Surface Area of 3-D Objects

The surface area of a 3-D object is the total area of all its outside surfaces. It tells you how much wrapping paper covers a gift, how much paint a wall or tank needs, or how much cardboard goes into a box. The trick is always the same: unfold the object into flat shapes, find the area of each one, and add them up.

Key ideas

Nets

A net is the flat pattern you’d get if you cut an object along some edges and unfolded it. The surface area of the object is the total area of its net. Surface area is measured in square units, such as $\text{cm}^2$ or $\text{m}^2$.

A box has three pairs of matching faces: top and bottom, front and back, and two sides.

Prisms

A prism has two identical, parallel ends (the bases) joined by rectangles.

• Rectangular prism (a box) with length $l$, width $w$ and height $h$: three pairs of matching rectangles, so

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

• Triangular prism: two identical triangles plus three rectangles. Each rectangle has the length of the prism as one side and one side of the triangle as the other. Add the five areas.

Cylinders

Unroll the curved side of a cylinder and you get a rectangle. Its height is the height of the cylinder, and its length is the distance around the circle: the circumference, $2\pi r$.

The net of a cylinder: two circles and a rectangle that wraps around them.

$$SA = \underbrace{2\pi r^2}_{\text{two circles}} + \underbrace{2\pi r h}_{\text{curved side}}$$

Pyramids, cones and slant height

A square-based pyramid has a square base and four identical triangles that meet at the top (the apex). A cone has a circular base and a curved surface that comes to a point.

For these objects, the measurement you need is the slant height $s$: the distance from the apex down the sloping surface to the edge of the base. It’s different from the height $h$, which goes straight down from the apex to the centre of the base. The height, the slant height and a distance across the base form a right triangle, so you can use the Pythagorean theorem to find whichever one you’re missing.

For a pyramid, $s^2 = h^2 + \left(\dfrac{b}{2}\right)^2$. For a cone, $s^2 = h^2 + r^2$.

• Square-based pyramid with base side $b$ and slant height $s$: one square plus four triangles, each with base $b$ and height $s$:

$$SA = b^2 + 4 \times \frac{1}{2} b s = b^2 + 2bs$$

• Cone with radius $r$ and slant height $s$: a circle for the base plus the curved surface, which has area $\pi r s$:

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

Composite objects

A composite object is made by joining simpler objects. Only count the surfaces you can see from the outside. Where two pieces touch, those faces are hidden, so leave them out. Sometimes the question also tells you to leave out a surface, like the bottom of a tank that sits on the ground, or the ends of a can when you only want the area of its label.

Rounding

When $\pi$ appears, use the $\pi$ key on your calculator and round only at the end (usually to one decimal place). You can also give an exact answer in terms of $\pi$, like $120\pi \text{ cm}^2$.

On the SAT

The SAT reference sheet gives area formulas for circles, rectangles, and triangles, and volume formulas, but no surface area formulas, so build surface area from the net, face by face. Answer choices are often in terms of $\pi$, so keep $\pi$ as a symbol: a cylinder with radius $3$ and height $5$ has surface area $2\pi(3)^2 + 2\pi(3)(5) = 18\pi + 30\pi = 48\pi$. If you do want a decimal, type pi in Desmos, and compare the decimal with the decimal values of the choices. See using Desmos on the SAT.

Worked examples

Example 1: Wrapping a gift box

A gift box is $30$ cm long, $20$ cm wide and $10$ cm high (the box in the first figure). What is the least amount of wrapping paper needed to cover it?

Solution. Find the area of each pair of faces:

Faces	Size	Area of one	Area of the pair
top and bottom	$30 \times 20$	$600$	$1200$
front and back	$30 \times 10$	$300$	$600$
two sides	$20 \times 10$	$200$	$400$

$$SA = 1200 + 600 + 400 = 2200 \text{ cm}^2$$

You need at least $2200 \text{ cm}^2$ of paper (in real life, a little more for the overlap).

Example 2: A tent (triangular prism)

A pup tent is a triangular prism. Each triangular end has a base of $1.6$ m, a height of $0.6$ m, and two sloping sides of $1.0$ m. The tent is $2.5$ m long. How much fabric is needed, including the floor?

Solution. The tent has five faces.

