Math 7 · 6.G.A.4 · 7.G.B.6

Surface area of prisms and cylinders

How much material does it take to cover a box, a tent or a tin exactly once, with nothing left bare and nothing doubled?

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Area of a circleMath 7 Nets of solidsMath 6

Where you are 3 in Surface area

Geometry11

Solids8

Surface area3

Surface area of a sphere

Surface area of prisms and cylinders

Surface area of pyramids and cones

When you reach for it

Paper to wrap a present, wood for a planter, metal for a tin, card for a tube: each asks how much area covers the outside of a solid. That number is the solid’s surface area, and this lesson finds it for prisms and cylinders, the two shapes behind most boxes, tins and buildings.

There is one idea underneath. Unfold the solid flat, and its surface becomes flat shapes whose areas you already know. For a prism or a cylinder that pattern is always the same: two bases, and one rectangle that wraps around them.

Real problems also ask which faces to cover. A room’s walls are painted but not its floor; a tube is open at both ends. So the lesson separates the lateral area, the sides alone, from the total surface area, everything.

This lesson is taught over two days. Part 1 is prisms: the net, boxes and triangular prisms, and lateral against total area. Part 2 is cylinders: the net, the formula, and problems with \pi in them.

The procedure

Part 1: prisms

What is being measured. The surface area of a solid is the total area of all its faces: the material that covers it once, with no gaps and no overlaps. It is an area, so it is measured in square units, such as \text{cm}^2 or \text{m}^2. For a solid with flat faces, find the area of each face and add. The real work is making sure every face is counted, and counted once.

The faces of a prism. A prism has two bases: polygons that are congruent (the same shape and size) and lie in parallel planes, with each corner of one joined by an edge to the matching corner of the other. In a right prism those joining edges are perpendicular to the bases, and every prism in this lesson is a right prism. Each of its other faces is then a rectangle, one for each edge of the base: three for a triangular prism, four for a box, six for a hexagonal prism. The length of the joining edges, the distance between the bases, is the prism’s height h.

The bases are not always at the bottom. A tent lying on the ground is a triangular prism whose bases are the two triangular ends; the floor is one of the rectangles. Look for the pair of congruent parallel faces that every slice of the prism copies. In a box, any opposite pair will do.

Unfold it. Cut along enough edges and the surface of a prism lies flat as one connected shape, its net. Unfolding moves each face without stretching it, so the area of the net is exactly the surface area.

Figure 1A box and its net. The four side rectangles lie in one strip 8 cm tall and 30 + 20 + 30 + 20 = 100 cm long, the perimeter of the base.

The box. In a box (a rectangular prism) l long, w wide and h tall, opposite faces are congruent rectangles, so the six faces come in three pairs: top and bottom, each lw; front and back, each lh; and the two ends, each wh. Counting each pair twice,

S = 2(lw + lh + wh)

A cube with edge a has six faces of area a^2, so S = 6a^2.

The side rectangles make one rectangle. In the net, the four side rectangles sit next to each other in a single strip. Every one is h tall, because each joins the two bases. Their widths are the edges of the base, taken in order round it, so the strip is as long as the base’s perimeter P. Its area is the lateral area:

L = Ph

In algebra: if the base has edges a, b and c, the side faces have areas ah, bh and ch, and ah + bh + ch = (a + b + c)h by the distributive property. That holds for any number of edges, so it holds for every right prism.

Total surface area. Add the two bases, each of area B:

S = 2B + Ph

This works for any base whose area and perimeter you can find. A base made of pieces is handled as in area of composite figures. The box formula is the case B = lw and P = 2(l + w), since 2lw + 2(l + w)h = 2(lw + lh + wh).

Lateral or total: the object decides. Which faces a problem wants is decided by the object, never by the formula:

the walls of a room are painted, but not the floor or the ceiling: lateral area, L = Ph;
a box with no lid, such as a planter, has only one base: B + Ph;
a tent with no floor has both bases, but one of its rectangles is missing: S minus that rectangle.

So before calculating, list the faces that really are covered. The formulas are a fast way of adding faces, and any face can still be added or taken away by hand.

