Surface Area

Measuring the total area of the outside of 3D shapes

Introduction

Surface area is the total area of all faces (or curved surfaces) of a 3D solid. At GCSE, you’ll need to calculate surface areas of cuboids, prisms, cylinders, cones, spheres, pyramids, and compound shapes. This skill is used in real life for wrapping presents, painting walls, designing packaging, or calculating how much material is needed to cover an object.

Key Vocabulary

• Surface area — total area of all outer faces of a solid.
• Face — a flat surface (e.g. side of a cube or prism).
• Lateral surface area — curved or side area excluding the base(s).
• Net — a 2D pattern that folds up to form a 3D solid.
• Curved surface — non-flat area, like the side of a cylinder or the surface of a sphere.
• Units — measured in squared units (cm², m², etc.).

Core Ideas

• Find the area of each face (rectangle, triangle, circle, etc.).
• Add them up to get the total surface area.
• Prisms → surface area = (perimeter of cross-section × length) + 2 × area of cross-section.
• Cylinder → curved area = \(2πrh\), total area = \(2πrh+2πr^2\).
• Cone → curved area = \(πrl\) (where \(l\) = slant height). Total = \(πrl+πr^2\).
• Sphere → surface area = \(4πr^2\).
• Pyramid → area of base + sum of triangular face areas.
• Compound solids → split into basic parts and calculate separately.

Step-by-Step Method

1. Identify the shape (cube, cuboid, prism, cylinder, cone, sphere, pyramid, compound).
2. Break into faces. Draw the net if needed.
3. Find dimensions of each face.
4. Calculate areas using 2D formulas (rectangle = lw, triangle = ½bh, circle = πr²).
5. Add areas to find the total surface area.
6. Attach units (cm², m²). If mixed, convert before working.

Worked Examples — Foundation

Example F1 — Cube

[diagram: cube with edge 4 cm]

Each face area = 4×4=16 cm². 6 faces → 6×16=96 cm² total.

Example F2 — Cuboid

[diagram: cuboid 5 cm × 3 cm × 2 cm]

SA = 2(lw+lh+wh)=2(5×3+5×2+3×2)=2(15+10+6)=2×31=62 cm².

Example F3 — Cylinder

[diagram: cylinder radius 5 cm, height 10 cm]

Curved area=2πrh=2π×5×10=100π. Two circles=2πr²=2π×25=50π. Total=150π≈471.2 cm².

Worked Examples — Higher

Example H1 — Cone

[diagram: cone radius 6 cm, slant height 10 cm]

Curved=πrl=π×6×10=60π. Base=πr²=π×36=36π. Total=96π≈301.6 cm².

Example H2 — Sphere

[diagram: sphere radius 7 cm]

SA=4πr²=4π×49=196π≈615.8 cm².

Example H3 — Compound shape

[diagram: cylinder radius 3 cm, height 8 cm, hemisphere on top]

Cylinder curved=2πrh=2π×3×8=48π. One circular base=πr²=9π. Hemisphere surface=2πr²=18π. Total=75π≈235.6 cm².

Common Mistakes & Fixes

• Forgetting top/bottom faces. Fix: Sketch a net and tick each face.
• Mixing radius and diameter. Fix: Remember r=d/2.
• Using height instead of slant height in cone. Fix: Label carefully; use Pythagoras if slant not given.
• Confusing surface area with volume. Fix: Surface = 2D total area, Volume = 3D space inside.
• Forgetting units. Fix: Always write cm², m².

Practice Questions — Foundation

1. Find SA of a cube with edge 7 cm.
2. A cuboid 8 cm × 5 cm × 2 cm. Work out its surface area.
3. A cylinder has radius 4 cm, height 12 cm. Find total SA.
4. A triangular prism has cross-section base 6 cm, height 5 cm, prism length 10 cm. Find surface area (include 3 rectangular sides).
5. A can has diameter 6 cm and height 10 cm. Find total SA (2 dp).

Practice Questions — Higher

1. Find SA of a cone with radius 7 cm, slant 10 cm (2 dp).
2. Sphere radius 12 cm. Find exact SA in terms of π.
3. A square-based pyramid has base edge 9 cm and slant height 13 cm. Find SA.
4. A cylinder radius 5 cm, height 20 cm, open top. Find SA (2 dp).
5. A compound solid has a cylinder radius 4 cm, height 10 cm with a cone (same radius) on top, slant height 6 cm. Find total SA.

Challenge Questions

1. A hemisphere of radius 10 cm is placed on a flat table. Find the curved SA and total SA (including base).
2. A closed cone (r=8 cm, h=15 cm) is painted. Find the surface area. (Hint: find slant with Pythagoras).
3. A cuboid 20 cm × 15 cm × 10 cm has a square hole of 5 cm × 5 cm through its height. Find external SA (ignore inner hole surfaces).
4. A football radius 11 cm. Find SA. Then estimate cost of leather covering at £0.05 per cm².
5. A silo is a cylinder radius 4 m, height 12 m, with a hemispherical roof. Find total SA (nearest m²).

Quick Revision Sheet

• Cuboid: \(SA=2(lw+lh+wh)\)
• Prism: \(SA=2×\text{cross-section area}+(\text{perimeter of cross-section}×\text{length})\)
• Cylinder: \(SA=2πrh+2πr^2\)
• Cone: \(SA=πrl+πr^2\)
• Sphere: \(SA=4πr^2\)
• Pyramid: base area + sum of triangle face areas
• Units: cm², m²; convert first

Conclusion & Next Steps

Surface area calculations connect 2D area formulas with 3D shapes. Mastering them prepares you for mixed problems in GCSE exams, where surface area, volume, and Pythagoras often combine.

• Next topics: 3D Pythagoras, density and mass, compound measures.
• Exam habits: Sketch nets, label radii/slants, double-check face counts.

Exam Tips

• Always draw the net if you’re unsure.
• Circle or underline “surface area” or “volume” in the question.
• Use slant height for cones, not vertical height.
• Keep π until last step for accuracy.
• Check final units: cm², m², etc.

Answer Key (Outline)

Foundation

1. 294 cm²
2. 172 cm²
3. \(2π×4×12+2π×16=96π+32π=128π≈402.1\) cm²
4. Cross-section=½×6×5=15; perimeter=6+5+√61≈23.8; side area=23.8×10≈238; total=2×15+238=268 cm² (approx)
5. \(2πrh+2πr^2=2π×3×10+2π×9=60π+18π=78π≈245.0\) cm²

Higher

1. \(πrl+πr^2=π×7×10+π×49=70π+49π=119π≈373.0\) cm²
2. \(4π×144=576π\) cm²
3. Base=81, triangles=4×½×9×13=234; total=315 cm²
4. \(2πrh+πr^2=2π×5×20+25π=200π+25π=225π≈706.9\) cm²
5. Cylinder=2πrh+πr^2=2π×4×10+16π=80π+16π=96π; cone=πrl=π×4×6=24π; total=120π≈377.0 cm²