Learn on PengiReveal Math, AcceleratedUnit 12: Area, Surface Area, and Volume

Lesson 12-1: Describe Cross Sections of Three-Dimensional Figures

In this Grade 7 lesson from Reveal Math, Accelerated, students learn how to identify and describe cross sections formed by slicing three-dimensional figures such as rectangular prisms, square pyramids, and cones. Students explore how the orientation of the slice — parallel, perpendicular, or angled — determines whether the resulting two-dimensional shape is a square, rectangle, triangle, or other polygon. The lesson also applies cross section area calculations using formulas for squares and triangles in real-world contexts.

Section 1

Cross Sections: Slicing 3D Shapes

Property

A cross section is the 2D shape formed when a flat plane slices through a 3D solid figure. The resulting 2D shape depends entirely on whether the slice is horizontal, vertical, or angled.

Examples

  • Cylinder: A horizontal slice parallel to the bases produces a circle. A vertical slice straight down produces a rectangle.
  • Square Pyramid: A horizontal slice parallel to the base produces a square. A vertical slice directly through the apex produces a triangle.
  • Cube: A slice parallel to a face produces a square, but an angled slice through a cube can produce a rectangle, a triangle, or even a hexagon!

Explanation

Imagine playing a game where you slice through flying fruit with a sword! When your flat sword slices straight through a 3D object, the flat interior surface exposed by the cut is the cross section. A common misconception is that slicing a shape always produces its base shape. But remember: changing the angle of your slice can create entirely different shapes!

Section 2

The Building Blocks: Review of Basic Formulas

Property

To find the area of composite figures, you must first be a master of the basic 2D shape formulas:

  • Rectangle / Parallelogram:
    A=bhA = bh
    (base x height)
  • Triangle:
    A=12bhA = \frac{1}{2}bh
    (half of base x height)
  • Trapezoid:
    A=12(b1+b2)hA = \frac{1}{2}(b_1 + b_2)h
    (half of the sum of the bases x height)
  • Circle:
    A=πr2A = \pi r^2
    (pi x radius squared)

Examples

  • Triangle: A triangle with a base of 8 cm and a height of 5 cm has an area of 1/2 x 8 x 5 = 20 square cm.
  • Trapezoid: A trapezoid with bases of 6 m and 10 m, and a height of 4 m has an area of 1/2 x (6 + 10) x 4 = 32 square m.
  • Circle: A circle with a radius of 7 inches has an area of pi x 7 squared = 49pi square inches.

Explanation

Before tackling a giant composite figure, remember that every complex shape is just a combination of simple shapes. Think of these formulas as your geometric toolbox. As long as you know how to find the area of a basic rectangle, triangle, or circle, you can solve any complex shape puzzle by breaking it down!

Section 3

Dimensions of Cross Sections Using Similarity

Property

When a pyramid or cone is sliced by a plane parallel to its base, the resulting cross section is similar to the base. The dimensions of the cross section can be found using the linear scale factor, which is the ratio of the distance from the vertex to the total height:

dcdb=hcht\frac{d_c}{d_b} = \frac{h_c}{h_t}

Section 4

Areas of Similar Cross Sections

Property

When a three-dimensional figure (such as a cone or pyramid) is sliced parallel to its base, the resulting cross section is similar to the base. If the scale factor between the corresponding linear dimensions of the two similar cross sections is kk, then the ratio of their areas is k2k^2.

Area of Cross Section 1Area of Cross Section 2=k2=(Linear Dimension 1Linear Dimension 2)2\frac{\text{Area of Cross Section 1}}{\text{Area of Cross Section 2}} = k^2 = \left(\frac{\text{Linear Dimension 1}}{\text{Linear Dimension 2}}\right)^2

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Unit 12: Area, Surface Area, and Volume

  1. Lesson 1Current

    Lesson 12-1: Describe Cross Sections of Three-Dimensional Figures

  2. Lesson 2

    Lesson 12-2: Understand and Use Square Roots

  3. Lesson 3

    Lesson 12-3: Solve Problems Involving Area and Surface Area

  4. Lesson 4

    Lesson 12-4: Solve Problems Involving Circumference of Circles

  5. Lesson 5

    Lesson 12-5: Solve Problems Involving Area of Circles

  6. Lesson 6

    Lesson 12-6: Understand and Use Cube Roots

  7. Lesson 7

    Lesson 12-7: Solve Problems Involving Volume of Cylinders and Cones

  8. Lesson 8

    Lesson 12-8: Solve Problems Involving Volume of Spheres

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Cross Sections: Slicing 3D Shapes

Property

A cross section is the 2D shape formed when a flat plane slices through a 3D solid figure. The resulting 2D shape depends entirely on whether the slice is horizontal, vertical, or angled.