Units. Every length must be in the same unit before anything is multiplied, and the answer is in that unit, squared. Changing \text{cm}^2 to \text{m}^2 is converting square units: 1\ \text{m}^2 = 10{,}000\ \text{cm}^2, not 100.

Part 2: cylinders

A cylinder has two bases that are congruent circles in parallel planes, joined by one curved surface. In a right cylinder, the only kind here, the line joining the centers of the two circles is perpendicular to them, and its length is the height h.

Unrolling the curved side. Cut the curved surface along a straight line from the top rim to the bottom rim, and unroll it. It lies flat as a rectangle, with no stretching and no tearing, because the side of a cylinder bends only one way: it is a flat sheet that has been rolled up, and unrolling reverses that. (A ball bends two ways at once, and no piece of it lies flat.)

Figure 2Cut the curved side straight down and unroll it. It becomes a rectangle h tall, and its length is the distance round the circle, 2\pi r.

The rectangle is h tall, because the cut ran from one rim to the other. Its long edge was the rim of a base circle, so its length is the Circumference 2\pi r. So the lateral area is

L = 2\pi r h

Total surface area. Each base is a circle of area \pi r^2, as in the area of a circle, so

S = 2\pi r^2 + 2\pi r h

Taking out the common factor 2\pi r gives S = 2\pi r(r + h), sometimes quicker to work with.

The same rule as a prism. The base area is B = \pi r^2 and the distance round the base is P = 2\pi r, so S = 2B + Ph again. That is no accident: a prism whose base is a regular polygon with very many sides looks almost exactly like a cylinder, and its perimeter is almost exactly the circle’s circumference.

Exact or rounded. Leaving \pi as a symbol gives an exact answer, such as 432\pi\ \text{cm}^2. A decimal answer is rounded, and should say how. Multiply by \pi at the very end, so that no rounding error is carried through the working. If a problem then asks for a whole number of packs or sheets, that last rounding goes up.

Radius or diameter. The formulas use the radius. If the diameter d is given, \displaystyle r = \frac{d}{2}, and the lateral area can also be written L = \pi d h.

Worked examples

Examples 1 to 4 go with Part 1. Examples 5 to 7 go with Part 2.

Example 1

Wrapping a gift box

Ulrika wraps a gift box 30 cm long, 20 cm wide and 8 cm tall. What is the least area of paper that covers it?

Solution

The paper must cover every face, so this is the total surface area. The three pairs of faces have areas

\begin{aligned} lw &= 30 \times 20 = 600 \\ lh &= 30 \times 8 = 240 \\ wh &= 20 \times 8 = 160 \end{aligned}

all in \text{cm}^2. Each pair counts twice:

S = 2(600 + 240 + 160) = 2 \times 1000 = 2000\ \text{cm}^2

Check with the other formula. The base is 30 by 20, so B = 600 and P = 30 + 20 + 30 + 20 = 100. Then 2B + Ph = 1200 + 100 \times 8 = 2000. The two ways agree, as they must: they add the same six rectangles in a different order.

Example 2

A tent made from a triangular prism

Yseult’s tent is a triangular prism lying on one of its rectangular faces. Each end is a triangle 2.4 m wide at the ground and 1.6 m tall, with two sloping sides each 2 m long. The tent is 3 m long. (a) How much fabric does the whole tent use, floor included? (b) How much if it has no floor?

Figure 3The tent in Example 2 and its net. The bases of this prism are the two triangular ends, not the floor.

Solution

The two triangular ends are congruent and parallel, so they are the bases, and the prism’s height is the tent’s length, h = 3 m. The rectangles are the two sloping sides and the floor.

(a) Each base is a triangle with base 2.4 and height 1.6:

B = \frac{1}{2} \times 2.4 \times 1.6 = 1.92\ \text{m}^2

The perimeter of the triangle is P = 2 + 2 + 2.4 = 6.4 m, so the three rectangles together have area Ph = 6.4 \times 3 = 19.2\ \text{m}^2. Then

S = 2B + Ph = 3.84 + 19.2 = 23.04\ \text{m}^2

Face by face, as the net shows: two sloping sides of 2 \times 3 = 6 each and a floor of 2.4 \times 3 = 7.2, giving 6 + 6 + 7.2 = 19.2.