Two triangular ends:

$$2 \times \frac{1}{2} \times 1.6 \times 0.6 = 0.96 \text{ m}^2$$

Two sloping sides (rectangles $1.0$ m by $2.5$ m):

$$2 \times 1.0 \times 2.5 = 5 \text{ m}^2$$

Floor (a rectangle $1.6$ m by $2.5$ m):

$$1.6 \times 2.5 = 4 \text{ m}^2$$

Total:

$$SA = 0.96 + 5 + 4 = 9.96 \text{ m}^2$$

Check the triangle: half the base is $0.8$ m, and $0.8^2 + 0.6^2 = 0.64 + 0.36 = 1.00 = 1.0^2$, so the sloping sides really are $1.0$ m. ✓

Example 3: A soup can

A soup can has a radius of $4$ cm and a height of $11$ cm.

• (a) Find the total surface area of the can, to one decimal place.
• (b) The paper label covers only the curved side. Find the area of the label.

Solution.

(a)

$$\begin{aligned} SA &= 2\pi r^2 + 2\pi r h \\ &= 2\pi(4)^2 + 2\pi(4)(11) \\ &= 32\pi + 88\pi \\ &= 120\pi \\ &\approx 377.0 \text{ cm}^2 \end{aligned}$$

(b) The label is just the curved side, a rectangle $2\pi(4) \approx 25.1$ cm long and $11$ cm high:

$$2\pi(4)(11) = 88\pi \approx 276.5 \text{ cm}^2$$

Example 4: A cone, using the height

A solid cone has a radius of $5$ cm and a height of $12$ cm. Find its surface area.

Solution. The formula needs the slant height, but we’re given the height. The height, the radius and the slant height form a right triangle:

$$\begin{aligned} s^2 &= h^2 + r^2 \\ &= 12^2 + 5^2 = 144 + 25 = 169 \\ s &= 13 \text{ cm} \end{aligned}$$

Now use the formula:

$$\begin{aligned} SA &= \pi r^2 + \pi r s \\ &= \pi(5)^2 + \pi(5)(13) \\ &= 25\pi + 65\pi \\ &= 90\pi \approx 282.7 \text{ cm}^2 \end{aligned}$$

Common mistakes

Using the height instead of the slant height. The triangles on a pyramid and the curved surface of a cone use the slant height $s$. If you’re given the height $h$, find $s$ with the Pythagorean theorem first, as in Example 4.

Forgetting faces, or counting some twice. Use a net or a table (as in Example 1) to list every face once. A box has $6$ faces, a triangular prism has $5$, and a square-based pyramid has $5$.

Counting hidden faces on composite objects. Where two objects are joined, the touching surfaces are inside, so they’re not part of the surface area. Also read the question carefully: an open-top box or a tank with no bottom has one fewer surface.

Using the diameter in place of the radius. The formulas use $r$. If a can is $8$ cm across, its radius is $4$ cm.

Mixing up circumference and area. The curved side of a cylinder is a rectangle whose length is the circumference $2\pi r$, not the area $\pi r^2$.

Wrong units. Surface area is an area, so the units are square units like $\text{cm}^2$ or $\text{m}^2$.

Practice

1. (Warm-up) Find the surface area of a cube with edges of $6$ cm.

Solution

A cube has $6$ identical square faces:

$$SA = 6 \times 6^2 = 6 \times 36 = 216 \text{ cm}^2$$

2. (Warm-up) Find the surface area of a rectangular prism $8$ cm long, $5$ cm wide and $3$ cm high.

Solution

$$\begin{aligned} SA &= 2(8)(5) + 2(8)(3) + 2(5)(3) \\ &= 80 + 48 + 30 \\ &= 158 \text{ cm}^2 \end{aligned}$$

3. (Core) A triangular prism has right-triangle ends with sides $6$ cm, $8$ cm and $10$ cm. The prism is $15$ cm long. Find its surface area.

Solution

The legs of each right triangle are $6$ and $8$ (the $10$ is the hypotenuse), so each end has area $\dfrac{1}{2} \times 6 \times 8 = 24 \text{ cm}^2$.

