Ratios of Lengths, Areas, and Volumes

Ratios of Lengths, Areas, and Volumes

How to solve the ratios of lengths, areas, and volumes problem: formula, examples, and their solutions.

Formula

If the ratio of the similar figures' lengths (1D) is (a/b), then the ratio of their areas (2D) is (a^2/b^2), and the ratio of their volumes (3D) is (a^3/b^3).

The purple figure and the brown figure are similar.

If the ratio of their lengths (1D) is a/b,
then the ratio of their areas (2D) is a2/b2,
and the ratio of their volumes (3D) is a3/b3.

The exponents are related to the concept's dimension:
2D → a2/b2
3D → a3/b3

Example 1

Two similar right cones are given below. Find the ratio of the lateral areas of the cones. The radius of the left cone's base: 3. The radius of the right cone's base: 5.

The ratio of the radius (1D) is 3/5.

And the lateral area (A) is a 2D concept.

So A/A' = 32/52.

Example 2

Two similar right cones are given below. Find the ratio of the volumes of the cones. The radius of the left cone's base: 3. The radius of the right cone's base: 5.

The ratio of the radius (1D) is 3/5.

And the volume (V) is a 3D concept.

So V/V' = 33/53.

Example 3

Two similar trapezoids are given below. Find the area of the big trapezoid. The little trapezoid's bases: 2, 4. The little trapezoid's height: 2. The big trapezoid's height: 3.

Find the area of the little trapezoid:
A = (1/2)⋅(3 + 4)⋅2 = 7

Area of a trapezoid

The ratio of the heights (1D) is 2/3.

And the area (A) is a 2D concept.

So A/A' = 22/32.

A = 7

So 7/A' = 4/9.