(b) With no floor, one of the rectangles is missing. It is not a base, so neither S nor L is the answer as it stands; take the floor away from the total:

23.04 - 7.2 = 15.84\ \text{m}^2

Example 3

A planter with no lid

A wooden planter is a box 60 cm long, 30 cm wide and 40 cm tall, with no lid. How much wood is it made of? Give the answer in \text{cm}^2 and in \text{m}^2.

Solution

The planter has a bottom and four walls but no top. Take the bottom as the base: it is covered once, and the four walls are the lateral area. So the wood is B + Ph.

\begin{aligned} B &= 60 \times 30 = 1800 \\ P &= 2(60 + 30) = 180 \\ Ph &= 180 \times 40 = 7200 \end{aligned}

The wood is 1800 + 7200 = 9000\ \text{cm}^2.

Check face by face: the front and back are each 60 \times 40 = 2400, the two ends are each 30 \times 40 = 1200, and 2400 + 2400 + 1200 + 1200 = 7200 for the walls.

A square meter is 100 cm by 100 cm, which is 10{,}000\ \text{cm}^2. So the wood is 9000 \div 10{,}000 = 0.9\ \text{m}^2.

Example 4

Cladding the walls of a shed

Waldemar’s shed has a flat roof and a rectangular floor 3 m by 2.5 m, and its walls are 2.2 m high. He covers the outside of the four walls with wooden cladding, all but the door, which is 0.8 m wide and 1.9 m tall. One pack of cladding covers 6\ \text{m}^2. How many packs must he buy?

Solution

The walls are the lateral faces of a box whose base is the floor, so they have area L = Ph. The roof and floor are bases, and they are not clad.

\begin{aligned} P &= 2(3 + 2.5) = 11\ \text{m} \\ L &= 11 \times 2.2 = 24.2\ \text{m}^2 \end{aligned}

The door is a rectangle inside one wall, so its area comes off:

24.2 - 0.8 \times 1.9 = 24.2 - 1.52 = 22.68\ \text{m}^2

That needs 22.68 \div 6 = 3.78 packs. Packs come whole, so round up: 3 packs would leave part of a wall bare, so he buys 4 packs.

Example 5

The metal in a paint tin

A paint tin is a closed cylinder with diameter 16 cm and height 19 cm. How much metal is its surface? Give the exact answer and the answer rounded to one decimal place.

Solution

The diameter is 16, so the radius is r = 8. A closed tin has both bases.

The two circles: 2\pi r^2 = 2 \times \pi \times 64 = 128\pi. The curved side unrolls to a rectangle 2\pi \times 8 = 16\pi long and 19 tall: 2\pi r h = 16\pi \times 19 = 304\pi. So

S = 128\pi + 304\pi = 432\pi\ \text{cm}^2

The factored form is a check: 2\pi r(r + h) = 16\pi \times 27 = 432\pi. Only now multiply by \pi: 432\pi = 1357.168\ldots, which is 1357.2\ \text{cm}^2 to one decimal place.

Example 6

A cardboard tube and an open bin

(a) A cardboard tube is 28 cm long, with diameter 5 cm, and open at both ends. What is the area of its outside surface, rounded to one decimal place?

(b) A round metal bin has diameter 0.6 m and height 0.9 m, and no lid. Its outside, bottom included, is painted. What area is painted, exactly and to two decimal places?

Solution

(a) A tube open at both ends has no bases, so its outside is the lateral area alone. With the diameter given, L = \pi d h saves halving and doubling:

L = \pi \times 5 \times 28 = 140\pi \approx 439.8\ \text{cm}^2

(b) The bin has its bottom but no top, so the painted area is one base plus the curved side, as with the planter. The radius is 0.3 m.

\begin{aligned} \pi r^2 &= \pi \times 0.09 = 0.09\pi \\ 2\pi r h &= 2\pi \times 0.3 \times 0.9 = 0.54\pi \end{aligned}

The painted area is 0.09\pi + 0.54\pi = 0.63\pi\ \text{m}^2, which is 1.979\ldots \approx 1.98\ \text{m}^2.