Examples

  • Cylinder: A horizontal slice parallel to the bases produces a circle. A vertical slice straight down produces a rectangle.
  • Square Pyramid: A horizontal slice parallel to the base produces a square. A vertical slice directly through the apex produces a triangle.
  • Cube: A slice parallel to a face produces a square, but an angled slice through a cube can produce a rectangle, a triangle, or even a hexagon!

Explanation

Imagine playing a game where you slice through flying fruit with a sword! When your flat sword slices straight through a 3D object, the flat interior surface exposed by the cut is the cross section. A common misconception is that slicing a shape always produces its base shape. But remember: changing the angle of your slice can create entirely different shapes!

Section 2

The Building Blocks: Review of Basic Formulas

Property

To find the area of composite figures, you must first be a master of the basic 2D shape formulas:

  • Rectangle / Parallelogram:
    A=bhA = bh
    (base x height)
  • Triangle:
    A=12bhA = \frac{1}{2}bh
    (half of base x height)
  • Trapezoid:
    A=12(b1+b2)hA = \frac{1}{2}(b_1 + b_2)h
    (half of the sum of the bases x height)
  • Circle:
    A=πr2A = \pi r^2
    (pi x radius squared)

Examples

  • Triangle: A triangle with a base of 8 cm and a height of 5 cm has an area of 1/2 x 8 x 5 = 20 square cm.
  • Trapezoid: A trapezoid with bases of 6 m and 10 m, and a height of 4 m has an area of 1/2 x (6 + 10) x 4 = 32 square m.
  • Circle: A circle with a radius of 7 inches has an area of pi x 7 squared = 49pi square inches.

Explanation

Before tackling a giant composite figure, remember that every complex shape is just a combination of simple shapes. Think of these formulas as your geometric toolbox. As long as you know how to find the area of a basic rectangle, triangle, or circle, you can solve any complex shape puzzle by breaking it down!

Section 3

Dimensions of Cross Sections Using Similarity

Property

When a pyramid or cone is sliced by a plane parallel to its base, the resulting cross section is similar to the base. The dimensions of the cross section can be found using the linear scale factor, which is the ratio of the distance from the vertex to the total height:

dcdb=hcht\frac{d_c}{d_b} = \frac{h_c}{h_t}

Section 4

Areas of Similar Cross Sections

Property

When a three-dimensional figure (such as a cone or pyramid) is sliced parallel to its base, the resulting cross section is similar to the base. If the scale factor between the corresponding linear dimensions of the two similar cross sections is kk, then the ratio of their areas is k2k^2.

Area of Cross Section 1Area of Cross Section 2=k2=(Linear Dimension 1Linear Dimension 2)2\frac{\text{Area of Cross Section 1}}{\text{Area of Cross Section 2}} = k^2 = \left(\frac{\text{Linear Dimension 1}}{\text{Linear Dimension 2}}\right)^2

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Unit 12: Area, Surface Area, and Volume

  1. Lesson 1Current

    Lesson 12-1: Describe Cross Sections of Three-Dimensional Figures

  2. Lesson 2

    Lesson 12-2: Understand and Use Square Roots

  3. Lesson 3

    Lesson 12-3: Solve Problems Involving Area and Surface Area

  4. Lesson 4

    Lesson 12-4: Solve Problems Involving Circumference of Circles

  5. Lesson 5

    Lesson 12-5: Solve Problems Involving Area of Circles

  6. Lesson 6

    Lesson 12-6: Understand and Use Cube Roots

  7. Lesson 7

    Lesson 12-7: Solve Problems Involving Volume of Cylinders and Cones

  8. Lesson 8

    Lesson 12-8: Solve Problems Involving Volume of Spheres