Right square-pyramid surface area
Use the square base and four congruent triangular faces. The height of a triangular face is its slant height
- If the slant height is not given Use the vertical height and half the base side \(\displaystyle s^2=h^2+(b/2)^2\)
- If the base is absent Use only the four triangular faces \(\displaystyle A_{\text{faces}}=2bs\)
- Use square units Convert all lengths to one unit before multiplying
Worked example
A closed square-based pyramid has base side length 6 cm, vertical height 4 cm and slant height 5 cm. Find its total surface area
- \(\displaystyle A_{\text{base}}=6^2=36\text{ cm}^2\)Calculate the area of the square base
- \(\displaystyle \begin{gathered}A_{\triangle}=\frac12\times6\times5\\{}=15\text{ cm}^2\end{gathered}\)Use the slant height as the perpendicular height of each triangular face
- \(\displaystyle \begin{gathered}A_{\text{faces}}=4\times15\\{}=60\text{ cm}^2\end{gathered}\)There are four congruent triangular faces
- \(\displaystyle SA=36+60=96\text{ cm}^2\)Add the base and triangular-face areas to obtain the total surface area
Common mistakes
Difficulty levels
Questions about this topic
What does surface area include?+
It includes the area of every face or curved surface that the question states is included
Why is slant height used on a pyramid face?+
The slant height is perpendicular to the base edge of each triangular face, so it is the height used in that triangle's area formula
Is this required Year 9 content?+
This is an optional extension. VCAA places right-pyramid surface-area problems in Level 10A
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 9 Mathematics, chapter 5: Measurement