Revision mat · Surface Area & Volume: Cones, Spheres & Pyramids

Calculator allowed16 marks

Question mix weighted by 68 real exam questions from 61 GCSE papers (2017–24), so the most common question types get the most space. Show your working.

Cone volume $V=\tfrac13\pi r^2 h$Cone surface $\pi r l + \pi r^2$Sphere $V=\tfrac43\pi r^3,\quad A=4\pi r^2$Pyramid volume $V=\tfrac13\times\text{base area}\times h$Slant height $l^2=r^2+h^2$
Surface area≈43% of real exam Qs
1A solid hemisphere has radius $12$ cm.
Work out the total surface area of the solid hemisphere.
Give your answer correct to 3 significant figures. (The curved surface area of a hemisphere is $2\pi r^2$.)
Solid ()12 cm
Answer: (3 marks)
Volume (cone, sphere, pyramid)≈28% of real exam Qs
2A hemisphere has a radius of $4$ cm.
Work out the volume of the hemisphere.
Give your answer correct to 3 significant figures.
Solid ()4 cm
Answer: (3 marks)
Composite solids≈22% of real exam Qs
3A cylindrical tube has internal diameter $9$ cm and height $32$ cm.
Balls of diameter $5$ cm are dropped in so they stack on top of one another.
Work out the greatest number of whole balls that fit inside the tube.
Answer: (3 marks)
Slant height & reverse≈3% of real exam Qs
4A pyramid has a square base of side $9$ cm and a volume of $270$ cm³.
The apex is directly above the centre of the base.
Work out the perpendicular height of the pyramid. (Volume of a pyramid $= \dfrac{1}{3} \times \text{base area} \times \text{height}$.)
Solid ()ABCDVX9 cmh
Answer: (3 marks)
★ Exam capstonemixed & other forms ≈4% of real exam Qs
5The diagram shows a frustum formed by removing a small cone from a larger cone.
The larger cone has base radius $15$ cm and slant height $39$ cm.
The frustum has a top radius of $5$ cm.
Work out the total surface area of the frustum.
Give your answer correct to 3 significant figures. (Curved surface area of a cone $= \pi r l$, where $l$ is the slant height.)
Solid ()5 cm15 cm39 cm
Answer: (4 marks)

Mark scheme · Surface Area & Volume: Cones, Spheres & Pyramids

Total: 16 marks

Award the marks shown for each correct step, then add up the total out of 16. A method mark counts even if the final answer is wrong.

1A solid hemisphere has radius $12$ cm.
Work out the total surface area of the solid hemisphere.
Give your answer correct to 3 significant figures. (The curved surface area of a hemisphere is $2\pi r^2$.)[3 marks]
Method
Curved surface $= 2\pi r^2 = 2 \times \pi \times 12^2 = 905$ cm².
Flat circular base $= \pi r^2 = \pi \times 12^2 = 452$ cm².
Total $= 2\pi r^2 + \pi r^2 = 3\pi r^2 = 1360$ cm² (3 s.f.).
Answer: $1360$
Marks
✔1 markCurved surface 2πr²
✔1 markAdd the base circle πr²
✔1 mark1360 cm²
2A hemisphere has a radius of $4$ cm.
Work out the volume of the hemisphere.
Give your answer correct to 3 significant figures.[3 marks]
Method
Volume of a full sphere $= \dfrac{4}{3}\pi r^3 = \dfrac{4}{3} \times \pi \times 4^3 = 268$ cm³.
Hemisphere $= \dfrac{1}{2} \times 268 = 134$ cm³ (3 s.f.).
Answer: $134$
Marks
✔1 markUse V = 4⁄3πr³
✔1 markSphere = 268, then halve
✔1 mark134 cm³
3A cylindrical tube has internal diameter $9$ cm and height $32$ cm.
Balls of diameter $5$ cm are dropped in so they stack on top of one another.
Work out the greatest number of whole balls that fit inside the tube.[3 marks]
Method
Each ball has diameter $5$ cm, which is at most the tube’s diameter ($9$ cm), so a ball fits across the tube.
Stacked up the height: $32 \div 5 = 6.40$.
Only whole balls count, so $6$ balls fit.
Answer: $6$
Marks
✔1 markA ball fits across the tube (ball diameter ≤ tube diameter)
✔1 markHeight ÷ ball diameter = 6.40
✔1 markRound down: 6 balls
4A pyramid has a square base of side $9$ cm and a volume of $270$ cm³.
The apex is directly above the centre of the base.
Work out the perpendicular height of the pyramid. (Volume of a pyramid $= \dfrac{1}{3} \times \text{base area} \times \text{height}$.)[3 marks]
Method
Base area $= 9^2 = 81$ cm².
$V = \dfrac{1}{3} \times \text{base area} \times h$, so $270 = \dfrac{1}{3} \times 81 \times h$.
Height $h = \dfrac{3V}{\text{base area}} = \dfrac{3 \times 270}{81} = 10$ cm.
Answer: $10$
Marks
✔1 markBase area = 81
✔1 markh = 3V ÷ base area
✔1 mark10 cm
5The diagram shows a frustum formed by removing a small cone from a larger cone.
The larger cone has base radius $15$ cm and slant height $39$ cm.
The frustum has a top radius of $5$ cm.
Work out the total surface area of the frustum.
Give your answer correct to 3 significant figures. (Curved surface area of a cone $= \pi r l$, where $l$ is the slant height.)[4 marks]
Method
The removed cone is similar, so its slant height $l = L \times \dfrac{r}{R} = 39 \times \dfrac{5}{15} = 13$ cm.
Curved surface of the frustum $= \pi R L - \pi r l = \pi(15\times39 - 5\times13) = 520π$ cm².
Add the base circle $\pi R^2 = 225π$ and the top circle $\pi r^2 = 25π$ cm².
Total $= 770π = 2420$ cm² (3 s.f.).
Answer: $2420$
Marks
✔1 markSimilar: small slant l = L·r/R = 13
✔1 markCurved = πRL − πrl = 520π
✔1 markAdd base πR² and top πr²
✔1 mark2420 cm² (3 s.f.)
mathedup.co.uk · sheet 2M7J