In this explainer, we will learn how to identify similar solids and use the similarity to find their dimensions, areas, and volumes.

### Definition: Similarity of Shapes

Two shapes or solids are similar if their corresponding sides are in the same proportion and their corresponding angles are equal.

Note that dilating a shape leads to a similar shape. Inversely, if two shapes are similar, there exists a dilation or compression that transforms one of the shapes into the other.

The important implication of the above definition is that the ratio of any two corresponding sides of a pair of similar shapes is constant.

We are going now to see how to use this to solve some question.

### Example 1: Using Similarity of Solids to Find an Unknown Length

Given that these two triangular prisms are similar, find the value of .

### Answer

Since the two prisms are similar, it means that the ratio of any two corresponding sides is constant (that is, it is the same for all pairs of corresponding sides). We are given one pair of corresponding sides: 29 mm and 14.5 mm. We see that the bigger shape, on the left, is obtained by enlarging the smaller shape, on the right, by a scale factor of . Any side in the bigger shape is twice the corresponding side in the smaller shape. Therefore, .

So, we have seen that any side in one of the shapes is equal to the corresponding side in the similar shape multiplied by a given factor. We can write this as: .

Let us think now of the **area** of one shape, for instance, a rectangle. Its area is
given by . If we have a similar rectangle to this one,
with for any pair of corresponding sides, we see that
.

For instance, in the diagram below, the bigger rectangle is obtained by enlarging the smaller
one by a factor of 1.5. The two rectangles are thus similar. The area of the smaller
rectangles is 4 small rectangles as shown. The area of the bigger rectangle is **not** 1.5
times the area of the smaller rectangle. We see that this is the area of a rectangle that we
get from the smaller rectangle by enlarging *only one* of its sides. The bigger
rectangle’s area is 9 small rectangles, and this is indeed small rectangles.

We can apply the same reasoning to the **volume** of a solid object. The volume is given
by multiplying an area (obtained by multiplying two lengths) by a third length. Let us look at
this example of a pentagonal prism.

The volume of the smaller prism is , and the volume of the bigger one is . Since the bigger one is obtained by enlarging the smaller one by a scale factor , we have and . Therefore, .

This can be easily visualized with a cuboid that is enlarged by, say, a factor of 3.

*Each* dimension is enlarged by a factor of 3, so the larger cuboid is made of 3 rows of
smaller cuboids. That is, in total, there are smaller cuboids in the bigger one.

When working with lengths, areas, and volumes, you need always to remember that if you are
working out a **length**, then you will maybe add lengths (for instance, to work out a
perimeter) or multiply a length by a factor. When instead you are calculating an **area**,
you will have to multiply **two lengths** together, and **three lengths** when it is a
**volume**. This will help you use the proper unit: for instance, **cm**,
**cm ^{2}** or

**cm**, remembering that the exponent indicates if it is a

^{3}**unit of length**: no exponent, or rather

**1**is the exponent, meaning the unit is not multiplied by itself, or a

**unit of area**: the exponent is

**2**(a length multiplied by a length implies the length unit multiplied by itself once), or a

**unit of volume**: the exponent is

**3**(a length multiplied by a length multiplied by a length).

Having this in mind is also very useful to convert units of area and volume properly. Look,
for instance, at this diagram, showing the factor between
1 mm^{3} (the smaller cube) and
1 cm^{3} (the bigger cube).

### Example 2: Finding the Scale Factor between Two Similar Solids Knowing Their Surface Areas

The surface areas of two similar solids are 64 square yards and 361 square yards. Find the ratio of their linear measures.

### Answer

We have here two similar solids, Shape 1 and Shape 2. This means that, for any pair of corresponding sides and , we have . It follows that, for any pair of corresponding areas and , we have , and, for any pair of corresponding volumes and , we have . Here, we are given a pair of corresponding areas, namely the surface areas of the two solids: 64 and 361 square yards. By looking at the possible answers for the ratio of the linear measures of the two shapes, , we see that is smaller than 1. It means that the bigger shape is Shape 1 and the smaller one is Shape 2. Therefore, here, and .

We can now find the ratio of their linear measures, , by using the relationship between and : . By plugging the values of and into the equation, we find Dividing both sides by 361: Taking the square root of both sides:

The ratio of the linear measures of the smaller shape to the bigger shape is .

### Example 3: Finding the Volume of a Solid Knowing the Volume of a Similar Solid

If the two given pyramids are similar and the volume of the larger pyramid is
160 m^{3}, find the volume of
the smaller one.

### Answer

We know that the pyramids are similar. Therefore, if is the scale factor between Pyramid 1 and Pyramid 2, we have .

We know here the volume of the larger pyramid. Let us say it is . We need to find the scale factor, , between the smaller pyramid and the larger one. We have then for any pair of corresponding sides and . We are given one pair of corresponding sides: , and . So, , and we find .

We can plug in this values of and into the equation . It gives Dividing each side of the equation by 8, we find

The volume of the smaller pyramid is
20 m^{3}.