Surface area of prisms & pyramids

Use the shape of every face.

Learning goal

Prisms have matching parallel bases

Calculate surface areas by adding the bases and lateral faces of prisms and pyramids.

For a right prism, SA = 2B + Ph, where B is the area of a base, P is its perimeter and h is the distance between bases. A net explains the lateral rectangles.

Predict · change · explain

Explore the dimensions

Predict first, then change one dimension. Diagrams are schematic; results use the entered values. Rounded values are marked ≈.

Pyramids meet at an apex

A pyramid has one base and triangular lateral faces. Sum each triangle’s area and the base. In a regular pyramid, each triangle’s slant height is perpendicular to its base edge.

Slant height and vertical height differ

Slant height lies on a face and is used for that triangle’s area; the vertical height runs from the apex perpendicular to the base.

Follow the reasoning

Worked examples

A triangular prism has base area 6 cm², base perimeter 12 cm and length 10 cm.

  1. SA = 2B + Ph = 2(6) + 12(10).
  2. Total surface area is 132 cm².

Square pyramid: base edge 4 cm, face slant height 5 cm.

  1. Base area = 16 cm²; each triangular face = 4×5÷2 = 10 cm².
  2. SA = 16 + 4(10) = 56 cm².

Guided Practice

A triangular prism has base area 12 cm², base perimeter 16 cm and length 5 cm. Find its surface area.

A first hint

Add two bases and the lateral rectangles.

Compare your reasoning

2(12) + 16(5) = 104 cm².

Practice

Fresh questions · useful feedback

Build fluency, one decision at a time.

Work through eight questions. Two independent correct answers introduce a harder version; supported answers bring the next question back to the foundation. You can use hints at any time.

0 of 8 questions completed

Enter a number, decimal or fraction such as −3/2. Leave out units and π. If asked for k in kπ, enter only k.

This round saves on this device. It is practice, not a school mark.

Connections

Explain why the lateral area of a right prism equals the perimeter of its base times its length.

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