π × 6.5 instead of squaring only the radius gives a much larger area. In πr², only r is squared.

+ Add to test✓ In your test7IGMC-045-01-02Circles, arcs and sectors Paper CEasy4 marks

Calculator

The circumference of a circular pond is 18.5 m.

(a)

Calculate the radius of the pond.

(2)

(b)

Calculate the area of the surface of the pond.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 18.5 ÷ (2π)	✓ 1
A1 2.94 m	✓ 1
Part (b)
M1 for π × (their radius)²	✓ 1
A1 27.2 m²	✓ 1

Exam tip: Keep the unrounded radius in your calculator memory for part (b). Using a rounded value can lose accuracy in the final answer.

+ Add to test✓ In your test8IGMC-045-01-03Circles, arcs and sectors Paper CEasy5 marks

Calculator

A sector of a circle has radius 9 cm and angle 40°.

(a)

Calculate the length of the arc of the sector.

(2)

(b)

Calculate the perimeter of the sector.

(1)

(c)

Calculate the area of the sector.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{40}{360}$ × 2 × π × 9	✓ 1
A1 6.28 cm	✓ 1
Part (b)
B1 24.3 cm, follow through their arc length + 18	✓ 1
Part (c)
M1 for $\frac{40}{360}$ × π × 9²	✓ 1
A1 28.3 cm²	✓ 1

Common error: The perimeter of a sector includes the two radii as well as the arc. Giving only the arc length for the perimeter loses the mark.

+ Add to test✓ In your test9IGMC-045-01-04Circles, arcs and sectors Paper CEasy4 marks

No calculator

A circle has radius 5 cm. Give each answer in terms of π.

(a)

Write down the area of the circle.

(1)

(b)

Write down the circumference of the circle.

(1)

(c)

The circle is cut in half along a diameter. Find the perimeter of one of the semicircles.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 25π cm²	✓ 1
Part (b)
B1 10π cm	✓ 1
Part (c)
M1 for half of their circumference, 5π	✓ 1
A1 5π + 10 cm	✓ 1

Common error: A semicircle's perimeter is not just half the circumference. The straight edge (the diameter, 10 cm) must be added.

+ Add to test✓ In your test10IGMC-045-01-05Circles, arcs and sectors Paper CEasy5 marks

Calculator

A bicycle wheel has a diameter of 64 cm.

(a)

Calculate the distance the bicycle travels when the wheel makes one complete revolution.

(2)

(b)

Calculate the number of revolutions the wheel makes when the bicycle travels 1 km. Use your unrounded answer to part (a), and give your final answer to the nearest whole number.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for π × 64	✓ 1
A1 201 cm (accept 201.0 to 201.1)	✓ 1
Part (b)
M1 for converting 1 km to 100 000 cm (or the circumference to 2.0106… m)	✓ 1
M1 for 100 000 ÷ their circumference	✓ 1
A1 497	✓ 1
Note: 498 from 100 000 ÷ 201 earns both M marks but not the A mark.	—

Exam tip: Keep the full calculator value of the circumference for the next step. Rounding to 201 first changes the final answer from 497 to 498. Also convert to the same units before dividing: 1 km = 100 000 cm.

+ Add to test✓ In your test11IGMC-045-02-01Surface area and volume of solids Paper CEasy5 marks

Calculator

A closed box is a cuboid measuring 12 cm by 8 cm by 5 cm.

(a)

Calculate the volume of the box.

(2)

(b)

Calculate the total surface area of the box.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 12 × 8 × 5	✓ 1
A1: 480 cm³	✓ 1
Part (b)
M1: at least two correct face areas from 96, 60, 40	✓ 1
M1: 2 × (96 + 60 + 40)	✓ 1
A1: 392 cm²	✓ 1

Exam tip: A cuboid has three pairs of identical faces. Find the three different face areas, add them, then double the total.

+ Add to test✓ In your test12IGMC-045-02-02Surface area and volume of solids Paper CEasy3 marks

Calculator

A cylindrical tin has radius 4.5 cm and height 11 cm.

(a)

Calculate the volume of the tin.

(2)

(b)

Write the capacity of the tin in litres.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: π × 4.5² × 11	✓ 1
A1: 700 cm³ (699.79…)	✓ 1
Part (b)
B1: 0.700 litres (ft their (a) ÷ 1000)	✓ 1

Common error: Using the diameter instead of the radius makes the volume four times too big. 1 litre = 1000 cm³.

+ Add to test✓ In your test13IGMC-045-02-03Surface area and volume of solids Paper CEasy6 marks

No calculator

A prism has length 15 cm. Its cross-section is a right-angled triangle whose two shorter sides are 6 cm and 8 cm.

(a)

Find the area of the cross-section.

(1)

(b)

Find the volume of the prism.

(2)

(c)

Find the total surface area of the prism.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 24 cm²	✓ 1
Part (b)
M1: 24 × 15	✓ 1
A1: 360 cm³	✓ 1
Part (c)
M1: hypotenuse = √(6² + 8²) = 10 cm	✓ 1
M1: 2 × 24 + (6 + 8 + 10) × 15	✓ 1
A1: 408 cm²	✓ 1

Exam tip: The three rectangular faces together have area (perimeter of cross-section) × length, which saves working out each rectangle separately.