The three rectangles are each $15$ cm long, with widths $6$, $8$ and $10$ cm.

$$\begin{aligned} SA &= 2(24) + 15(6) + 15(8) + 15(10) \\ &= 48 + 90 + 120 + 150 \\ &= 408 \text{ cm}^2 \end{aligned}$$

4. (Core) A square-based pyramid has a base side of $10$ m and a height of $12$ m.

• (a) Find the slant height.
• (b) Find the surface area.

Solution

(a) The height, half the base side ($5$ m) and the slant height form a right triangle:

$$s^2 = 12^2 + 5^2 = 144 + 25 = 169 \quad\Rightarrow\quad s = 13 \text{ m}$$

(b)

$$\begin{aligned} SA &= b^2 + 4 \times \frac{1}{2} b s \\ &= 10^2 + 4 \times \frac{1}{2} \times 10 \times 13 \\ &= 100 + 260 \\ &= 360 \text{ m}^2 \end{aligned}$$

5. (Core) A cone has a radius of $3$ cm and a height of $4$ cm.

• (a) Find the surface area of the solid cone, to one decimal place.
• (b) An ice-cream cone of the same size has no base. Find the area of its outside surface.

Solution

First find the slant height: $s^2 = 4^2 + 3^2 = 25$, so $s = 5$ cm.

(a) $SA = \pi(3)^2 + \pi(3)(5) = 9\pi + 15\pi = 24\pi \approx 75.4 \text{ cm}^2$.

(b) Only the curved surface: $\pi(3)(5) = 15\pi \approx 47.1 \text{ cm}^2$.

6. (Core) A cylindrical water tank stands on the ground. It has a radius of $1.5$ m and a height of $4$ m. You’ll paint the top and the curved side, but not the bottom.

• (a) Find the area to be painted, to one decimal place.
• (b) One litre of paint covers $10 \text{ m}^2$. Paint is sold in whole litres. How many litres should you buy?

Solution

(a) Top: $\pi(1.5)^2 = 2.25\pi \approx 7.07 \text{ m}^2$. Curved side: $2\pi(1.5)(4) = 12\pi \approx 37.70 \text{ m}^2$.

$$A = 2.25\pi + 12\pi = 14.25\pi \approx 44.8 \text{ m}^2$$

(b) $44.8 \div 10 = 4.48$ L. You can’t buy part of a litre, and $4$ L isn’t enough, so buy $5$ L.

7. (Core) A display stand is made of a cube with $10$ cm edges, with a smaller cube with $4$ cm edges glued to the centre of its top. Find the surface area of the stand, including the bottom.

Solution

Big cube: $6 \times 10^2 = 600 \text{ cm}^2$, but a $4 \times 4 = 16 \text{ cm}^2$ patch of its top is covered by the small cube. So it shows $600 - 16 = 584 \text{ cm}^2$.

Small cube: its bottom is hidden, so only $5$ faces show: $5 \times 4^2 = 80 \text{ cm}^2$.

$$SA = 584 + 80 = 664 \text{ cm}^2$$

8. (Challenge) A grain silo is a cylinder with a cone-shaped roof. The cylinder has a radius of $3$ m and a height of $10$ m. The roof is $4$ m tall from the top of the cylinder to its peak. You’ll paint the outside wall and the roof (not the floor). Find the area to paint, to one decimal place.

Solution

Curved wall of the cylinder: $2\pi(3)(10) = 60\pi \text{ m}^2$.

Roof: find the slant height first. $s^2 = 4^2 + 3^2 = 25$, so $s = 5$ m. The roof is the curved surface of a cone (no base, since it sits on the cylinder): $\pi(3)(5) = 15\pi \text{ m}^2$.

The top of the cylinder and the base of the cone are hidden where they join, so neither is counted.

$$A = 60\pi + 15\pi = 75\pi \approx 235.6 \text{ m}^2$$

9. (Challenge) A cube has a surface area of $294 \text{ cm}^2$. Find the length of each edge and the volume of the cube.

Solution

A cube has $6$ equal square faces, so each face has area $294 \div 6 = 49 \text{ cm}^2$.

Each face is a square, so the edge is $\sqrt{49} = 7$ cm.

Volume: $7^3 = 343 \text{ cm}^3$.

Check: $6 \times 7^2 = 6 \times 49 = 294$. ✓