The curved side is six times the bottom here, since \displaystyle \frac{2\pi r h}{\pi r^2} = \frac{2h}{r} = 6. In a tall, narrow container the side, not the ends, is most of the material.

Example 7

Working backwards from the area

(a) A closed tin has radius 6 cm and total surface area 168\pi\ \text{cm}^2. How tall is it?

(b) A cylinder is 7 m tall and its lateral area is 70\pi\ \text{m}^2. What is its radius?

Solution

(a) Put what is known into S = 2\pi r^2 + 2\pi r h and solve for h:

\begin{aligned} 2\pi(6^2) + 2\pi(6)h &= 168\pi \\ 72\pi + 12\pi h &= 168\pi \\ 12\pi h &= 96\pi \\ h &= 8 \end{aligned}

The tin is 8 cm tall. Subtracting 72\pi from both sides keeps them equal, and dividing by 12\pi is allowed because it is not zero; every term has a factor \pi, so the \pi cancels. Check: 72\pi + 2\pi(6)(8) = 72\pi + 96\pi = 168\pi.

(b) Here only the curved side is given, so use L = 2\pi r h:

\begin{aligned} 2\pi r \times 7 &= 70\pi \\ 14\pi r &= 70\pi \\ r &= 5 \end{aligned}

The radius is 5 m. Check: 2\pi \times 5 \times 7 = 70\pi.

Finding the radius from a total area is a harder problem, because r then appears squared as well, and it waits for later algebra.

Boundaries

Right prisms and right cylinders only. In an oblique prism, leaning over, the side faces are parallelograms, and their heights are not the distance between the bases, so L = Ph does not hold. An oblique cylinder’s side does not unroll into a rectangle either. For a leaning solid, find each face’s area separately.

A face has to be measured before it can be added. If a triangular base gives its sides but not its height, or its height but not its sloping sides, the missing length comes first, often from the Pythagorean theorem.

The formulas count the outside once. A hollow pipe with a real thickness has an inside surface as well, a second lateral area with the inner radius. Where two solids are joined, the faces pressed together are hidden, so the surface area of a combined solid is not the sum of the two surface areas.

Surface area is the exact covering, nothing more. Wrapping paper overlaps, a label is glued over itself, paint goes on in coats. Those extra amounts are for the problem to add.

Surface area is not volume. A cube of edge 2 cm and a box 1 cm by 1 cm by 5.5 cm both have surface area 24\ \text{cm}^2, but the cube holds 8\ \text{cm}^3 and the long box only 5.5\ \text{cm}^3.

Where this goes next

How much a prism or cylinder holds is the volume of prisms and cylinders, built on the same base area B, multiplied by the height instead of wrapped in a band.

Pointed solids change the side faces: pyramids and cones have triangles, or a sector of a circle, in place of the wrapping rectangle. A ball cannot be unrolled, yet a sphere’s surface area equals the lateral area of the cylinder that fits snugly round it.

Real objects are rarely one solid, and modeling with solids builds them from prisms, cylinders and cones. And when every length is multiplied by k, the surface area is multiplied by k^2, the rule in scale factor and area.

Exercises

Part 1: boxes and triangular prisms

Find the total surface area of a box 7 cm long, 4 cm wide and 3 cm tall.
Find the total surface area of a cube with edge 9 cm.
A prism’s bases are right triangles with legs 6 cm and 8 cm and longest side 10 cm. The prism is 15 cm long. Find its total surface area.
A prism’s bases are triangles with sides 13 cm, 13 cm and 10 cm; the height of each triangle, measured to the 10 cm side, is 12 cm. The prism is 20 cm long. Find its total surface area.

