CDS Mathematics · Mensuration 3D

Frustums & Combined Solids

The frustum of a cone (buckets, glasses, lamp shades), a cone cut parallel to its base, and solids built from a cylinder, a cone and a hemisphere.

All Mensuration 3D notes CDS Mathematics strategy

Why this matters

The largest page of the chapter after the cuboid pair, and almost all MODERATE. Tents and buildings made of joined solids recur every few papers; the only judgement needed is which surfaces are on the outside.

Concept 1 of 3: The frustum

A frustum is a cone with its top sliced off. Its volume formula looks like a cone's with the radius replaced by an 'average' of the two radii; its slant height is Pythagoras on the height and the DIFFERENCE of the radii.

Definition

For end radii $R > r$ and height $h$ :

Volume $= \dfrac13\pi h\left(R^2 + Rr + r^2\right)$ .
Slant height $l = \sqrt{h^2 + (R - r)^2}$ .
Curved surface $= \pi(R + r)l$ ; add $\pi R^2 + \pi r^2$ for the whole surface (an open bucket adds only the bottom).
Completing the cone: the missing top cone has height $H$ with $\dfrac{H}{H + h} = \dfrac rR$ .

Frustum

V = \tfrac13\pi h(R^2 + Rr + r^2), \qquad l = \sqrt{h^2 + (R - r)^2}

Worked example

A bucket has end radii

14

cm and

7

cm and height

12

cm. Find its capacity.

(\pi = \tfrac{22}{7})

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (I) 2019 — Elementary Mathematics · Q68Moderate

Example 1 · Mensuration 3D · Frustums and Combined Solids

A bucket is in the form of a truncated cone. The diameters of the base and top of the bucket are 6 cm and 12 cm respectively. If the height of the bucket is 7 cm, what is the capacity of the bucket ?

Slant height uses the difference of the radii

l = \sqrt{h^2 + (R - r)^2}

, not

\sqrt{h^2 + R^2}

. Using the full radius gives the slant of the whole cone, which is longer.

Drill 6 more on the frustum

Concept 2 of 3: A cone cut parallel to its base

The piece cut off the top is a small cone similar to the whole. Every length scales by the same factor

k

, surfaces by

k^2

and volumes by

k^3

. The frustum is what is left.

Definition

If the top cone has $k$ times the height of the whole:

Its volume is $k^3$ of the whole; the frustum's is $1 - k^3$ .
Its curved surface is $k^2$ of the whole; the frustum's is $1 - k^2$ .
The cut is $kH$ below the apex, i.e. $(1 - k)H$ above the base.

Similar top cone

\frac{V_{\text{top}}}{V} = k^3, \qquad \frac{S_{\text{top}}}{S} = k^2

Worked example

A cone

40

cm high is cut parallel to its base so that the top cone has

\dfrac18

of the volume. How far above the base is the cut?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2022 · CDS (II) 2022 — Elementary Mathematics · Q81Moderate

Example 2 · Mensuration 3D · Frustums and Combined Solids

The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume is

\frac{1}{27}

of the volume of the given cone, then what is the height of the frustum of the cone ?

Height from the base, or from the apex?

k

measures the top cone from the APEX. A question asking for the height of the cut above the base wants

1 - k

of the height, and the

k

value is the planted option.

Drill 3 more on a cone cut parallel to its base

Concept 3 of 3: Solids joined together

Volumes of joined solids simply add. Surfaces do not: where two solids meet, the joining circle is hidden inside, so count only what the outside air touches.

Definition

Tent (cylinder with a cone on top): canvas $= 2\pi rh + \pi rl$ ; no floor, no joining circle.
Building (cylinder with a hemispherical dome): air $= \pi r^2h + \dfrac23\pi r^3$ .
Cone on a hemisphere: surface $= \pi rl + 2\pi r^2$ .
A hemisphere on a cube's face: the cube loses $\pi r^2$ and gains $2\pi r^2$ , a net $+\pi r^2$ .
A cone hollowed out of a cylinder: the cavity adds the cone's curved surface.

Cylinder-and-cone tent

\text{canvas} = 2\pi r h + \pi r l

Worked example

A tent is a cylinder of radius

14

m and height

3

m, topped by a cone of the same radius with slant height

25

m. Find the canvas needed.

(\pi = \tfrac{22}{7})

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2021 · CDS (II) 2021 — Elementary Mathematics · Q69Moderate

Example 3 · Mensuration 3D · Frustums and Combined Solids

A cone of height 16 cm and diameter 14 cm is mounted on a hemisphere of same diameter. What is the volume of the solid thus formed ? (take

\pi = \frac{22}{7}

)

The joining circle is not a surface

Where the cone meets the hemisphere, or the dome meets the cylinder, the circle is inside the solid. Adding

\pi r^2

for it is the commonest overcount.

Drill 10 more on solids joined together

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The frustum
Frustum
$V = \tfrac13\pi h(R^2 + Rr + r^2), \qquad l = \sqrt{h^2 + (R - r)^2}$
A cone cut parallel to its base
Similar top cone
$\frac{V_{\text{top}}}{V} = k^3, \qquad \frac{S_{\text{top}}}{S} = k^2$
Solids joined together
Cylinder-and-cone tent
$\text{canvas} = 2\pi r h + \pi r l$

Watch out for (3)

Slant height uses the difference of the radii
→ The frustum
Height from the base, or from the apex?
→ A cone cut parallel to its base
The joining circle is not a surface
→ Solids joined together

Test yourself on Mensuration 3D

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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