+ Add to test✓ In your test14IGMC-045-02-04Surface area and volume of solids Paper CEasy4 marks

Calculator

A solid ball is a sphere of radius 6 cm.
[Volume of sphere = $\frac{4}{3}\pi r^3$, surface area of sphere = $4\pi r^2$]

(a)

Calculate the volume of the ball.

(2)

(b)

Calculate the surface area of the ball.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\frac{4}{3}$ × π × 6³	✓ 1
A1: 905 cm³ (904.77…)	✓ 1
Part (b)
M1: 4 × π × 6²	✓ 1
A1: 452 cm² (452.38…)	✓ 1

Common error: Cubing the radius for surface area (or squaring it for volume) mixes up the formulas. Area units are squared, volume units are cubed.

+ Add to test✓ In your test15IGMC-045-02-05Surface area and volume of solids Paper CEasy3 marks

No calculator

A cube has a volume of 343 cm³.

(a)

Find the length of one edge of the cube.

(1)

(b)

Find the total surface area of the cube.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 7 cm	✓ 1
Part (b)
M1: 6 × 7²	✓ 1
A1: 294 cm²	✓ 1

Exam tip: Know the cubes 1³ to 10³ so you can find cube roots quickly without a calculator.

+ Add to test✓ In your test16IGMC-045-00-06Perimeter and area of plane shapes Paper CMedium6 marks

Calculator

A lawn is made from a rectangle 20 m long and 14 m wide. A rectangle 8 m long and 6 m wide has been removed from one corner, leaving an L-shape.

(a)

Calculate the area of the lawn.

(2)

(b)

Calculate the perimeter of the lawn.

(2)

(c)

Turf costs $4.50 per square metre.
Calculate the cost of covering the whole lawn with turf.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 20 × 14 − 8 × 6	✓ 1
A1 232 m²	✓ 1
Part (b)
M1 complete sum of all six edges, e.g. 20 + 14 + 12 + 6 + 8 + 8	✓ 1
A1 68 m	✓ 1
Part (c)
M1 their 232 × 4.50	✓ 1
A1 $1044	✓ 1

Exam tip: Removing a rectangle from a corner does not change the perimeter: the two new inner edges are exactly as long as the two pieces of edge that were removed. Check by adding all six edges.

+ Add to test✓ In your test17IGMC-045-00-07Perimeter and area of plane shapes Paper CMedium5 marks

Calculator

A window is made from a rectangle 1.2 m wide and 1.5 m high, with a semicircle on top. The diameter of the semicircle is the top edge of the rectangle.

(a)

Calculate the total area of the window.

(3)

(b)

Calculate the perimeter of the window.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1.2 × 1.5 (= 1.8)	✓ 1
M1 $\frac{1}{2}$ × π × 0.6² (= 0.5654…)	✓ 1
A1 2.37 m²	✓ 1
Part (b)
M1 1.5 + 1.5 + 1.2 + $\frac{1}{2}$ × π × 1.2	✓ 1
A1 6.08 m	✓ 1

Common error: The top edge of the rectangle is inside the window, so it is not part of the perimeter. Include only the two sides, the base and the curved arc.

+ Add to test✓ In your test18IGMC-045-00-10Perimeter and area of plane shapes Paper CMedium5 marks

Calculator

A circular pond has an area of 50 m².

(a)

Calculate the radius of the pond.

(3)

(b)

A fence is put around the edge of the pond.
Calculate the length of the fence.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 πr² = 50	✓ 1
M1 r = √(50 ÷ π)	✓ 1
A1 3.99 m	✓ 1
Part (b)
M1 2 × π × their r	✓ 1
A1 25.1 m	✓ 1

Common error: Dividing 50 by π gives r², not r. You must take the square root to find the radius.

+ Add to test✓ In your test19IGMC-045-00-11Perimeter and area of plane shapes Paper CMedium6 marks

No calculator

The diagonals of a rhombus are 24 cm and 10 cm long. The diagonals of a rhombus bisect each other at right angles.

(a)

Work out the area of the rhombus.

(2)

(b)

Work out the perimeter of the rhombus.

(2)

(c)

One side of the rhombus is taken as the base.
Work out the perpendicular height of the rhombus, giving your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{1}{2}$ × 24 × 10 or 4 × $\frac{1}{2}$ × 12 × 5	✓ 1
A1 120 cm²	✓ 1
Part (b)
M1 side = √(12² + 5²) = 13	✓ 1
A1 52 cm	✓ 1
Part (c)
M1 13 × h = 120	✓ 1
A1 9.23 cm	✓ 1

Exam tip: A rhombus is a parallelogram, so its area is also base × perpendicular height. Linking the two area methods lets you find the height.

+ Add to test✓ In your test20IGMC-045-01-06Circles, arcs and sectors Paper CMedium5 marks

Calculator

A sector of a circle has radius 12 cm. The arc length of the sector is 15 cm.

(a)

Calculate the angle of the sector.

