Surface Area

Surface area is the total area of all the outer faces of a three-dimensional shape. If you could peel the surface off a solid and lay it flat, the surface area is how much flat space it would cover.

Surface area tells you how much material is needed to cover, wrap, paint, or fabricate a 3D object. It is always measured in square units — square inches (in$^2$), square feet (ft$^2$), square meters (m$^2$), etc. This page covers five common solids: rectangular prisms, cylinders, spheres, cones, and square pyramids.

Surface Area Formulas at a Glance

Rectangular Prism

A rectangular prism (box) has six faces — three pairs of identical rectangles. The surface area is the sum of the areas of all six faces.

$SA = 2lw + 2lh + 2wh$

Think of it as: two top-and-bottom faces ($lw$), two front-and-back faces ($lh$), and two left-and-right faces ($wh$).

Example 1: Find the surface area of a box that is 10 in long, 6 in wide, and 4 in tall.

$SA = 2(10)(6) + 2(10)(4) + 2(6)(4)$

$SA = 120 + 80 + 48 = 248 \text{ in}^2$

Answer: The surface area is 248 square inches.

Example 2: A room is 14 ft long, 12 ft wide, and 8 ft tall. How many square feet of wall space is there (four walls only, no ceiling or floor)?

For just the four walls, we need the lateral surface area. The walls consist of two pairs:

$\text{Wall area} = 2(14)(8) + 2(12)(8) = 224 + 192 = 416 \text{ ft}^2$

Answer: There are 416 square feet of wall space.

Cylinder

A cylinder has three surfaces: two circular ends (the top and bottom) and one curved side that unrolls into a rectangle.

$SA = 2\pi r^2 + 2\pi rh$

• $2\pi r^2$ = area of the two circular ends
• $2\pi rh$ = area of the curved side (called the lateral surface area)

If you need only the lateral (side) surface area — for example, when wrapping a label around a can — use:

$\text{Lateral SA} = 2\pi rh$

Example 3: Find the total surface area of a cylinder with radius 5 cm and height 12 cm.

$SA = 2\pi (5)^2 + 2\pi (5)(12)$

$SA = 2\pi (25) + 2\pi (60) = 50\pi + 120\pi = 170\pi \approx 534.07 \text{ cm}^2$

Answer: The surface area is approximately 534.07 cm$^2$.

Sphere

A sphere has no edges and no flat faces — just one continuous curved surface.

$SA = 4\pi r^2$

Example 4: Find the surface area of a sphere with a diameter of 10 inches.

The radius is half the diameter: $r = 5$ in.

$SA = 4\pi (5)^2 = 4\pi (25) = 100\pi \approx 314.16 \text{ in}^2$

Answer: The surface area is approximately 314.16 in$^2$.

Cone

A cone has two surfaces: a flat circular base and a curved lateral surface that tapers to a point (the apex).

$SA = \pi r^2 + \pi r l$

• $\pi r^2$ = area of the circular base
• $\pi r l$ = lateral surface area (the curved side)

Here $l$ is the slant height — the distance measured along the surface from the base edge to the apex. The slant height is not the same as the vertical height $h$. They are related by the Pythagorean theorem:

$l = \sqrt{r^2 + h^2}$

If you need only the lateral surface area — for example, when making a paper cone or a funnel — use:

$\text{Lateral SA} = \pi r l$

Example 5: Find the total surface area of a cone with radius 6 cm and slant height 10 cm.

$SA = \pi (6)^2 + \pi (6)(10)$

$SA = 36\pi + 60\pi = 96\pi \approx 301.59 \text{ cm}^2$

Answer: The total surface area is approximately 301.59 cm$^2$.

Example 6: A cone has a radius of 4 inches and a vertical height of 7 inches. Find the lateral surface area.

First, find the slant height using the Pythagorean theorem:

$l = \sqrt{r^2 + h^2} = \sqrt{4^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65} \approx 8.06 \text{ in}$

Now calculate the lateral surface area:

$\text{Lateral SA} = \pi r l = \pi (4)(\sqrt{65}) = 4\sqrt{65}\,\pi \approx 101.31 \text{ in}^2$

Answer: The lateral surface area is approximately 101.31 in$^2$.

Square Pyramid

A square pyramid has five faces: one square base and four identical triangular faces that meet at the apex.