Part 1: which faces are covered

A prism’s base is a regular hexagon with sides 4 cm, and the prism is 11 cm tall. Find its lateral area.
A box with no lid is 25 cm long, 18 cm wide and 10 cm tall. How much card is it made of, ignoring the flaps that hold it together?
A room is 5 m by 4 m, and its walls are 2.5 m high. The walls are painted, apart from a window 1.5 m by 1.2 m and a door 0.9 m by 2 m. What area is painted?
A wooden wedge is a prism whose ends are right triangles 12 cm along the bottom, 5 cm tall at the back and 13 cm along the slope. The wedge is 8 cm wide. Every face is painted except the bottom. What area is painted?

Part 1: a base made of pieces, and working backwards

Figure 4The prism in Exercise 9. Its base is an L shape, and the two edges inside the corner are not marked.

The prism in the figure has an L-shaped base and is 4 cm deep. Find its total surface area.
A box has a square base with sides 5 cm, and its total surface area is 250\ \text{cm}^2. How tall is it?

Part 1: reasoning

Explain why the side faces of a right prism, laid next to each other in a net, form one rectangle whose length is the perimeter of the base.
The edge of a cube is doubled. By what number is its surface area multiplied? Show why.

Part 2: whole cylinders

Give each answer exactly, in terms of \pi, and rounded as asked.

A closed cylinder has radius 3 cm and height 10 cm. Find its total surface area, rounded to one decimal place.
A closed cylinder has diameter 14 cm and height 20 cm. Find its total surface area, rounded to one decimal place.

Part 2: only part of the surface

Find the lateral area of a cylinder with radius 1.5 m and height 4.5 m, rounded to two decimal places.
A round post has diameter 18 cm. A poster 22 cm tall is wrapped once round it with 1.5 cm of overlap. Find the length of the poster and its area, each rounded to one decimal place.
A round water tank stands on the ground. It has diameter 3 m and height 1.2 m. Its top and its curved side are painted, but not its bottom. What area is painted, rounded to two decimal places?

Part 2: comparing and working backwards

A cylindrical gift tin has radius 5 cm and height 16 cm. A box 10 cm by 10 cm by 16 cm fits exactly round it. Which needs more wrapping paper, and what fraction of the box’s surface area is the tin’s?
A closed cylinder has radius 4 cm and total surface area 120\pi\ \text{cm}^2. How tall is it?
A cylinder is 4 m tall and its lateral area is 72\pi\ \text{m}^2. Find its radius.
A closed cylinder has radius 20 cm and height 1.5 m. Find its total surface area in square meters, exactly and rounded to two decimal places.

Part 2: reasoning

Explain why the curved surface of a cylinder unrolls into a rectangle, and why the length of that rectangle is 2\pi r.
A closed cylinder has radius 3 cm and height 10 cm. (a) Show that doubling both the radius and the height multiplies its surface area by 4. (b) Show that doubling only the height does not double the surface area.
Explain why S = 2B + Ph gives the surface area of a cylinder as well as of a prism, and what B and P are for a cylinder.