(3)

(b)

Calculate the area of the sector.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{\theta}{360}$ × 2 × π × 12 = 15	✓ 1
M1 for θ = 15 × 360 ÷ (24π)	✓ 1
A1 71.6°	✓ 1
Part (b)
M1 for $\frac{\text{their }\theta}{360}$ × π × 12² or $\frac{1}{2}$ × 12 × 15	✓ 1
A1 90.0 cm²	✓ 1

Exam tip: Set up the arc length formula as an equation and rearrange for the angle. Using the unrounded angle in part (b) gives exactly 90.

+ Add to test✓ In your test21IGMC-045-01-07Circles, arcs and sectors Paper CMedium5 marks

Calculator

A sector of a circle has an angle of 75° and an area of 50 cm².

(a)

Calculate the radius of the sector.

(3)

(b)

Calculate the perimeter of the sector.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{75}{360}$ × π × r² = 50	✓ 1
M1 for r² = 50 × 360 ÷ (75π) [= 76.39…]	✓ 1
A1 8.74 cm	✓ 1
Part (b)
M1 for $\frac{75}{360}$ × 2 × π × their r + 2 × their r	✓ 1
A1 28.9 cm	✓ 1

Common error: After finding r², remember to take the square root. Giving 76.4 as the radius is a frequent slip.

+ Add to test✓ In your test22IGMC-045-01-08Circles, arcs and sectors Paper CMedium5 marks

Calculator

A circular flower bed has radius 7 m. It is surrounded by a path of width 1.5 m, so that the outer edge of the path is also a circle with the same centre.

(a)

Calculate the area of the path.

(3)

(b)

A fence is put around the outer edge of the path. Fencing costs $12.40 per metre. Calculate the cost of the fence, giving your answer to the nearest dollar.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 outer radius 8.5 m	✓ 1
M1 for π × 8.5² − π × 7²	✓ 1
A1 73.0 m²	✓ 1
Part (b)
M1 for 2 × π × 8.5 × 12.40	✓ 1
A1 $662	✓ 1

Common error: Squaring the path width (1.5²) does not give the area of the path. Subtract the inner circle's area from the outer circle's area.

+ Add to test✓ In your test23IGMC-045-01-11Circles, arcs and sectors Paper CMedium5 marks

Calculator

The minute hand of a clock is 11 cm long.

(a)

Calculate the distance travelled by the tip of the minute hand in 20 minutes.

(2)

(b)

Calculate the area swept out by the minute hand between 3:10 pm and 3:45 pm.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{120}{360}$ × 2 × π × 11	✓ 1
A1 23.0 cm	✓ 1
Part (b)
B1 angle 210° seen or implied	✓ 1
M1 for $\frac{210}{360}$ × π × 11²	✓ 1
A1 222 cm²	✓ 1

Exam tip: The minute hand turns 6° every minute, so the angle is the number of minutes × 6°.

+ Add to test✓ In your test24IGMC-045-02-06Surface area and volume of solids Paper CMedium6 marks

Calculator

A solid cone has base radius 5 cm and vertical height 12 cm.
[Curved surface area of cone = $\pi r l$, volume of cone = $\frac{1}{3}\pi r^2 h$]

(a)

Show that the slant height of the cone is 13 cm.

(2)

(b)

Calculate the curved surface area of the cone.

(2)

(c)

Calculate the volume of the cone.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 5² + 12² = 169	✓ 1
A1: √169 = 13 shown	✓ 1
Part (b)
M1: π × 5 × 13	✓ 1
A1: 204 cm² (204.2…)	✓ 1
Part (c)
M1: $\frac{1}{3}$ × π × 5² × 12	✓ 1
A1: 314 cm³ (314.15…)	✓ 1

Common error: The curved surface area uses the slant height l; the volume uses the vertical height h. Swapping them loses the accuracy mark.

+ Add to test✓ In your test25IGMC-045-02-07Surface area and volume of solids Paper CMedium5 marks

Calculator

A cuboid tank has a base measuring 80 cm by 50 cm. It contains water to a depth of 30 cm.

(a)

Find the volume of water in the tank, giving your answer in litres.

(2)

(b)

All the water is poured into an empty cylindrical tank of radius 20 cm. No water is spilled.
Calculate the depth of water in the cylindrical tank.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 80 × 50 × 30 = 120 000 cm³	✓ 1
A1: 120 litres	✓ 1
Part (b)
M1: π × 20² × h = 120 000	✓ 1
M1: h = 120 000 ÷ (π × 400)	✓ 1
A1: 95.5 cm (95.49…)	✓ 1

Exam tip: The volume of water stays the same when it is poured. Set the two volume expressions equal and solve for the unknown depth.

+ Add to test✓ In your test26IGMC-045-02-10Surface area and volume of solids Paper CMedium6 marks

Calculator

A solid metal cube of edge 8 cm is melted down. The metal is used to make as many solid spheres of radius 1.5 cm as possible.
[Volume of sphere = $\frac{4}{3}\pi r^3$]

(a)

Calculate the volume of one sphere.

(2)

(b)

Find the number of complete spheres that can be made.

