Resources For Teachers For Tutors For Students & Parents AI New Pricing
Year 10 Maths - 10 Measurement

Surface Area of Prisms and Cylinders

20 practice questions 2 video lessons Theory + worked examples
Practice 20 questions
Practice questions

Every question with a fully worked solution.

Start practising
Watch 2 video(s)
  • Surface Area of Prisms and Cylinders - Video - Surface Area of Right Prisms Watch
  • Surface Area of Prisms and Cylinders - Video - Surface Area Of A Cylinder Watch
Create a free accountTrack your progress and save your work as you go.
Create free account

Theory

The surface area of a solid is the total area of all its faces, the area of its net. This Year 10 maths page revises Year 9: cubes, rectangular and triangular prisms, trapezium prisms, open boxes, cylinders (closed, open at one end and pipes), exact answers in \(\pi\), practical painting and covering problems, and working backwards to a missing length.

A right prism is a solid with two identical, parallel ends joined by rectangles. The shape of the end (the cross-section) names the prism: rectangular, triangular, trapezium and so on.

The surface area (SA) is the total area of all the faces. If you unfold the prism flat you get its net; the surface area is the area of the net.

A rectangular prism has three pairs of matching faces. A triangular prism has two triangular ends and three rectangles. The rectangles around any prism fit together into one long rectangle whose length is the perimeter of the end and whose width is the length of the prism.

In real problems some faces are missing or hidden: a pool or planter has no top, a shed floor is not painted, and where two solids are glued together the touching faces are no longer on the surface.

A cylinder has two identical circular ends joined by a curved surface. Cut the curved surface straight down and unroll it: it is a rectangle whose length is the circumference \(2\pi r\) and whose width is the height \(h\). A closed cylinder has both circles, a cup or open tank has one, and a pipe has none.

Net of a rectangular prismThe net of a rectangular prism: a column of four rectangles (top, front, bottom, back) with an end rectangle on each side of the front. Length l, width w and height h are marked. top front bottom back end end l h w
Unfold a rectangular prism and you get six rectangles in three matching pairs: \(SA = 2(lw + lh + wh)\).
Triangular prism with its two triangular endsA triangular prism lying on its side. The two triangular ends are outlined in red. Three rectangular faces join the ends and run the length of the prism. base length end
Any prism has two equal ends. The rectangles around the sides together measure perimeter of the end \(\times\) length.
Net of a closed cylinderThe net of a closed cylinder: two circles of radius r and a rectangle. The rectangle's width is the height h and its length is the circumference 2 pi r. r curved surface length = 2Ο€r h
Unroll the curved surface and it is a rectangle: length \(2\pi r\) (the circumference), width \(h\). So its area is \(2\pi rh\).
PrismFacesSurface area
Cube, edge \(s\)6 equal squares\(6s^{2}\)
Rectangular prism3 pairs of rectangles\(2(lw + lh + wh)\)
Any right prism2 ends + rectangles\(2 \times\) end area \(+\) perimeter of end \(\times\) length
\[SA = 2(lw + lh + wh) \qquad SA = 2A_{\text{end}} + P_{\text{end}} \times l\]
SA=2Aend+Pend×l
Sides you cannot see. The rectangles need every side of the end shape. Use Pythagoras' theorem for a missing sloping side, such as the hypotenuse of a right-angled triangle or the slanted side of a trapezium.
Composite solids. A hole cut through a solid removes two openings but adds the inside walls. Gluing one solid on another hides the area where they touch, once from each solid.

Cylinders:

\[SA = 2\pi r^{2} + 2\pi rh \qquad \text{curved surface} = 2\pi rh\]
SA=2πr2+2πrh
CylinderSurfacesArea
Closed (solid)2 circles + curved surface\(2\pi r^{2} + 2\pi rh\)
Open at one end1 circle + curved surface\(\pi r^{2} + 2\pi rh\)
Open at both ends (tube, pipe)curved surface only\(2\pi rh\)
Diameter given? Halve it first: \(r = \dfrac{d}{2}\). The curved surface can also be written \(\pi dh\).
Circumference given? Find the radius first: \(r = \dfrac{C}{2\pi}\). A circumference of \(10\pi\) cm means \(r = 5\) cm.

