Learn on PengiBig Ideas Math, Course 2, AcceleratedChapter 5: Volume and Similar Solids

Lesson 4: Surface Areas and Volumes of Similar Solids

In this Grade 7 lesson from Big Ideas Math Course 2 Accelerated, students explore how surface area and volume of similar solids change when dimensions are scaled by a factor of k, discovering that surface area scales by k² and volume scales by k³. Students learn to identify similar solids using proportional corresponding dimensions and apply these relationships to solve problems involving cylinders and pyramids. The lesson directly supports Standard 8.G.9 and builds fluency with scale factor reasoning across three-dimensional figures.

Section 1

Identifying Similar Solids

Property

Two solids of the same type are similar if the ratios of their corresponding linear measures (such as heights, radii, or side lengths) are equal. This common ratio is called the scale factor.

lengthAlengthB=widthAwidthB=heightAheightB=scale factor \frac{\text{length}_A}{\text{length}_B} = \frac{\text{width}_A}{\text{width}_B} = \frac{\text{height}_A}{\text{height}_B} = \text{scale factor}

Section 2

Finding Missing Dimensions in Similar Solids

Property

If two solids are similar, then the ratio of their corresponding linear measures is constant. For any two corresponding lengths, a1a_1 and a2a_2 from the first solid, and b1b_1 and b2b_2 from the second solid, the following proportion is true:

a1b1=a2b2\frac{a_1}{b_1} = \frac{a_2}{b_2}

Examples

Section 3

Ratio of Surface Areas in Similar Solids

Property

If two similar solids have a ratio of corresponding linear measures of a:ba:b, then the ratio of their surface areas is a2:b2a^2:b^2.

Surface Area of Solid ASurface Area of Solid B=(ab)2\frac{{\text{Surface Area of Solid A}}}{{\text{Surface Area of Solid B}}} = \left(\frac{{a}}{{b}}\right)^2

Section 4

Volume Principle for Similar Objects

Property

If we multiply each dimension of a three-dimensional object by kk, then:

  1. The new object is similar to the original object, and
  2. The volume of the new object is k3k^3 times the volume of the original object.

Examples

  • A cube with a side length of 2 cm has a volume of 8 cm3^3. If you scale it by a factor of k=4k=4, the new side is 8 cm and the new volume is 83=5128^3 = 512 cm3^3, which is 43×8=64×84^3 \times 8 = 64 \times 8.
  • A model car is built at a 1:201:20 scale. The volume of the model is (120)3=18000(\frac{1}{20})^3 = \frac{1}{8000} times the volume of the actual car.

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Identifying Similar Solids

Property

Two solids of the same type are similar if the ratios of their corresponding linear measures (such as heights, radii, or side lengths) are equal. This common ratio is called the scale factor.

lengthAlengthB=widthAwidthB=heightAheightB=scale factor \frac{\text{length}_A}{\text{length}_B} = \frac{\text{width}_A}{\text{width}_B} = \frac{\text{height}_A}{\text{height}_B} = \text{scale factor}

Section 2

Finding Missing Dimensions in Similar Solids

Property

If two solids are similar, then the ratio of their corresponding linear measures is constant. For any two corresponding lengths, a1a_1 and a2a_2 from the first solid, and b1b_1 and b2b_2 from the second solid, the following proportion is true:

a1b1=a2b2\frac{a_1}{b_1} = \frac{a_2}{b_2}

Examples

Section 3

Ratio of Surface Areas in Similar Solids

Property

If two similar solids have a ratio of corresponding linear measures of a:ba:b, then the ratio of their surface areas is a2:b2a^2:b^2.

Surface Area of Solid ASurface Area of Solid B=(ab)2\frac{{\text{Surface Area of Solid A}}}{{\text{Surface Area of Solid B}}} = \left(\frac{{a}}{{b}}\right)^2

Section 4

Volume Principle for Similar Objects

Property

If we multiply each dimension of a three-dimensional object by kk, then:

  1. The new object is similar to the original object, and
  2. The volume of the new object is k3k^3 times the volume of the original object.

Examples

  • A cube with a side length of 2 cm has a volume of 8 cm3^3. If you scale it by a factor of k=4k=4, the new side is 8 cm and the new volume is 83=5128^3 = 512 cm3^3, which is 43×8=64×84^3 \times 8 = 64 \times 8.
  • A model car is built at a 1:201:20 scale. The volume of the model is (120)3=18000(\frac{1}{20})^3 = \frac{1}{8000} times the volume of the actual car.