Home / United States / Math Classes / 8th Grade Math / Surface Area and Volume of Similar Solids

Similar solids have the same shape but a different size. We can compare the surface area and volume of similar solids and find unknown values using the basic equation of surface area and volume. Check out some solved examples to learn how changes in dimensions affect the value of surface area and volume. ...Read MoreRead Less

Similar solids have the same shape but differ in size, implying that corresponding dimensions are proportional and corresponding faces of the solid are similar polygons.

In the below figures the two cylinders will be similar if the ratios of their corresponding radii and heights are the same. Similarly, the two cubes will be similar if the ratio of their corresponding side lengths is the same.

We can identify similar solids by observing their corresponding dimensions. If the corresponding dimensions are proportional then the solids are said to be similar.

Follow the given steps to identify similar solids.

**Step 1**: Determine the dimensions that correspond.**Step 2**: Calculate the length ratio of the corresponding dimensions.**Step 3**: Check to see if all of the ratios are the same; if this is so, the solids are similar.

Let us take an example to understand this better. The radius of Cylinder A is 2 units and the height is 3 units. The radius of Cylinder B is 4 units, and the height is 5 units. Check to see if these cylinders are similar.

**Solution:**

**Step 1**: The height and radius of the cylinders are the corresponding dimensions.

- The radius of Cylinder A is 2 and the height is 3.
- The radius of Cylinder B is 4 and the height is 5.

**Step 2**: Calculate the length ratio of the corresponding dimensions.

- The radii have a ratio of \(\frac{2}{4}=\frac{1}{2}\).
- The height ratio is \(\frac{3}{5}\).

**Step 3**: The dimensions have different ratios.

\(\frac{1}{2}\neq\frac{3}{5}\)

As a result, these cylinders are not similar .

We can use the concept of proportion or equal ratio of the corresponding dimensions of similar solids to find the unknown measures in any of the given similar solids.

Let’s take an example and see how to measure the missing measures in such cases.

We have two similar figures as shown. Find the missing height \(h\).

**Solution**:

Two solids are similar if the length ratio of the corresponding dimensions are same, that is,

\(\frac{1.5}{h}=\frac{3}{1}\)

\(h=0.5\) cm

So the value of \(h\) is \(0.5\) cm.

The surface area of a solid object is a measurement of the total area occupied by the object’s surface.

When two solids are similar, the square of the ratio of their corresponding linear measures equals the value of the ratio of their corresponding surface areas.

Let us consider solid A and solid B to be similar solids. Solid A and solid B have the radius as \(a\) units, and \(b\) units, respectively.

So the ratio of their corresponding surface areas will be given by,

\(\frac{\text{Surface area of A}}{\text{Surface area of B}}=\left(\frac{a}{b}\right)^2\)

The volume of each of the similar solids is a measurement of the quantity of space occupied by each of the solids. When two solids are similar, the cube of the ratio of their corresponding linear measures equals the value of the ratio of their volumes.

Let us take the same set of similar solids that are two similar cylinders, solid A and solid B. Solid A and solid B have the radius a units and b units, respectively.

So the ratio of their corresponding volumes will be given by,

\(\frac{\text{Volume of A}}{\text{Volume of B}}=\left(\frac{a}{b}\right)^3\)

**Example 1**:

Are the given two rectangular prisms similar? How will you figure this out?

**Solution**:

The two prisms are similar if the measure of corresponding heights, widths, and lengths are all in proportion.

\(\frac{\text{small prism}}{\text{large prism}}=\frac{3}{4.5}=\frac{1}{\frac{4.5}{3}}=\frac{1}{1.5}\)

\(\frac{\text{small prism}}{\text{large prism}}=\frac{4}{6}=\frac{1}{\frac{6}{4}}=\frac{1}{1.5}\) [Simplify]

\(\frac{\text{small prism}}{\text{large prism}}=\frac{5}{7.5}=\frac{1}{\frac{7.5}{5}}=\frac{1}{1.5}\) [Simplify]

The ratio of all three corresponding sides is equal to \(\frac{1}{1.5}\). This indicates the two rectangular prisms are proportional, owing to a constant ratio.

**Example 2**:

The following figures are similar. What is \(x\)?

**Solution**:

\(\frac{\text{Height of smaller triangular prism}}{\text{Height of larger triangular prism}} =\frac{\text{length of smaller triangular prism}}{\text{Length of larger triangular prism}}\)

\(\frac{2}{4}=\frac{4}{x}\) Substitute.

\(2\times x=4\times4\) Cross Multiply

\(2x=16\) Simplify.

\(2x\div2=16\div2\) Divide both sides by 2.

\(x=8\)

The length that is missing is 8 centimetres.

**Example 3:**

The following pyramids are similar and the larger pyramid has a surface area of 392 square centimeters. What is the surface area of the smaller pyramid?

**Solution**:

\(\frac{\text{Surface area of the smaller pyramid}}{\text{Surface area of the larger pyramid}} =\left(\frac{\text{Base of the smaller pyramid}}{\text{Base of the larger pyramid}}\right)^2\)

\(\frac{S}{392}\ =\left(\frac{10}{14}\right)^2\) Substitute

\(\frac{S}{392}=\frac{100}{196}\) Evaluate

\(S=200\) Multiply each side by 392.

Hence, the surface area of smaller pyramid is 200 square centimeters.

**Example 4**:

The two cylinders are similar. What is the volume of the larger cylinder if the volume of the smaller cylinder is 40 cubic feet?

**Solution:**

\(\frac{\text{Volume of the larger cylinder}}{\text{Volume of the smaller cylinder}} =\left(\frac{\text{Radius of the larger cylinder}}{\text{Radius of the smaller cylinder}}\right)^3\)

\(\frac{V}{40}\ =\left(\frac{3}{2}\right)^3\) Substitute

\(\frac{V}{40}=\frac{27}{8}\) Evaluate

\(V=135\) Multiply each side by 40

Hence, the volume of larger cylinder is 135 cubic feet.

**Example 5:**

The cylindrical pitcher below has a surface area of around square inches. The pitcher and creamer are similar solids. Determine the creamer’s surface area.

**Solution**:

\(\frac{\text{Surface area of the creamer}}{\text{Surface area of the pitcher}} =\left(\frac{\text{Diameter of the creamer}}{\text{Diameter of the pitcher}}\right)^2\)

\(\frac{S}{90}\ =\left(\frac{3}{6}\right)^2\) Substitute

\(\frac{S}{90}=\frac{9}{36}\) Evaluate

\(S=22.5\) Multiply each side by 90

Hence, the surface area of creamer is 22.5 square inches.

**Example 6:**

The cylindrical pitcher below has a volume of approximately cubic inches. Both the pitcher and the creamer are similar solids. Calculate the volume of the creamer.

**Solution:**

\(\frac{\text{Volume of the creamer}}{\text{Volume of the pitcher}} =\left(\frac{\text{Diameter of the creamer}}{\text{Diameter of the pitcher}}\right)^3\)

\(\frac{V}{157}\ =\left(\frac{3}{6}\right)^3\) Substitute

\(\frac{V}{157}=\frac{27}{216}\) Evaluate

\(V=19.625\) Multiply each side by 157

Hence, the volume of the creamer is 19.625 cubic inches.

Frequently Asked Questions on Similar Solids

Similar solids have the same shape but have different sizes. The measure of their corresponding dimensions are proportional, and the shape of their corresponding faces are the same.

Two solids are similar if the ratio of length of their corresponding radii, heights, base lengths, widths, or any other dimensions are equal.