Challenge · Surface Area & Volume: Cones, Spheres & Pyramids

Calculator allowed14 marks

These problems mix the volume and surface-area formulas, slant height (Pythagoras) and composite solids. Give answers to 3 s.f. unless asked for an exact value. A worked example shows how to start; hints are at the foot. Calculator; use the $\pi$ button.

Worked example: how to start
A solid is a cone of radius $3$ cm and height $4$ cm sitting on a hemisphere of radius $3$ cm. Work out its total volume, to 3 s.f.
Cone: $V=\tfrac13\pi(3)^2(4)=12\pi$.
Hemisphere: $V=\tfrac12\cdot\tfrac43\pi(3)^3=18\pi$.
Total $=12\pi+18\pi=30\pi=94.2$ cm$^3$.
1Work out the volume of a sphere of radius $6$ cm, to 3 s.f.[2 marks]
Answer:
2A solid cone has radius $5$ cm and perpendicular height $12$ cm. Work out its curved surface area, to 3 s.f.[3 marks]
Answer:
3A square-based pyramid has a base of side $6$ cm and perpendicular height $10$ cm. Work out its volume.[2 marks]
Answer:
4A sphere has a volume of $500\text{ cm}^3$. Work out its radius, to 3 s.f.[3 marks]
Answer:
5A toy is a cone of radius $3$ cm and height $4$ cm on top of a hemisphere of radius $3$ cm. Work out the total volume, giving your answer in terms of $\pi$.[4 marks]
Answer:
★Extension: A cone and a sphere have the same radius $r$ and the same volume. Show that the height of the cone is $4r$.
Stuck? Hints (don't peek unless you need to)1. Use $V=\tfrac43\pi r^3$.2. Find the slant height $l$ with Pythagoras first, then use $\pi r l$.3. Volume $=\tfrac13\times\text{base area}\times h$.4. Put the volume equal to $\tfrac43\pi r^3$ and rearrange for $r$.5. Work out the cone and hemisphere volumes separately, then add.

Solutions & mark scheme · Surface Area & Volume: Cones, Spheres & Pyramids

Total: 14 marks

Award the marks shown for each correct step; many of these have more than one valid route, so give method marks for any correct working.

1Work out the volume of a sphere of radius $6$ cm, to 3 s.f.[2]
Model solution
$V=\tfrac43\pi(6)^3=288\pi$.
$=905$ cm$^3$ (3 s.f.).
Answer: $905\text{ cm}^3$
Marks
✔1$V=288\pi$
✔1$=905$ cm$^3$
2A solid cone has radius $5$ cm and perpendicular height $12$ cm. Work out its curved surface area, to 3 s.f.[3]
Model solution
$l^2=5^2+12^2=169$, so $l=13$ cm.
Curved SA $=\pi r l=\pi(5)(13)=65\pi$.
$=204$ cm$^2$ (3 s.f.).
Answer: $204\text{ cm}^2$
Marks
✔1Slant height $l=13$
✔1$\pi r l=65\pi$
✔1$=204$ cm$^2$
3A square-based pyramid has a base of side $6$ cm and perpendicular height $10$ cm. Work out its volume.[2]
Model solution
Base area $=6\times6=36$ cm$^2$.
$V=\tfrac13\times36\times10=120$ cm$^3$.
Answer: $120\text{ cm}^3$
Marks
✔1Base area $=36$
✔1$V=120$ cm$^3$
4A sphere has a volume of $500\text{ cm}^3$. Work out its radius, to 3 s.f.[3]
Model solution
$\tfrac43\pi r^3=500$, so $r^3=\dfrac{3\times500}{4\pi}=119.4\ldots$
$r=\sqrt[3]{119.4}=4.92$ cm (3 s.f.).
Answer: $4.92\text{ cm}$
Marks
✔1Rearranges to $r^3=\dfrac{1500}{4\pi}$
✔1$r^3=119\ldots$
✔1$r=4.92$ cm
5A toy is a cone of radius $3$ cm and height $4$ cm on top of a hemisphere of radius $3$ cm. Work out the total volume, giving your answer in terms of $\pi$.[4]
Model solution
Cone: $\tfrac13\pi(3)^2(4)=12\pi$.
Hemisphere: $\tfrac12\cdot\tfrac43\pi(3)^3=18\pi$.
Total $=12\pi+18\pi=30\pi$ cm$^3$.
Answer: $30\pi\text{ cm}^3$
Marks
✔1Cone $=12\pi$
✔1Hemisphere $=18\pi$
✔1Adds the two parts
✔1$=30\pi$ cm$^3$
★Extension
Equal volumes: $\tfrac13\pi r^2 h=\tfrac43\pi r^3$.
Divide both sides by $\tfrac13\pi r^2$: $h=4r$.
mathedup.co.uk · sheet Q937