Mensuration

Compound solids and frustums

3D Upright And Conical Paper 1 + 2 8 min read

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In this note
Formula bank
Joined solids
Frustums
Melting, pouring and rising water
Ways they test it
Mistakes to avoid
Summary
Practice
Can I do this?
Quiz

In this note

Syllabus 5.5: compound shapes and solids

5.5Find volumes and surface areas of compound solids made of the shapes you know.
5.5Find the volume of a frustum as big cone − small cone.
5.5Solve 'melted and recast' and 'water rises' problems by setting two volumes equal.

Before you start: Prisms and cylinders and Pyramids, cones and spheres.

Formula bank

Compound volumeLearn it\(V=V_1+V_2\) (or \(V_1-V_2\) for a hole)split into shapes you know.Use it when joined or hollowed solids.

FrustumLearn it\(V=V_{\text{big cone}}-V_{\text{small cone}}\)the small cone was cut off the top.Use it when a bucket, lampshade or cut cone.

Same volumeLearn it\(V_{\text{before}}=V_{\text{after}}\)melting, pouring or reshaping keeps the volume.Use it when 'melted', 'recast', 'poured'.

Water riseLearn it\(\text{rise}=\dfrac{\text{volume added}}{\text{base area of tank}}\)the water forms a thin prism.Use it when 'the water level rises by'.

Joined solids

Cone + hemisphere: add the two volumes.

A compound solid is two (or more) solids you know stuck together. Find each volume and add. For SURFACE area, only count the outside: where two solids join, the joined faces are hidden and are NOT counted.

General methodHow to answer any compound-solid question

Split the solid into shapes you know; mark each one's measurements.
Volume: add the parts (subtract any hole).
Surface area: list only the outside faces; leave out the joined circles or faces.
Keep everything in terms of \(\pi\) until the end, then round.

Example A solid is a cylinder (radius 3 cm, height 8 cm) with a hemisphere of radius 3 cm on top. Find its volume.

Cylinder + half a sphere.

cylinder \(=\pi(9)(8)=72\pi\)
hemisphere \(=\tfrac23\pi(27)=18\pi\)
total \(=90\pi\approx282.74\)

Answer 282.74 cm³

Example The same solid: find its total surface area.

Outside faces: base circle, curved cylinder, curved hemisphere. The circle where they join is hidden.

base \(\pi(9)=9\pi\)
curved cylinder \(2\pi(3)(8)=48\pi\)
curved hemisphere \(2\pi(9)=18\pi\)
total \(=75\pi\approx235.62\)

Answer 235.62 cm²

Try it yourself

A cuboid 5 cm × 5 cm × 2 cm has a square-based pyramid of height 6 cm on its top face. Volume (cm³)?

Show answer

Answer 100

Cylinder r 4 cm, h 10 cm with a hemisphere r 4 cm on one end. Total surface area in terms of π? Type the number in front of π.

Hint

Base circle + curved cylinder + curved hemisphere.

Show answer

Answer 128

Frustums

Frustum = big cone − small cone.

A frustum is what is left when the top of a cone is cut off parallel to the base. You are usually given (or can find) the heights of both cones.

Example A cone of radius 6 cm and height 12 cm has a cone of radius 3 cm and height 6 cm cut off its top. Find the volume of the frustum.

Big − small.

big \(=\tfrac13\pi(36)(12)=144\pi\)
small \(=\tfrac13\pi(9)(6)=18\pi\)
frustum \(=126\pi\approx395.84\)

Answer 395.84 cm³

Melting, pouring and rising water

General methodHow to answer any 'same volume' question

Find the volume you know (the old shape, or the water added).
Write the formula for the new shape with the unknown in it.
Set them equal and solve.

Example A metal sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the height of the cylinder.

Sphere volume = cylinder volume.

\(\tfrac43\pi(216)=288\pi\)
\(\pi(16)h=288\pi\)
\(h=18\)

Answer 18 cm

Example 30 litres of water are poured into an empty tank with a rectangular base 50 cm by 40 cm. How deep is the water?

30 litres = 30 000 cm³.

depth \(=\dfrac{30\,000}{50\times40}=15\)

Answer 15 cm

Try it yourself

A sphere of radius 3 cm has the same volume as a cylinder of radius 4 cm. Height of the cylinder (cm)?

Show answer

Answer 2.25

30 litres of water go into a tank with base 50 cm by 40 cm. Depth (cm)?

Show answer

Answer 15

Ways they test it

Type 1 Joined solid volume

"The solid is a cone joined to a hemisphere. Find its volume." (3 marks)

Spot it a diagram of two solids.

