Volume of frustum radii 5 and 10, heights 6 and 12 Step by Step

Step by step

Question: a frustum is a cone with radius 10 and height 12, with a small cone of radius 5 and height 6 cut off the top: find its volume.

A frustum is a cone with its top cut off.

So its volume is the big cone minus the small cone.

The cone rule: one third times π r squared times the height.

10 squared is 100.

One third of 100 times 12 is 400π.

big cone = 13 × π × 10² × 12 = 400π

The same for the small cone: 50π.

small cone = 13 × π × 5² × 6 = 50π

Careful!

We cannot take away the radii first.

We work out the two whole cones, then take away.

Careful: not 13 × π × (10 − 5)² × 6 ✗

400π minus 50π is 350π.

That is about 1099.56, to 2 decimal places.

frustum = 400π − 50π = 350π ≈ 1099.56

Let's check: the frustum is smaller than the big cone and bigger than the small cone, as it should be.

And 5/10 equals 6/12: the cones are similar.

So it fits!

small cone 157.08 < 1099.56 < big cone 1256.64

Let's look back.

First: big cone.

Then: small cone.

Then: take away.

Last: check.

Answer

frustum = 400π − 50π = 350π ≈ 1099.56