Volume and Surface Area of a Regular Prism

For a prism whose base is a regular polygon, finds the volume as base area × height and the surface area as twice the base plus the sides. The base area comes from the regular polygon formula. It suits hexagonal bolts and polygonal columns.

A regular prism has matching regular polygons top and bottom, joined by vertical sides. The head of a hexagonal bolt, a pencil and a polygonal column are all this shape.

Both the volume and the surface area follow once the base area is known.

A=ns24tanπnV=AhA = \dfrac{n s^2}{4 \tan\dfrac{\pi}{n}} \qquad V = A h

The sides are nn upright rectangles, so the lateral area is nshn s h. The total surface adds the top and bottom, giving 2A+nsh2A + n s h.

Where the base formula comes from

Drawing lines from the centre of a regular polygon to each vertex cuts it into nn identical isosceles triangles. Each has the side ss as its base, and its height is the distance from the centre to that side.

That distance comes from halving the central angle. The central angle is 2π÷n2\pi \div n, and the tangent of half of it gives s2tan(π/n)\dfrac{s}{2\tan(\pi / n)}. One triangle therefore has area s2×s2tan(π/n)\dfrac{s}{2} \times \dfrac{s}{2\tan(\pi / n)}, and multiplying by nn produces the formula above.

Example

The default input is a prism of height 10 on a regular hexagon of side 4.

The base area is 6×16÷(4tan30°)6 \times 16 \div (4\tan 30°), about 41.5692. The volume multiplies that by the height, about 415.6922. The lateral area is 6×4×106 \times 4 \times 10, exactly 240, and the total surface is 2×41.5692+2402 \times 41.5692 + 240, about 323.1384.

Setting the number of sides to 4 describes a prism on a square of side 4, in other words a cuboid, and the calculator duly returns a base area of 16 and a volume of 160.

Points to watch

The number of sides must be a whole number of at least 3. Fewer than that is not a polygon.

Adding sides moves the base towards a circle. Holding the side length fixed while adding sides makes the prism steadily fatter, so comparisons with a cylinder should keep the perimeter fixed instead.

A pyramid on the same base with the same height holds exactly one third of this volume, and there is a separate tool here for that.