(2)

(c)

Calculate the volume of metal left over.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\frac{4}{3}$ × π × 1.5³	✓ 1
A1: 14.1 cm³ (14.137…)	✓ 1
Part (b)
M1: 8³ ÷ their (a), i.e. 512 ÷ 14.137… (= 36.2…)	✓ 1
A1: 36	✓ 1
Part (c)
M1: 512 − 36 × 14.137…	✓ 1
A1: 3.06 cm³ (3.06…)	✓ 1

Common error: The number of complete spheres must be rounded down, not to the nearest whole number. Use the unrounded sphere volume in part (c).

+ Add to test✓ In your test27IGMC-045-00-08Perimeter and area of plane shapes Paper EMedium5 marks

No calculator

A sector of a circle has radius 9 cm and angle 140°.
Give each answer in terms of π.

(a)

Find the arc length of the sector.

(2)

(b)

Find the perimeter of the sector.

(1)

(c)

Find the area of the sector.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{140}{360}$ × 2 × π × 9	✓ 1
A1 7π cm	✓ 1
Part (b)
B1 18 + 7π cm	✓ 1
Part (c)
M1 $\frac{140}{360}$ × π × 9²	✓ 1
A1 $\frac{63}{2}$π or 31.5π cm²	✓ 1

Exam tip: Simplify the fraction of the circle first: $\frac{140}{360} = \frac{7}{18}$. Then $\frac{7}{18}$ × 18π = 7π, which keeps the working short without a calculator.

+ Add to test✓ In your test28IGMC-045-00-09Perimeter and area of plane shapes Paper EMedium7 marks

Calculator

In triangle PQR, PQ = 8.5 cm, PR = 11.2 cm and angle QPR = 47°.

(a)

Calculate the area of triangle PQR.

(2)

(b)

Calculate the length QR.

(3)

(c)

Calculate the perimeter of triangle PQR.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{1}{2}$ × 8.5 × 11.2 × sin 47°	✓ 1
A1 34.8 cm²	✓ 1
Part (b)
M1 QR² = 8.5² + 11.2² − 2 × 8.5 × 11.2 × cos 47°	✓ 1
A1 QR² = 67.8(4)…	✓ 1
A1 QR = 8.24 cm	✓ 1
Part (c)
M1 8.5 + 11.2 + their QR	✓ 1
A1 27.9 cm	✓ 1

Exam tip: Use the unrounded value of QR (8.2363…) from your calculator in part (c). Rounding early can change the final 3 significant figure answer.

+ Add to test✓ In your test29IGMC-045-01-09Circles, arcs and sectors Paper EMedium6 marks

No calculator

A sector of a circle has radius 6 cm and angle 120°. Give your answers to parts (a) and (b) in terms of π.

(a)

Find the area of the sector.

(2)

(b)

Find the perimeter of the sector.

(2)

(c)

The sector is rolled up so that its two straight edges meet, forming the curved surface of a cone. Find the radius of the base of the cone.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{120}{360}$ × π × 6²	✓ 1
A1 12π cm²	✓ 1
Part (b)
M1 for $\frac{120}{360}$ × 2 × π × 6 [= 4π]	✓ 1
A1 4π + 12 cm	✓ 1
Part (c)
M1 for 2πr = 4π	✓ 1
A1 2 cm	✓ 1

Exam tip: When a sector forms a cone, the arc length becomes the circumference of the cone's base, and the sector's radius becomes the slant height.

+ Add to test✓ In your test30IGMC-045-02-08Surface area and volume of solids Paper EMedium7 marks

Calculator

A paperweight is a solid hemisphere of radius 9 cm.
[Volume of sphere = $\frac{4}{3}\pi r^3$, surface area of sphere = $4\pi r^2$]

(a)

Calculate the volume of the paperweight.

(2)

(b)

The paperweight is made of a metal with density 2.7 g/cm³.
Calculate its mass in kilograms.

(2)

(c)

Calculate the total surface area of the paperweight.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\frac{1}{2}$ × $\frac{4}{3}$ × π × 9³	✓ 1
A1: 1530 cm³ (1526.8…)	✓ 1
Part (b)
M1: their (a) × 2.7, giving 4122.4… g (or 4131 g from 1530)	✓ 1
A1: ÷ 1000 to give 4.12 kg (4.122…); accept 4.13 kg from using 1530	✓ 1
Part (c)
M1: curved surface $\frac{1}{2}$ × 4 × π × 9²	✓ 1
M1: adding flat circle π × 9²	✓ 1
A1: 763 cm² (763.4…)	✓ 1

Common error: A solid hemisphere has a flat circular face as well as its curved surface. Leaving out the circle gives only 2πr².

+ Add to test✓ In your test31IGMC-045-02-09Surface area and volume of solids Paper EMedium6 marks

No calculator

A solid pyramid has a square base of side 10 cm. The vertex is vertically above the centre of the base, and the vertical height is 12 cm.
[Volume of pyramid = $\frac{1}{3}$ × area of base × height]

(a)

Find the volume of the pyramid.

(2)

(b)

Find the height of one triangular face, measured from the midpoint of a base edge to the vertex.

(2)

(c)

Find the total surface area of the pyramid.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\frac{1}{3}$ × 10² × 12	✓ 1
A1: 400 cm³	✓ 1
Part (b)
M1: √(12² + 5²)	✓ 1
A1: 13 cm	✓ 1
Part (c)
M1: 10² + 4 × $\frac{1}{2}$ × 10 × 13	✓ 1
A1: 360 cm²	✓ 1

Exam tip: For the triangular faces you need the slant height of a face, not the vertical height of the pyramid. Use half the base side in Pythagoras.

