Learn: covering a pyramid-shaped model
You are covering a regular square-pyramid model with paper. Its square base has side length 6 cm. Each of its four triangular faces has a base of 6 cm and a perpendicular face height of 5 cm. How much paper covers the four sides? How much covers the sides and bottom?
Vocabulary and units
- The base is the polygon at the bottom; here it is a square.
- The apex is the vertex where the side faces meet.
- Lateral area means the area of the triangular side faces, excluding the base.
- Total surface area includes the lateral area and the base.
- The slant height ℓ of a regular pyramid is the altitude of a triangular face, measured along that face perpendicular to its base edge.
- The vertical height h runs perpendicularly from the apex to the base plane. It is not the slant height or a sloping edge.
Lengths are in cm; areas are in cm². A regular pyramid has a regular-polygon base and its apex directly above the base’s centre, giving equal slant heights for its side faces.
Worked example: add four triangles and a square
- One triangular face: ½ × 6 × 5 = 15 cm².
- Four faces: 4 × 15 = 60 cm² lateral area.
- Square base: 6 × 6 = 36 cm².
- Total: 60 + 36 = 96 cm².
- Interpret: covering only the sides needs 60 cm²; including the bottom needs 96 cm², before allowing for overlaps or waste.
For a regular pyramid with base perimeter P, base area B and common slant height ℓ, adding its triangles gives lateral area = ½Pℓ and total area = B + ½Pℓ. For a regular square pyramid with base side s, P = 4s and B = s², so total area = s² + 2sℓ.
Explore: unfold the faces into a net
The square is the base; the four blue triangles are the sides. The orange segment marks a face altitude of 5 cm. It meets the square’s edge at a right angle. This flat net uses a scale of 20 drawing units per centimetre.
Total surface area: 4 × 15 + 36 = 96 cm².
If the problem gives vertical height
For this regular square pyramid, the vertical height h, half the base side s/2, and slant height ℓ form a right triangle: ℓ² = h² + (s/2)². If h = 4 cm and s = 6 cm, then ℓ = √(16 + 9) = 5 cm. Using 4 cm as the triangular face’s height would underestimate the surface area.
Another base: an equilateral triangle
A regular triangular pyramid has base edges of 4 cm and common slant height 5 cm. The base perimeter is 12 cm, so lateral area = ½ × 12 × 5 = 30 cm². The base triangle’s altitude is √(4² − 2²) = 2√3 cm; its area is ½ × 4 × 2√3 = 4√3 cm². Total surface area is 30 + 4√3 cm², approximately 36.93 cm².
If a pyramid is not regular, calculate each triangular face using its own base and perpendicular face height, then add the base area. Do not assume all slant heights are equal.
Practice: decide which height and faces count
1. A regular square pyramid has side 8 cm and slant height 5 cm. Find lateral and total area.
P = 32 cm. Lateral area = ½ × 32 × 5 = 80 cm². Base = 64 cm², so total = 144 cm².
2. A regular square pyramid has side 10 m and vertical height 12 m. Find total area.
First ℓ = √(12² + 5²) = 13 m. Lateral area = ½ × 40 × 13 = 260 m². Add the 100 m² base: 360 m².
3. The side faces total 42 cm² and the base is 18 cm². Find total surface area.
Add both parts: 42 + 18 = 60 cm².
4. Why not use the apex-to-corner edge as the face height?
A triangle’s area uses a height perpendicular to its base. The sloping edge generally does not meet that base at a right angle.
Review: sketch the net before calculating
Identify the base, the triangular faces and the correct face heights. State whether the base is included. Use square units and reserve ½Pℓ for a shared slant height. Continue with right triangles or geometry.
Optional reference: Math Is Fun: pyramids and their surface area.