Surface Area of a Right Prism

\( S=\text{perimeter of cross-section}\times\text{length}+2\times\text{area of cross-section} \)
Geometry GCSE
Question 11 of 20

\( Triangular prism: perimeter=36, area=54, length=25. Find S. \)

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
\( S=perimeter×length+2×area \)

Explanation

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Statement

The surface area of a right prism is the total area of all its faces. A right prism has two identical parallel cross-sections (front and back) and rectangular faces joining them. The general formula is:

\[ S = \text{perimeter of cross-section} \times \text{length} + 2 \times \text{area of cross-section} \]

The first part \(\text{perimeter} \times \text{length}\) gives the total area of the rectangular side faces. The second part \(2 \times \text{area of cross-section}\) accounts for the front and back ends.

Why it’s true

  • A prism is built by extending a 2D shape (cross-section) along a straight length.
  • The two ends are congruent copies of the cross-section, giving \(2 \times \text{area of cross-section}\).
  • The side faces are rectangles. Each side of the cross-section, when stretched through the prism’s length, creates a rectangle of area “side × length”. Adding all these rectangles together gives \(\text{perimeter} \times \text{length}\).
  • Adding both contributions gives the total surface area.

Recipe (how to use it)

  1. Identify the shape of the cross-section (triangle, trapezium, hexagon, etc.).
  2. Calculate its perimeter.
  3. Calculate its area.
  4. Multiply perimeter by the prism’s length.
  5. Add \(2 \times \text{area of cross-section}\).
  6. Label the final answer with square units.

Spotting it

Use this formula whenever you have a prism (a solid with identical parallel cross-sections along its length). Common exam prisms include triangular, trapezoidal, or hexagonal prisms.

Common pairings

  • Volume of a prism: \(V = \text{area of cross-section} \times \text{length}\).
  • 2D area and perimeter formulas for polygons and circles.

Mini examples

  1. Given: Triangular prism with base triangle area \(12\,cm^2\), perimeter \(18\,cm\), length \(10\,cm\). Answer: \(S = 18 \times 10 + 2 \times 12 = 180 + 24 = 204\,cm^2\).
  2. Given: Rectangular prism (cross-section 6 by 4). Area=24, Perimeter=20, length=15. Answer: \(S = 20 \times 15 + 2 \times 24 = 300 + 48 = 348\,cm^2\).

Pitfalls

  • Forgetting to double the cross-section area.
  • Mixing volume with surface area.
  • Calculating perimeter incorrectly when the cross-section is irregular.

Exam strategy

  • Write the cross-section dimensions clearly and check your perimeter and area.
  • Always separate the two parts of the formula before adding.
  • If it’s a triangular prism, be especially careful with triangle area (½ base × height).

Summary

Right prisms are solids formed by extending a 2D shape along a length. To find the surface area, calculate all side rectangles (\(\text{perimeter} \times \text{length}\)) and the two ends (\(2 \times \text{area of cross-section}\)). This formula is essential for GCSE questions involving packaging, engineering, and 3D modelling.

Worked examples

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  1. \( Triangular prism: cross-section perimeter=18 cm, area=12 cm², length=10 cm. Find S. \)
    1. \( Perimeter×length=180 \)
    2. \( 2×area=24 \)
    3. \( Total=204 \)
    Answer: \( 204\,cm^2 \)
  2. \( Rectangular prism: cross-section 6×4, perimeter=20, area=24, length=15. Find S. \)
    1. \( 20×15=300 \)
    2. \( 2×24=48 \)
    3. \( Total=348 \)
    Answer: \( 348\,cm^2 \)
  3. \( Triangular prism: perimeter=24, area=18, length=8. Find S. \)
    1. \( 24×8=192 \)
    2. \( 2×18=36 \)
    3. \( Total=228 \)
    Answer: \( 228\,cm^2 \)
  4. \( Pentagonal prism: perimeter=30, area=25, length=12. Find surface area. \)
    1. \( 30×12=360 \)
    2. \( 2×25=50 \)
    3. \( Total=410 \)
    Answer: \( 410\,cm^2 \)
  5. \( Hexagonal prism: perimeter=42, area=72, length=9. Find S. \)
    1. \( 42×9=378 \)
    2. \( 2×72=144 \)
    3. \( Total=522 \)
    Answer: \( 522\,cm^2 \)
  6. \( Triangular prism: perimeter=40, area=60, length=20. Find S. \)
    1. \( 40×20=800 \)
    2. \( 2×60=120 \)
    3. \( Total=920 \)
    Answer: \( 920\,cm^2 \)
  7. \( Trapezoidal prism: perimeter=50, area=80, length=15. Find surface area. \)
    1. \( 50×15=750 \)
    2. \( 2×80=160 \)
    3. \( Total=910 \)
    Answer: \( 910\,cm^2 \)
  8. \( Hexagonal prism: perimeter=60, area=155, length=10. Find S. \)
    1. \( 60×10=600 \)
    2. \( 2×155=310 \)
    3. \( Total=910 \)
    Answer: \( 910\,cm^2 \)
  9. \( Triangular prism: perimeter=36, area=54, length=25. Find S. \)
    1. \( 36×25=900 \)
    2. \( 2×54=108 \)
    3. \( Total=1008 \)
    Answer: \( 1008\,cm^2 \)
  10. \( Octagonal prism: perimeter=80, area=300, length=12. Find S. \)
    1. \( 80×12=960 \)
    2. \( 2×300=600 \)
    3. \( Total=1560 \)
    Answer: \( 1560\,cm^2 \)