+ Add to test✓ In your test32IGMC-045-02-11Surface area and volume of solids Paper EMedium4 marks

No calculator

Two bottles are mathematically similar. The smaller bottle has height 12 cm and the larger bottle has height 18 cm.

(a)

The surface area of the smaller bottle is 240 cm².
Find the surface area of the larger bottle.

(2)

(b)

The volume of the larger bottle is 1350 cm³.
Find the volume of the smaller bottle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 240 × (18 ÷ 12)² or 240 × 2.25	✓ 1
A1: 540 cm²	✓ 1
Part (b)
M1: 1350 ÷ (1.5)³ or 1350 × (2/3)³	✓ 1
A1: 400 cm³	✓ 1

Exam tip: If the length scale factor is k, areas scale by k² and volumes by k³. Check that the smaller solid gets the smaller answer.

+ Add to test✓ In your test33IGMC-045-00-12Perimeter and area of plane shapes Paper EHard6 marks

Calculator

A circular flower bed of radius r metres is surrounded by a path of width 1.5 m. The outer edge of the path is also a circle with the same centre. The area of the path is 33 m².

(a)

Calculate the value of r.

(4)

(b)

Calculate the length of the outer edge of the path.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π(r + 1.5)² − πr² = 33	✓ 1
M1 expands to π(3r + 2.25) = 33	✓ 1
M1 r = (33 ÷ π − 2.25) ÷ 3	✓ 1
A1 r = 2.75	✓ 1
Part (b)
M1 2 × π × (their r + 1.5)	✓ 1
A1 26.7 m	✓ 1

Common error: (r + 1.5)² is r² + 3r + 2.25, not r² + 2.25. Missing the middle term makes the equation impossible to solve correctly.

+ Add to test✓ In your test34IGMC-045-00-13Perimeter and area of plane shapes Paper EHard7 marks

No calculator

A rectangle has length (2x + 3) cm and width (x − 1) cm. The area of the rectangle is 42 cm².

(a)

Show that 2x² + x − 45 = 0.

(2)

(b)

Solve 2x² + x − 45 = 0 and explain why only one solution is valid.

(3)

(c)

Work out the perimeter of the rectangle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (2x + 3)(x − 1) = 42	✓ 1
A1 2x² − 2x + 3x − 3 = 42 leading to 2x² + x − 45 = 0 with no errors	✓ 1
Part (b)
M1 (2x − 9)(x + 5) = 0, or correct substitution into the quadratic formula, e.g. x = $\frac{-1 \pm \sqrt{1 + 360}}{4}$	✓ 1
A1 x = 4.5 and x = −5	✓ 1
B1 x = −5 rejected because the width x − 1 = −6 would be negative (or length negative), so x = 4.5	✓ 1
Part (c)
M1 length 12 and width 3.5, or 2 × (12 + 3.5)	✓ 1
A1 31 cm	✓ 1

Exam tip: In a 'show that' question, write every line of the expansion and finish with the given equation exactly. Check your dimensions at the end: 12 × 3.5 = 42.

+ Add to test✓ In your test35IGMC-045-00-14Perimeter and area of plane shapes Paper EHard7 marks

Calculator

A and B are points on a circle with centre O and radius 10 cm. Angle AOB = 75°. The chord AB divides the circle into two segments.

(a)

Calculate the area of the minor segment.

(4)

(b)

Calculate the perimeter of the minor segment.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sector area $\frac{75}{360}$ × π × 10² (= 65.4…)	✓ 1
M1 triangle area $\frac{1}{2}$ × 10 × 10 × sin 75° (= 48.2…)	✓ 1
M1 sector area − triangle area	✓ 1
A1 17.2 cm²	✓ 1
Part (b)
M1 arc length $\frac{75}{360}$ × 2 × π × 10 (= 13.09…)	✓ 1
M1 chord AB = 2 × 10 × sin 37.5° or by the cosine rule (= 12.17…)	✓ 1
A1 arc + chord = 25.3 cm	✓ 1

Common error: The perimeter of a segment is the arc plus the straight chord. Adding two radii instead gives the perimeter of the sector, not the segment.

+ Add to test✓ In your test36IGMC-045-01-10Circles, arcs and sectors Paper EHard8 marks

Calculator

A and B are points on a circle, centre O, radius 8 cm. Angle AOB = 110°. The chord AB divides the circle into a minor segment and a major segment.

(a)

Calculate the area of the minor sector OAB.

(2)

(b)

Calculate the area of triangle OAB.

(2)

(c)

Calculate the area of the minor segment.