Answers

122\ \text{cm}^2. 2(7 \times 4 + 7 \times 3 + 4 \times 3), which is 2(28 + 21 + 12) = 2 \times 61.
486\ \text{cm}^2. Six faces of 9^2 = 81, and 6 \times 81 = 486.
408\ \text{cm}^2. \displaystyle B = \frac{1}{2} \times 6 \times 8 = 24, so 2B = 48; P = 24, so Ph = 24 \times 15 = 360; 48 + 360 = 408.
840\ \text{cm}^2. \displaystyle B = \frac{1}{2} \times 10 \times 12 = 60, so 2B = 120; P = 36, so Ph = 36 \times 20 = 720; 120 + 720 = 840.
264\ \text{cm}^2. P = 6 \times 4 = 24, and L = Ph = 24 \times 11. The base’s area is not needed for the lateral area.
1310\ \text{cm}^2. One base and the walls: B = 25 \times 18 = 450, P = 86, Ph = 86 \times 10 = 860, and 450 + 860 = 1310.
41.4\ \text{m}^2. The walls: P = 18 and L = 18 \times 2.5 = 45. Take off the window, 1.8, and the door, 1.8: 45 - 3.6 = 41.4.
204\ \text{cm}^2. The two ends: \displaystyle 2 \times \frac{1}{2} \times 12 \times 5 = 60. The slope: 13 \times 8 = 104. The back: 5 \times 8 = 40. Total 60 + 104 + 40 = 204, which is the full surface area 300 less the bottom, 12 \times 8 = 96.
124\ \text{cm}^2. The unmarked edges are 6 - 2 = 4 and 5 - 2 = 3. The base is a 6 by 2 rectangle and a 2 by 3 rectangle, so B = 12 + 6 = 18. Its perimeter is 6 + 2 + 4 + 3 + 2 + 5 = 22. S = 2 \times 18 + 22 \times 4 = 36 + 88.
10 cm. 2 \times 25 + 20h = 250, so 20h = 200.
Each side face joins an edge of one base to the matching edge of the other, and in a right prism it is a rectangle whose other side is the height h. So all the side faces are h tall, and in the net they stand side by side with their tops in one straight line and their bottoms in another, making one rectangle h tall. Its length is the sum of their widths, the base’s edges taken once each going round, and that sum is the perimeter P.
By 4. With edge a the surface area is 6a^2; with edge 2a it is 6(2a)^2 = 24a^2, which is 4 \times 6a^2. Each face’s area is multiplied by 2 \times 2 = 4, so the total is too.
78\pi \approx 245.0\ \text{cm}^2. 2\pi(9) + 2\pi(3)(10) = 18\pi + 60\pi.
378\pi \approx 1187.5\ \text{cm}^2. r = 7: 2\pi(49) + 2\pi(7)(20) = 98\pi + 280\pi.
13.5\pi \approx 42.41\ \text{m}^2. 2\pi \times 1.5 \times 4.5 = 13.5\pi.
Length 18\pi + 1.5 \approx 58.0 cm; area 22(18\pi + 1.5) = 396\pi + 33 \approx 1277.1\ \text{cm}^2. The poster goes once round, which is the circumference \pi d = 18\pi, plus the overlap.
5.85\pi \approx 18.38\ \text{m}^2. r = 1.5: the top is \pi(1.5)^2 = 2.25\pi and the side is 2\pi(1.5)(1.2) = 3.6\pi.
The box needs more. Tin: 2\pi(25) + 2\pi(5)(16) = 50\pi + 160\pi, which is 210\pi \approx 659.7\ \text{cm}^2. Box: 2(100 + 160 + 160) = 840\ \text{cm}^2. The fraction is \displaystyle \frac{210\pi}{840} = \frac{\pi}{4} \approx 0.785.
11 cm. 32\pi + 8\pi h = 120\pi, so 8\pi h = 88\pi.
9 m. 2\pi r \times 4 = 72\pi, so 8\pi r = 72\pi.
0.68\pi \approx 2.14\ \text{m}^2. In meters r = 0.2: 2\pi(0.04) + 2\pi(0.2)(1.5) = 0.08\pi + 0.6\pi.
The curved side bends in only one direction: every line on it from the top rim straight down to the bottom rim is straight. Cutting along one of these and flattening the surface stretches nothing, the way a rolled sheet of paper unrolls. The cut edges are h long and the rims become straight edges, so the flat shape is a rectangle h tall. Its long edge was the rim of a base circle, so its length is that circle’s circumference, 2\pi r.
(a) The surface area is 78\pi\ \text{cm}^2. With radius 6 and height 20 it is 2\pi(36) + 2\pi(6)(20), which is 72\pi + 240\pi = 312\pi, and 312\pi = 4 \times 78\pi. In general 2\pi(2r)(2r + 2h) = 4 \times 2\pi r(r + h). (b) With radius 3 and height 20 it is 18\pi + 120\pi = 138\pi, not 156\pi. The side doubles but the two circles do not change.
Both solids are two congruent bases joined by a side surface that unrolls into one rectangle, h tall and as long as the distance round the base. So in both the area is two bases plus that rectangle, 2B + Ph. For a cylinder the base is a circle, so B = \pi r^2, and the distance round it is its circumference, P = 2\pi r; that gives 2\pi r^2 + 2\pi r h.