📐 Pythagoras & Mensuration
IGCSE Year 8 — Intermediate Level
Instructions: Answer all questions. Show your working clearly. Use a calculator where needed — give answers to 3 significant figures unless stated otherwise. Leave units in every final answer.
📖 Key Concepts & Formulas
- Pythagoras' Theorem: In a right-angled triangle, a2 + b2 = c2, where c is the hypotenuse (longest side).
- Finding hypotenuse: c = √(a2 + b2)
- Finding a shorter side: a = √(c2 − b2)
- Volume of cone: V = 13πr2h (use perpendicular height h)
- Volume of pyramid: V = 13 × base area × perpendicular height
- Surface area of cone: SA = πr2 + πrl (where l is slant height)
SECTION A
3D Pythagoras
[14 marks]
💡 Tip: To find the space diagonal of a cuboid with length l, width w, and height h:
First find the base diagonal: d = √(l² + w²), then use that to find the space diagonal: D = √(d² + h²)
First find the base diagonal: d = √(l² + w²), then use that to find the space diagonal: D = √(d² + h²)
Q1 [2 marks]
A cuboid has dimensions 6 cm × 8 cm × 10 cm. Find the length of the space diagonal (from one corner to the opposite corner through the centre).
Q2 [2 marks]
A room is 4 m long, 3 m wide, and 2.5 m high. A spider is in one corner on the floor and a fly is in the opposite corner on the ceiling. What is the shortest straight-line distance between them?
Q3 [2 marks]
A rectangular box has a base of 9 cm × 12 cm. The space diagonal is 17 cm. Find the height of the box.
Q4 [2 marks]
A cuboid has dimensions 7 cm × 24 cm × h cm. The space diagonal is 26 cm. Find h.
Q5 [2 marks]
A cuboid has dimensions 15 cm × 20 cm × 8 cm. Find the length of the space diagonal, giving your answer to 3 significant figures.
Q6 [2 marks]
A storage container is 3 m long, 2 m wide, and 4 m high. Find the longest straight line that fits inside the container.
Q7 [2 marks]
The base of a cuboid measures 10 cm by 24 cm. If the space diagonal is 29 cm, find the height.
SECTION B
Mensuration with Pythagoras — Cones & Pyramids
[26 marks]
💡 Key idea: When you're given the slant height of a cone or pyramid but need the perpendicular height, use Pythagoras' theorem! The perpendicular height, radius (or half-base), and slant height form a right-angled triangle.
Q8 [3 marks]
A cone has radius 6 cm and slant height 10 cm. Find its volume.
Q9 [3 marks]
A cone has radius 9 cm and perpendicular height 12 cm. Find its total surface area.
Q10 [3 marks]
A square-based pyramid has base sides of 8 cm and perpendicular height of 6 cm. Find its volume.
Q11 [3 marks]
A square-based pyramid has base sides of 10 cm and slant height (of each triangular face) of 13 cm. Find its volume.
Q12 [2 marks]
A cone has volume 377 cm3 and radius 5 cm. Find its perpendicular height.
Q13 [3 marks]
A cone has diameter 16 cm and slant height 10 cm. Find its volume.
Q14 [3 marks]
A square-based pyramid has base sides of 12 cm and perpendicular height of 8 cm. Find the slant height of each triangular face.
Q15 [3 marks]
A cone has radius 7 cm and slant height 25 cm. Find its volume, giving your answer in terms of π.
Q16 [3 marks]
A square-based pyramid has base sides of 24 cm and slant height of 13 cm. Find its volume.
SECTION C
Mixed Challenge Problems
[25 marks]Q17 [3 marks]
A tent is in the shape of a triangular prism. The triangular face has base 4 m and height 1.5 m. The length of the tent is 6 m.
(a) Find the volume of air inside the tent.
(b) Find the length of each sloping side of the triangular face.
(a) Find the volume of air inside the tent.
(b) Find the length of each sloping side of the triangular face.
Q18 [3 marks]
A cone has radius 15 cm and perpendicular height 20 cm.
(a) Find the slant height of the cone.
(b) Hence find its total surface area.
(a) Find the slant height of the cone.
(b) Hence find its total surface area.
Q19 [4 marks]
A square-based pyramid has base sides of 16 cm and perpendicular height of 15 cm.
(a) Find the volume of the pyramid.
(b) Find the slant height of each triangular face.
(a) Find the volume of the pyramid.
(b) Find the slant height of each triangular face.
Q20 [3 marks]
A cone has radius 8 cm and perpendicular height 15 cm.
(a) Find the slant height of the cone.
(b) Hence find its total surface area, giving your answer to 3 significant figures.
(a) Find the slant height of the cone.
(b) Hence find its total surface area, giving your answer to 3 significant figures.
Q21 [4 marks]
A square-based pyramid has base sides of 10 cm. The perpendicular height is 12 cm.
(a) Find the volume of the pyramid.