(1)

(d)

Calculate the perimeter of the minor segment.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{110}{360}$ × π × 8²	✓ 1
A1 61.4 cm² (61.43…)	✓ 1
Part (b)
M1 for $\frac{1}{2}$ × 8 × 8 × sin 110°	✓ 1
A1 30.1 cm² (30.07…)	✓ 1
Part (c)
B1 31.4 cm² from unrounded values (31.36…), or 31.3 cm² from 61.4 − 30.1; follow through their (a) − their (b)	✓ 1
Part (d)
M1 for chord AB = 2 × 8 × sin 55° or AB² = 8² + 8² − 2 × 8 × 8 × cos 110° [AB = 13.1]	✓ 1
M1 for $\frac{110}{360}$ × 2 × π × 8 [arc = 15.36] + their AB	✓ 1
A1 28.5 cm (28.46…)	✓ 1

Exam tip: A segment's perimeter is the arc plus the chord, not the two radii. Find the chord with trigonometry or the cosine rule, and keep unrounded values until the final answer.

+ Add to test✓ In your test37IGMC-045-01-13Circles, arcs and sectors Paper EHard7 marks

Calculator

OAB is a sector of a circle, centre O, radius 10 cm and angle 60°. C lies on OA and D lies on OB such that OC = OD = r cm. OCD is a sector of a circle, centre O, with the same angle of 60°. The region ABDC, between the two arcs, has an area of 30 cm².

(a)

Calculate the value of r.

(4)

(b)

Calculate the perimeter of the region ABDC.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for area of sector OAB = $\frac{60}{360}$ × π × 10² [= 52.36]	✓ 1
M1 for $\frac{60}{360}$ × π × r² = their 52.36 − 30	✓ 1
M1 for r² = 42.7 or r² = 100 − $\frac{180}{\pi}$	✓ 1
A1 r = 6.53	✓ 1
Part (b)
M1 for arc AB = $\frac{60}{360}$ × 2π × 10 [= 10.47] or arc CD = $\frac{60}{360}$ × 2π × their r [= 6.84]	✓ 1
M1 for arc AB + arc CD + 2 × (10 − their r)	✓ 1
A1 24.2 cm (accept 24.25 or 24.3 from using r = 6.53)	✓ 1

Common error: The straight edges of region ABDC are AC and BD, each 10 − r, not the full radius of 10 cm.

+ Add to test✓ In your test38IGMC-045-01-14Circles, arcs and sectors Paper EHard7 marks

No calculator

A square of side 8 cm is drawn so that all four of its vertices lie on a circle.

(a)

Find the exact radius of the circle, giving your answer as a surd in its simplest form.

(2)

(b)

Find the exact area of the region inside the circle but outside the square.

(3)

(c)

A second circle is drawn inside the square so that it touches all four sides. Find the ratio of the area of the larger circle to the area of the smaller circle, in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for diagonal² = 8² + 8² or diagonal = 8√2	✓ 1
A1 4√2 cm	✓ 1
Part (b)
M1 for π × (4√2)² [= 32π]	✓ 1
M1 for their 32π − 64	✓ 1
A1 32π − 64 cm²	✓ 1
Part (c)
B1 area of smaller circle = 16π	✓ 1
B1 2 : 1	✓ 1

Exam tip: The diagonal of the square is a diameter of the outer circle, while the side of the square is a diameter of the inner circle.

+ Add to test✓ In your test39IGMC-045-02-12Surface area and volume of solids Paper EHard7 marks

Calculator

A solid cone has base radius 9 cm and vertical height 24 cm. A small cone of vertical height 8 cm is cut from the top by a cut parallel to the base. The remaining solid is a frustum.
[Volume of cone = $\frac{1}{3}\pi r^2 h$]

(a)

Find the radius of the base of the small cone.

(2)

(b)

Calculate the volume of the frustum.

(3)

(c)

A hollow container has exactly the same shape and size as the frustum. It is completely filled with water. All the water is then poured into an empty cylinder of radius 6 cm.
Calculate the depth of water in the cylinder.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: r ÷ 9 = 8 ÷ 24, or 9 × $\frac{8}{24}$, or 9 ÷ 3, or equivalent	✓ 1
A1: 3 cm	✓ 1
Part (b)
M1: large cone $\frac{1}{3}$ × π × 9² × 24 (= 648π)	✓ 1
M1: subtracting small cone $\frac{1}{3}$ × π × 3² × 8 (= 24π)	✓ 1
A1: 1960 cm³ (624π = 1960.35…)	✓ 1
Part (c)
M1: their (b) ÷ (π × 6²), e.g. 624π ÷ 36π	✓ 1
A1: 17.3 cm (17.33…)	✓ 1

Exam tip: Keep the volumes in terms of π: 648π − 24π = 624π. This avoids rounding errors, and the π cancels in part (c).

+ Add to test✓ In your test40IGMC-045-02-13Surface area and volume of solids Paper EHard8 marks

No calculator

A sphere of radius r cm fits exactly inside a closed cylinder. The cylinder has radius r cm and height 2r cm.
[Volume of sphere = $\frac{4}{3}\pi r^3$, surface area of sphere = $4\pi r^2$]

(a)

Show that the volume of the sphere is $\frac{2}{3}$ of the volume of the cylinder.

(3)

(b)

Show that the curved surface area of the cylinder is equal to the surface area of the sphere.