Volume of each part.
Add.

Type 2 Outside surface

"Find the total surface area of the solid." (4 marks)

Spot it two solids joined.

List the outside faces only.
Add their areas.

Type 3 Frustum

"Find the volume of the frustum." (3 marks)

Spot it a cut cone or a bucket.

Big cone − small cone.

Type 4 Melted / recast

"The sphere is melted and made into a cylinder. Find its height." (3 marks)

Spot it 'melted', 'recast', 'poured'.

Old volume = new volume.
Solve.

For example Sphere r 6 into cylinder r 4

\(288\pi=16\pi h\)

Answer h = 18 cm

Type 5 Water rises

"A ball is dropped into the tank. By how much does the water rise?" (3 marks)

Spot it an object put into water.

Volume of the object.
÷ base area of the tank.

Mistakes to avoid

✗ Counting the joined circle in the surface area✓ leave out faces that are glued togetherThey are inside the solid.
✗ Frustum as a cone of the frustum's height✓ big cone − small cone
✗ Litres straight into cm✓ 1 litre = 1000 cm³Convert first.
✗ Rounding each part✓ keep \(\pi\) until the endRounding early loses accuracy marks.

Summary

Compound volume = add the parts (subtract holes).
Compound surface = outside faces only.
Frustum = big cone − small cone.
Melting or pouring keeps the volume the same.

If you see …

two solids joinedsplit and add.
'melted' or 'poured'set volumes equal.
'rises by'volume ÷ tank base area.

Practice

Practice: easy, medium, hard

Easy

Cylinder r 3, h 8 with a hemisphere r 3 on top: volume in terms of π? Type the number in front of π.

Show answer

Answer 90

How many cm³ in 30 litres?

Show answer

Answer 30000

A cone has volume \(144\pi\) and a smaller cone cut from it has volume \(18\pi\). Frustum volume in terms of π? Type the number in front of π.

Show answer

Answer 126

Medium

Sphere radius 6 cm recast as a cylinder of radius 4 cm. Height (cm)?

Guide me

What is it asking? a height

What do we know? sphere \(288\pi\)

Plan \(16\pi h=288\pi\)

Show working

Answer 18

A cuboid 5 × 5 × 2 cm with a pyramid of height 6 cm on the 5 × 5 face. Volume (cm³)?

Guide me

What is it asking? a volume

What do we know? 50 + pyramid

Plan \(\tfrac13(25)(6)=50\)

Show working

Answer 100

30 litres into a 50 cm by 40 cm tank. Depth (cm)?

Guide me

What is it asking? a depth

What do we know? 30 000 cm³

Plan ÷ 2000

Show working

Answer 15

Hard

Frustum: big cone radius 6 cm height 12 cm, small cone radius 3 cm height 6 cm removed. Volume (2 d.p.)?

Guide me

What is it asking? a volume

What do we know? \(144\pi\) and \(18\pi\)

Plan subtract

Show working

\(126\pi\)

Answer 395.84

Part	Answer / working	Marks
	\(\tfrac13\pi(36)(12)-\tfrac13\pi(9)(6)\)	M2
	395.84	A1
Total: 3 marks

Cylinder radius 3 cm, height 8 cm with a hemisphere of radius 3 cm on top. Total surface area (2 d.p.)?

Guide me

What is it asking? an area

What do we know? outside faces: base, curved cylinder, curved hemisphere

Plan \(9\pi+48\pi+18\pi\)

Show working

\(75\pi\)

Answer 235.62

Part	Answer / working	Marks
	\(2\pi(3)(8)\) and \(2\pi(3)^2\)	M1
	+ base \(\pi(3)^2\)	M1
	235.62	A1
Total: 3 marks

Cone, radius 4 cm, slant height 10 cm, with a hemisphere of radius 4 cm on its base. Total surface area (2 d.p.)?

Guide me

What is it asking? an area

What do we know? cone curved \(\pi rl=40\pi\)

Plan hemisphere curved \(2\pi r^2=32\pi\)

Show working

\(72\pi\)

Answer 226.19

Part	Answer / working	Marks
	\(\pi(4)(10)\)	M1
	\(2\pi(4)^2\)	M1
	226.19	A1
Total: 3 marks

Can I do this?

I can find the volume and outside surface area of a joined solid.
I can find the volume of a frustum.
I can solve melting and rising-water problems.

Anything unticked: revise that section first. All ticked: take the quiz.

Quiz

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