$SA = s^2 + 2sl$

• $s^2$ = area of the square base
• $2sl$ = total area of the four triangular faces (the lateral surface area)

Each triangular face has base $s$ and height $l$ (the slant height of the triangular face), so each triangle has area $\frac{1}{2}sl$. With four triangles: $4 \times \frac{1}{2}sl = 2sl$.

The slant height $l$ is not the height of the pyramid. It is the distance from the midpoint of a base edge straight up to the apex, measured along the face. If you know the pyramid height $h$ and the base side $s$:

$l = \sqrt{h^2 + \left(\frac{s}{2}\right)^2}$

If you need only the lateral surface area (the four triangular faces):

$\text{Lateral SA} = 2sl$

Example 7: Find the total surface area of a square pyramid with base side 8 m and slant height 13 m.

$SA = (8)^2 + 2(8)(13)$

$SA = 64 + 208 = 272 \text{ m}^2$

Answer: The total surface area is 272 m$^2$.

Example 8: A square pyramid has a base side of 10 ft and a vertical height of 12 ft. Find the total surface area.

First, find the slant height. The distance from the center of the base to the midpoint of a base edge is $\frac{s}{2} = \frac{10}{2} = 5$ ft.

$l = \sqrt{h^2 + \left(\frac{s}{2}\right)^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ ft}$

Now calculate the total surface area:

$SA = (10)^2 + 2(10)(13) = 100 + 260 = 360 \text{ ft}^2$

Answer: The total surface area is 360 ft$^2$.

Real-World Application: HVAC — Fabricating a Cylindrical Duct Section

An HVAC technician needs to fabricate a section of round ductwork from flat sheet metal. The duct has a diameter of 12 inches and needs to be 4 feet (48 inches) long. How much sheet metal is needed for the curved body of the duct?

Since the duct is open at both ends (it connects to other duct sections), only the lateral surface area is needed — not the full surface area.

Step 1 — Identify the radius:

$r = \frac{12}{2} = 6 \text{ in}$

Step 2 — Calculate the lateral surface area:

$\text{Lateral SA} = 2\pi rh = 2\pi (6)(48) = 576\pi \approx 1{,}809.56 \text{ in}^2$

Step 3 — Convert to square feet (since sheet metal is often sold in square-foot sheets):

$\frac{1{,}809.56}{144} \approx 12.57 \text{ ft}^2$

Step 4 — Account for seam overlap. Duct fabrication requires a seam where the sheet metal edges join. Adding 1 inch of overlap along the 48-inch length adds:

$48 \times 1 = 48 \text{ in}^2 \approx 0.33 \text{ ft}^2$

$\text{Total} \approx 12.57 + 0.33 = 12.90 \text{ ft}^2$

Answer: The technician needs approximately 12.9 square feet of sheet metal. In practice, the technician would cut a flat rectangle that is $2\pi r + 1 = 2\pi(6) + 1 \approx 38.7$ inches wide and 48 inches long, then roll it into a cylinder and join the seam.

Lateral vs. Total Surface Area

Many real-world problems only need the lateral (side) surface area — not the full surface area. Ask yourself: do I need to cover the top and bottom, or just the sides?

• Painting the walls of a room: lateral only (not the floor or ceiling)
• Wrapping a gift box: total surface area (all six faces)
• Sheet metal for a duct: lateral only (both ends are open)
• Painting a storage tank: depends on whether the top is exposed

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Surface area is the total outer area of a 3D shape, measured in square units
• Rectangular prism: $SA = 2lw + 2lh + 2wh$ — three pairs of rectangular faces
• Cylinder: $SA = 2\pi r^2 + 2\pi rh$ — two circles plus the curved side
• Sphere: $SA = 4\pi r^2$ — one continuous curved surface
• Cone: $SA = \pi r^2 + \pi r l$ — circular base plus curved side; slant height $l = \sqrt{r^2 + h^2}$
• Square pyramid: $SA = s^2 + 2sl$ — square base plus four triangular faces; slant height $l = \sqrt{h^2 + (s/2)^2}$
• Lateral surface area (sides only) is often more useful than total surface area in real-world applications
• For cones and pyramids, the slant height is always measured along the surface, not the vertical height through the center

Return to Geometry for more topics in this section.