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G06 • Lesson 47 of 105

Volume & Surface Area of 3D Solids

Extends 2D area concepts to three dimensions. Students learn to calculate the volume and surface area of prisms, cylinders, pyramids, cones, and spheres. Emphasizes understanding formulas through net diagrams and cross-sectional reasoning rather than pure memorization.

Middle School Geometry & Data • 6-8

Prerequisites: E09, G05

Key Concepts

  • volume of prisms and cylinders (V = Bh)
  • volume of pyramids, cones (V = 1/3 Bh), and spheres (V = 4/3 pi r^3)
  • surface area using nets and formulas
  • choosing appropriate units (square vs cubic)

Volume & Surface Area of 3D Solids

Three-dimensional shapes occupy volume (measured in cubic units like cm³) and have surface area (measured in square units like cm²). Volume tells you how much space is inside; surface area tells you how much material covers the outside. This lesson covers the major formulas.

Prisms and Cylinders (Two Identical Bases)

Both prisms and cylinders share the same volume idea: take the area of the base and multiply by the height.

V = B × h    (B = area of the base)

Rectangular Prism

Base = rectangle: B = l × w

V = l × w × h

SA = 2lw + 2lh + 2wh

Cylinder

Base = circle: B = pi × r²

V = pi × r² × h

SA = 2 × pi × r² + 2 × pi × r × h

Pyramids and Cones (One Base, Comes to a Point)

These shapes taper to a vertex, so their volume is exactly one-third of the corresponding prism or cylinder:

V = (1/3) × B × h

Pyramid (any polygon base)

V = (1/3) × B × h

SA = B + (1/2) × perimeter × slant height

Cone (circular base)

V = (1/3) × pi × r² × h

SA = pi × r² + pi × r × l   (l = slant height)

Spheres

V = (4/3) × pi × r³     SA = 4 × pi × r²

Net Diagrams

A net is a 2D pattern that folds into a 3D shape. To find surface area, imagine "unfolding" the shape flat. A cylinder's net is two circles plus a rectangle (whose width is the circumference 2 × pi × r). Drawing nets helps you see why the SA formulas work.

Example 1 — Volume of a Cylinder

A cylindrical water tank has radius 5 m and height 10 m. Find its volume.

  • B = pi × r² = 3.14 × 25 = 78.5 m².
  • V = B × h = 78.5 × 10 = 785 m³.
  • Example 2 — Volume of a Cone

    An ice cream cone has radius 3 cm and height 12 cm. Find its volume.

  • V = (1/3) × pi × r² × h = (1/3) × 3.14 × 9 × 12.
  • V = (1/3) × 339.12 = 113.04 cm³.
  • Example 3 — Surface Area of a Rectangular Prism

    A gift box measures 8 in × 5 in × 3 in. How much wrapping paper is needed (surface area)?

  • SA = 2lw + 2lh + 2wh
  • SA = 2(8)(5) + 2(8)(3) + 2(5)(3) = 80 + 48 + 30 = 158 in².
  • Common Mistake: Square vs. Cubic Units

    Surface area uses square units (cm², in²) because it measures flat area. Volume uses cubic units (cm³, in³) because it measures 3D space. Mixing them up is a frequent error. Always label your answer with the correct unit type.

    Practice Problems

    1. Find the volume of a rectangular prism: length 6 cm, width 4 cm, height 10 cm.

    Show Solution

    V = 6 × 4 × 10 = 240 cm³

    2. A sphere has radius 6 cm. Find its volume. (Use pi ≈ 3.14.)

    Show Solution

    V = (4/3) × 3.14 × 6³ = (4/3) × 3.14 × 216 = (4/3) × 678.24 = 904.32 cm³

    3. A pyramid has a square base with side 10 m and height 15 m. Find the volume.

    Show Solution

    B = 10² = 100 m². V = (1/3) × 100 × 15 = 500 m³

    4. Find the surface area of a cylinder with radius 3 cm and height 7 cm.

    Show Solution

    SA = 2 × pi × r² + 2 × pi × r × h = 2(3.14)(9) + 2(3.14)(3)(7) = 56.52 + 131.88 = 188.4 cm²

    5. A cone and a cylinder have the same base radius (4 cm) and same height (9 cm). How do their volumes compare?

    Show Solution

    Cylinder: V = pi × 16 × 9 = 144pi. Cone: V = (1/3) × pi × 16 × 9 = 48pi. The cone's volume is exactly one-third of the cylinder's.

    Lesson Summary

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