# Ratios of Lengths, Areas, and Volumes

How to find the ratios of the lengths, areas, and volumes of the similar figures: formulas, examples, and their solutions.

## Formulas

Here are two similar figures.

If the ratio of their lengths is
a/b,

then the ratio of the other lengths (1D) is also
a/b (= a1/b1).

Lengths: sides, edges, diagonals, etc.

Ratio

If the ratio of their lengths is
a/b,

then the ratio of their areas (2D) is
a2/b2.

Areas: surface area, faces, etc.

If the ratio of their lengths is
a/b,

then the ratio of their volumes (3D) is
a3/b3.

Volumes: total volume, sliced volume, etc.

As you can see,
the dimension determines the exponents.

1D → a1/b1
2D → a2/b2
3D → a3/b3

## Example 1

It says these two cones are similar.

The ratio of their radii (1D) is
3/5.

So the ratio of their lateral areas (2D) is
A/A' = 32/52.

32 = 9
52 = 25

So A/A' = 9/25.

## Example 2

It says these two cones are similar.

The ratio of their radii (1D) is
3/5.

So the ratio of their volumes (3D) is
V/V' = 33/53.

33 = 27
53 = 125

So V/V' = 27/125.

## Example 3

To use the ratios of lengths and areas,
first find the area of the left trapezoid.
(The ratio of the heights is given: 2/3.)

b1 = 3
b2 = 4
h = 2

So A = (1/2)(3 + 4)⋅2.

Area of a trapezoid

Cancel (1/2) and 2.

3 + 4 = 7

So A = 7.

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

A = 7

So the ratio of their areas (2D) is
7/A' = 22/32.

22 = 4
32 = 9

So 7/A' = 4/9.

Solve the proportion.

Then 4A' is equal to, 7⋅9, 63.

Proportion

Divide both sides by 4.

Then A' = 63/4.