Surface Area of Solids

Surface area is how much material would wrap a solid. The reliable method is to imagine unfolding it into a flat net and adding the pieces.

• Prism: two bases plus the lateral faces. Lateral area = perimeter of base × height.
• Cylinder: two circles plus a rectangle that wraps around. SA = 2πr² + 2πrh.
• Pyramid: base plus triangular faces. Lateral area = ½ × base perimeter × slant height.
• Cone: circle plus a sector. SA = πr² + πrℓ.
• Sphere: SA = 4πr², which is exactly four of its great circles.

The trap is slant height. For pyramids and cones the *slant* height goes on the sloping face, not the vertical height. If you are given the vertical height, get the slant height with the Pythagorean theorem first.

Surface area is computed by decomposition into a net. For a prism or cylinder, the lateral surface unrolls into a rectangle of width equal to the base perimeter (or circumference) and height h, giving LA = Ph and LA = 2πrh.

For a regular pyramid, the lateral faces are congruent triangles of base s and height ℓ (the slant height), so LA = ½Pℓ. For a cone the lateral surface unrolls into a circular sector, giving LA = πrℓ.

The slant height relates to the vertical height by ℓ = √(h² + r²) for a cone, and ℓ = √(h² + a²) for a pyramid, where a is the base's apothem. Substituting h for ℓ is the dominant error in the topic.

The sphere's SA = 4πr² has no elementary net, since a sphere is not developable — the reason all flat world maps distort. It satisfies dV/dr = SA, mirroring the circle's dA/dr = C.

Under a similarity of ratio k, surface areas scale by k² and volumes by k³.