(b) Find the slant height of each triangular face.
(c) Hence find the total surface area.
(a) Find the volume of the pyramid.
(b) Find the slant height of each triangular face.
(c) Hence find the total surface area.
Q22 [4 marks]
A cone has radius 10 cm and slant height 26 cm.
(a) Find the perpendicular height of the cone.
(b) Hence find its volume, giving your answer in terms of π.
(a) Find the perpendicular height of the cone.
(b) Hence find its volume, giving your answer in terms of π.
Q23 [4 marks]
A square-based pyramid has base sides of 18 cm and slant height of 15 cm.
(a) Find the perpendicular height of the pyramid.
(b) Hence find its volume.
(a) Find the perpendicular height of the pyramid.
(b) Hence find its volume.
ANSWERS
Answer Key
Q1: Base diagonal = √(6² + 8²) = √100 = 10 cm. Space diagonal = √(10² + 10²) = √200 ≈ 14.1 cm
Q2: Base diagonal = √(4² + 3²) = √25 = 5 m. Space diagonal = √(5² + 2.5²) = √31.25 ≈ 5.59 m
Q3: Base diagonal = √(9² + 12²) = √225 = 15 cm. h = √(17² − 15²) = √(289−225) = √64 = 8 cm
Q4: Base diagonal = √(7² + 24²) = √(49+576) = √625 = 25 cm. h = √(26² − 25²) = √(676−625) = √51 ≈ 7.14 cm
Q5: Base diagonal = √(15² + 20²) = √625 = 25 cm. Space diagonal = √(25² + 8²) = √689 ≈ 26.3 cm
Q6: D = √(3² + 2² + 4²) = √(9+4+16) = √29 ≈ 5.39 m
Q7: Base diagonal = √(10² + 24²) = √676 = 26 cm. h = √(29² − 26²) = √(841−676) = √165 ≈ 12.8 cm
Q8: h = √(10² − 6²) = √64 = 8 cm. V = (1/3)π × 6² × 8 = 96π ≈ 302 cm³
Q9: l = √(9² + 12²) = √225 = 15 cm. SA = π × 9² + π × 9 × 15 = 81π + 135π = 216π ≈ 679 cm²
Q10: V = (1/3) × 8² × 6 = (1/3) × 64 × 6 = 128 cm³
Q11: h = √(13² − 5²) = √144 = 12 cm. V = (1/3) × 10² × 12 = 400 cm³
Q12: 377 = (1/3)π × 5² × h → h = 377 ÷ ((1/3)π × 25) ≈ 14.4 cm
Q13: r = 8 cm. h = √(10² − 8²) = √36 = 6 cm. V = (1/3)π × 8² × 6 = 128π ≈ 402 cm³
Q14: l = √(6² + 8²) = √100 = 10 cm
Q15: h = √(25² − 7²) = √576 = 24 cm. V = (1/3)π × 7² × 24 = 392π cm³
Q16: h = √(13² − 12²) = √25 = 5 cm. V = (1/3) × 24² × 5 = 960 cm³
Q17(a): V = (½ × 4 × 1.5) × 6 = 18 m³
(b) Sloping side = √(2² + 1.5²) = √6.25 = 2.5 m
(b) Sloping side = √(2² + 1.5²) = √6.25 = 2.5 m
Q18(a): l = √(15² + 20²) = √625 = 25 cm
(b) SA = π × 15² + π × 15 × 25 = 225π + 375π = 600π ≈ 1885 cm²
(b) SA = π × 15² + π × 15 × 25 = 225π + 375π = 600π ≈ 1885 cm²
Q19(a): V = (1/3) × 16² × 15 = 1280 cm³
(b) Slant height = √(8² + 15²) = √305 ≈ 17.5 cm
(b) Slant height = √(8² + 15²) = √305 ≈ 17.5 cm
Q20(a): l = √(8² + 15²) = √289 = 17 cm
(b) SA = π × 8² + π × 8 × 17 = 64π + 136π = 200π ≈ 628 cm²
(b) SA = π × 8² + π × 8 × 17 = 64π + 136π = 200π ≈ 628 cm²
Q21(a): V = (1/3) × 10² × 12 = 400 cm³
(b) l = √(5² + 12²) = √169 = 13 cm
(c) SA = 10² + 4 × (½ × 10 × 13) = 100 + 260 = 360 cm²
(b) l = √(5² + 12²) = √169 = 13 cm
(c) SA = 10² + 4 × (½ × 10 × 13) = 100 + 260 = 360 cm²
Q22(a): h = √(26² − 10²) = √576 = 24 cm
(b) V = (1/3)π × 10² × 24 = 800π cm³
(b) V = (1/3)π × 10² × 24 = 800π cm³
Q23(a): h = √(15² − 9²) = √144 = 12 cm
(b) V = (1/3) × 18² × 12 = 1296 cm³
(b) V = (1/3) × 18² × 12 = 1296 cm³