Lesson Explainer: Similarity of Solids Mathematics • 8th Grade

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 π‘₯.


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 2914.5=2. Any side in the bigger shape is twice the corresponding side in the smaller shape. Therefore, π‘₯=2Γ—15.5=31mm.

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: 𝑙=π‘“Γ—π‘™οŠ¨οŠ§.

Let us think now of the area of one shape, for instance, a rectangle. Its area is given by 𝐴=π‘ŽΓ—π‘οŠ§οŠ§οŠ§. If we have a similar rectangle to this one, with 𝑙=π‘“Γ—π‘™οŠ¨οŠ§ for any pair of corresponding sides, we see that 𝐴=(π‘“Γ—π‘Ž)Γ—(𝑓×𝑏)=π‘“Γ—π‘“Γ—π΄οŠ¨οŠ§οŠ§οŠ§.

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 1.5Γ—1.5Γ—4 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 𝑉=π΄Γ—β„ŽοŠ§οŠ§οŠ§, and the volume of the bigger one is 𝑉=π΄Γ—β„ŽοŠ¨οŠ¨οŠ¨. Since the bigger one is obtained by enlarging the smaller one by a scale factor 𝑓, we have β„Ž=π‘“Γ—β„ŽοŠ¨οŠ§ and 𝐴=π‘“Γ—π‘“Γ—π΄οŠ¨οŠ§. Therefore, 𝑉=π‘“Γ—π‘“Γ—π΄Γ—π‘“Γ—β„Ž=π‘“Γ—π‘“Γ—π‘“Γ—π‘‰οŠ¨οŠ§οŠ§οŠ§.

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 3Γ—3=9 smaller cuboids. That is, in total, there are 3Γ—3Γ—3=27 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, cm2 or cm3, remembering that the exponent indicates if it is a 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 mm3 (the smaller cube) and 1 cm3 (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.

  1. 827
  2. 1927
  3. 819
  4. 64361
  5. 64425


We have here two similar solids, Shape 1 and Shape 2. This means that, for any pair of corresponding sides π‘™οŠ§ and π‘™οŠ¨, we have 𝑙=π‘“Γ—π‘™οŠ¨οŠ§. It follows that, for any pair of corresponding areas 𝐴 and 𝐴, we have 𝐴=π‘“Γ—π΄οŠ¨οŠ¨οŠ§, and, for any pair of corresponding volumes π‘‰οŠ§ and π‘‰οŠ¨, we have 𝑉=π‘“Γ—π‘‰οŠ¨οŠ©οŠ§. 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, 𝑓, we see that 𝑓 is smaller than 1. It means that the bigger shape is Shape 1 and the smaller one is Shape 2. Therefore, here, 𝐴=361squareyards and 𝐴=64squareyards.

We can now find the ratio of their linear measures, 𝑓, by using the relationship between 𝐴 and 𝐴: 𝐴=π‘“Γ—π΄οŠ¨οŠ¨οŠ§. By plugging the values of 𝐴 and 𝐴 into the equation, we find 64=𝑓×361. Dividing both sides by 361: 64361=𝑓×36136164361=𝑓. Taking the square root of both sides: ο„ž64361=βˆšπ‘“βˆš64√361=𝑓𝑓=819.

The ratio of the linear measures of the smaller shape to the bigger shape is 819.

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 m3, find the volume of the smaller one.


We know that the pyramids are similar. Therefore, if 𝑓 is the scale factor between Pyramid 1 and Pyramid 2, we have 𝑉=𝑓×𝑓×𝑓×𝑉=π‘“Γ—π‘‰οŠ¨οŠ§οŠ©οŠ§.

We know here the volume of the larger pyramid. Let us say it is 𝑉160ο…οŠ¨οŠ©m. We need to find the scale factor, 𝑓, between the smaller pyramid and the larger one. We have then 𝑙=π‘“Γ—π‘™οŠ¨οŠ§ for any pair of corresponding sides π‘™οŠ§ and π‘™οŠ¨. We are given one pair of corresponding sides: 𝑙=5m, and 𝑙=10m. So, 10=𝑓×5, and we find 𝑓=2.

We can plug in this values of 𝑓 and π‘‰οŠ¨ into the equation 𝑉=π‘“Γ—π‘‰οŠ¨οŠ©οŠ§. It gives 160=2×𝑉160=8×𝑉. Dividing each side of the equation by 8, we find 160Γ·8=8×𝑉÷820=𝑉.

The volume of the smaller pyramid is 20 m3.

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