(2)

(c)

The total surface area of the cylinder is 54π cm².
Find the volume of the sphere, giving your answer in terms of π.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: cylinder volume = πr² × 2r = 2πr³	✓ 1
M1: $\frac{4}{3}\pi r^3$ ÷ 2πr³	✓ 1
A1: = $\frac{2}{3}$ correctly shown	✓ 1
Part (b)
M1: curved surface area = 2πr × 2r	✓ 1
A1: = 4πr², the same as the sphere	✓ 1
Part (c)
M1: 4πr² + 2πr² = 54π, so 6πr² = 54π	✓ 1
A1: r = 3	✓ 1
A1: 36π cm³	✓ 1

Common error: The total surface area of a closed cylinder includes the two circular ends. Using only the curved surface gives r² = 13.5.

+ Add to test✓ In your test41IGMC-045-02-14Surface area and volume of solids Paper EHard8 marks

Calculator

A water trough is a prism of length 2.5 m. Its cross-section is an isosceles trapezium. The bottom edge is 40 cm, the top edge is 60 cm and the depth is 30 cm. The trough stands on its 40 cm face, with the 60 cm edge at the top.

(a)

Calculate the area of the cross-section.

(2)

(b)

Calculate the capacity of the trough in litres.

(2)

(c)

The trough contains 200 litres of water. Let the depth of the water be h cm.
Form an equation in h and solve it to find the depth of water.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\frac{1}{2}$ × (60 + 40) × 30	✓ 1
A1: 1500 cm²	✓ 1
Part (b)
M1: 1500 × 250 (= 375 000 cm³)	✓ 1
A1: 375 litres	✓ 1
Part (c)
M1: width of water surface = 40 + $\frac{2h}{3}$	✓ 1
M1: 250 × $\frac{1}{2}$(40 + 40 + $\frac{2h}{3}$)h = 200 000	✓ 1
M1: simplifies to h² + 120h − 2400 = 0, with correct use of the quadratic formula (or completing the square)	✓ 1
A1: h = 17.5 cm (17.459…), with the negative root rejected	✓ 1

Exam tip: The depth of water is not proportional to its volume, because the trough widens upwards. The width increases by 20 cm over 30 cm of depth, so it increases by $\frac{2}{3}$ cm for each 1 cm of depth.

Marking point	Mark	Notes
Part (a)
M1 \(\frac{1}{2}\) × 9 × 6.4	✓ 1
A1 28.8 cm²	✓ 1
Part (b)
M1 \(\frac{1}{2}\) × 12 × h = 28.8 or h = 28.8 × 2 ÷ 12	✓ 1
A1 4.8 cm	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{1}{2}\) × (7 + 13) × 5	✓ 1
A1 50 cm²	✓ 1
Part (b)
B1 1 m² = 10 000 cm²	✓ 1
A1 200 tiles	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{40}{360}\) × 2 × π × 9	✓ 1
A1 6.28 cm	✓ 1
Part (b)
B1 24.3 cm, follow through their arc length + 18	✓ 1
Part (c)
M1 for \(\frac{40}{360}\) × π × 9²	✓ 1
A1 28.3 cm²	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(\frac{4}{3}\) × π × 6³	✓ 1
A1: 905 cm³ (904.77…)	✓ 1
Part (b)
M1: 4 × π × 6²	✓ 1
A1: 452 cm² (452.38…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{1}{2}\) × 24 × 10 or 4 × \(\frac{1}{2}\) × 12 × 5	✓ 1
A1 120 cm²	✓ 1
Part (b)
M1 side = √(12² + 5²) = 13	✓ 1
A1 52 cm	✓ 1
Part (c)
M1 13 × h = 120	✓ 1
A1 9.23 cm	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{\theta}{360}\) × 2 × π × 12 = 15	✓ 1
M1 for θ = 15 × 360 ÷ (24π)	✓ 1
A1 71.6°	✓ 1
Part (b)
M1 for \(\frac{\text{their }\theta}{360}\) × π × 12² or \(\frac{1}{2}\) × 12 × 15	✓ 1
A1 90.0 cm²	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{75}{360}\) × π × r² = 50	✓ 1
M1 for r² = 50 × 360 ÷ (75π) [= 76.39…]	✓ 1
A1 8.74 cm	✓ 1
Part (b)
M1 for \(\frac{75}{360}\) × 2 × π × their r + 2 × their r	✓ 1
A1 28.9 cm	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{120}{360}\) × 2 × π × 11	✓ 1
A1 23.0 cm	✓ 1
Part (b)
B1 angle 210° seen or implied	✓ 1
M1 for \(\frac{210}{360}\) × π × 11²	✓ 1
A1 222 cm²	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(\frac{4}{3}\) × π × 1.5³	✓ 1
A1: 14.1 cm³ (14.137…)	✓ 1
Part (b)
M1: 8³ ÷ their (a), i.e. 512 ÷ 14.137… (= 36.2…)	✓ 1
A1: 36	✓ 1
Part (c)
M1: 512 − 36 × 14.137…	✓ 1
A1: 3.06 cm³ (3.06…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{140}{360}\) × 2 × π × 9	✓ 1
A1 7π cm	✓ 1
Part (b)
B1 18 + 7π cm	✓ 1
Part (c)
M1 \(\frac{140}{360}\) × π × 9²	✓ 1
A1 \(\frac{63}{2}\)π or 31.5π cm²	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{1}{2}\) × 8.5 × 11.2 × sin 47°	✓ 1
A1 34.8 cm²	✓ 1
Part (b)
M1 QR² = 8.5² + 11.2² − 2 × 8.5 × 11.2 × cos 47°	✓ 1
A1 QR² = 67.8(4)…	✓ 1
A1 QR = 8.24 cm	✓ 1
Part (c)
M1 8.5 + 11.2 + their QR	✓ 1
A1 27.9 cm	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{120}{360}\) × π × 6²	✓ 1
A1 12π cm²	✓ 1
Part (b)
M1 for \(\frac{120}{360}\) × 2 × π × 6 [= 4π]	✓ 1
A1 4π + 12 cm	✓ 1
Part (c)
M1 for 2πr = 4π	✓ 1
A1 2 cm	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(\frac{1}{2}\) × \(\frac{4}{3}\) × π × 9³	✓ 1
A1: 1530 cm³ (1526.8…)	✓ 1
Part (b)
M1: their (a) × 2.7, giving 4122.4… g (or 4131 g from 1530)	✓ 1
A1: ÷ 1000 to give 4.12 kg (4.122…); accept 4.13 kg from using 1530	✓ 1
Part (c)
M1: curved surface \(\frac{1}{2}\) × 4 × π × 9²	✓ 1
M1: adding flat circle π × 9²	✓ 1
A1: 763 cm² (763.4…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(\frac{1}{3}\) × 10² × 12	✓ 1
A1: 400 cm³	✓ 1
Part (b)
M1: √(12² + 5²)	✓ 1
A1: 13 cm	✓ 1
Part (c)
M1: 10² + 4 × \(\frac{1}{2}\) × 10 × 13	✓ 1
A1: 360 cm²	✓ 1

Marking point	Mark	Notes
Part (a)
M1 sector area \(\frac{75}{360}\) × π × 10² (= 65.4…)	✓ 1
M1 triangle area \(\frac{1}{2}\) × 10 × 10 × sin 75° (= 48.2…)	✓ 1
M1 sector area − triangle area	✓ 1
A1 17.2 cm²	✓ 1
Part (b)
M1 arc length \(\frac{75}{360}\) × 2 × π × 10 (= 13.09…)	✓ 1
M1 chord AB = 2 × 10 × sin 37.5° or by the cosine rule (= 12.17…)	✓ 1
A1 arc + chord = 25.3 cm	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{110}{360}\) × π × 8²	✓ 1
A1 61.4 cm² (61.43…)	✓ 1
Part (b)
M1 for \(\frac{1}{2}\) × 8 × 8 × sin 110°	✓ 1
A1 30.1 cm² (30.07…)	✓ 1
Part (c)
B1 31.4 cm² from unrounded values (31.36…), or 31.3 cm² from 61.4 − 30.1; follow through their (a) − their (b)	✓ 1
Part (d)
M1 for chord AB = 2 × 8 × sin 55° or AB² = 8² + 8² − 2 × 8 × 8 × cos 110° [AB = 13.1]	✓ 1
M1 for \(\frac{110}{360}\) × 2 × π × 8 [arc = 15.36] + their AB	✓ 1
A1 28.5 cm (28.46…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for area of sector OAB = \(\frac{60}{360}\) × π × 10² [= 52.36]	✓ 1
M1 for \(\frac{60}{360}\) × π × r² = their 52.36 − 30	✓ 1
M1 for r² = 42.7 or r² = 100 − \(\frac{180}{\pi}\)	✓ 1
A1 r = 6.53	✓ 1
Part (b)
M1 for arc AB = \(\frac{60}{360}\) × 2π × 10 [= 10.47] or arc CD = \(\frac{60}{360}\) × 2π × their r [= 6.84]	✓ 1
M1 for arc AB + arc CD + 2 × (10 − their r)	✓ 1
A1 24.2 cm (accept 24.25 or 24.3 from using r = 6.53)	✓ 1

Marking point	Mark	Notes
Part (a)
M1: r ÷ 9 = 8 ÷ 24, or 9 × \(\frac{8}{24}\), or 9 ÷ 3, or equivalent	✓ 1
A1: 3 cm	✓ 1
Part (b)
M1: large cone \(\frac{1}{3}\) × π × 9² × 24 (= 648π)	✓ 1
M1: subtracting small cone \(\frac{1}{3}\) × π × 3² × 8 (= 24π)	✓ 1
A1: 1960 cm³ (624π = 1960.35…)	✓ 1
Part (c)
M1: their (b) ÷ (π × 6²), e.g. 624π ÷ 36π	✓ 1
A1: 17.3 cm (17.33…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(\frac{1}{2}\) × (60 + 40) × 30	✓ 1
A1: 1500 cm²	✓ 1
Part (b)
M1: 1500 × 250 (= 375 000 cm³)	✓ 1
A1: 375 litres	✓ 1
Part (c)
M1: width of water surface = 40 + \(\frac{2h}{3}\)	✓ 1
M1: 250 × \(\frac{1}{2}\)(40 + 40 + \(\frac{2h}{3}\))h = 200 000	✓ 1
M1: simplifies to h² + 120h − 2400 = 0, with correct use of the quadratic formula (or completing the square)	✓ 1
A1: h = 17.5 cm (17.459…), with the negative root rejected	✓ 1