How to find the surface area of a prism

  1. Sketch the prism or its net and list every face. Check which faces are really there (a lid? a floor?).
  2. Find any missing side lengths, using Pythagoras' theorem if needed. Put all lengths in the same unit.
  3. Calculate each face area. For matching faces, find one and multiply.
  4. Add the face areas and give square units.

Problems

  1. Paint, tiles or material: find the area that is covered, then multiply by the price per square metre.
  2. Working backwards: write the surface area with a pronumeral for the unknown length, set it equal to the given area and solve.
  3. Hole through a solid: total \(=\) outside surface \(-\) the two openings \(+\) the inside walls.
  4. Stacked or glued solids: add both surface areas, then subtract the touching area twice (once from each solid).
  5. Cylinders: \(\pi r^{2}\) for each end that is there plus \(2\pi rh\) for the curved surface. Halve a diameter first. A painted tank standing on the ground has no painted base.
  6. Working backwards with a prism: a closed box with a \(6\) cm square base and \(SA = 240\text{ cm}^2\) gives \(2 \times 36 + 4 \times 6h = 240\), so \(24h = 168\) and \(h = 7\) cm.
Example 1 β€” Triangular prism
A prism is \(20\) cm long. Its ends are right-angled triangles with shorter sides of \(9\) cm and \(12\) cm. Find its surface area.
Solution

The rectangles need all three sides of the triangle, so find the sloping side \(x\) first.

Triangular prism 20 cm long with right-angled ends 9 cm by 12 cmA triangular prism 20 cm long. Each end is a right-angled triangle with shorter sides 9 cm and 12 cm and an unknown sloping side x. 12 cm 9 cm x 20 cm
\(x^{2}\)\(=\)\(9^{2} + 12^{2} = 225\)
\(x\)\(=\)\(15\text{ cm}\)
\(\text{two ends}\)\(=\)\(2 \times \dfrac{1}{2} \times 12 \times 9 = 108\)
\(\text{rectangles}\)\(=\)\((9 + 12 + 15) \times 20 = 720\)
\(SA\)\(=\)\(108 + 720 = \mathbf{828}\text{ cm}^2\)
SA=108+(9+12+15)×20=828
Example 2 β€” No top, mixed units
A planter box with no top is \(1.2\) m long, \(50\) cm wide and \(40\) cm deep. Find the area of timber needed for the floor and the four sides, in square metres.
Solution

Change every length to metres: \(50\text{ cm} = 0.5\text{ m}\) and \(40\text{ cm} = 0.4\text{ m}\). There is no top, so count the floor once.

\(\text{floor}\)\(=\)\(1.2 \times 0.5 = 0.6\)
\(\text{long sides}\)\(=\)\(2 \times 1.2 \times 0.4 = 0.96\)
\(\text{short sides}\)\(=\)\(2 \times 0.5 \times 0.4 = 0.4\)
\(A\)\(=\)\(0.6 + 0.96 + 0.4\)
\(\)\(=\)\(\mathbf{1.96}\text{ m}^2\)
A=0.6+0.96+0.4=1.96

To cost the timber, multiply by the price per square metre. At \(\$35\) per square metre: \(1.96 \times 35 = \$68.60\).

Example 3 β€” Closed cylinder
Find the surface area of a closed cylinder with radius \(5\) cm and height \(8\) cm, as an exact value and correct to two decimal places.
Solution
Closed cylinder radius 5 cm height 8 cmA closed cylinder with radius 5 cm and height 8 cm. 5 cm 8 cm
\(SA\)\(=\)\(2\pi r^{2} + 2\pi rh\)
\(\)\(=\)\(2\pi \times 5^{2} + 2\pi \times 5 \times 8\)
\(\)\(=\)\(50\pi + 80\pi\)
\(\)\(=\)\(\mathbf{130\pi}\text{ cm}^2\)
\(\)\(=\)\(408.4070\ldots\)
\(\)\(\approx\)\(\mathbf{408.41}\text{ cm}^2\)
SA=2π×52+2π×5×8=130π

Collect the \(\pi\) terms before you use the calculator. The two circles give \(50\pi\); the curved surface gives \(80\pi\).

Example 4 β€” Working backwards
A closed cylinder has a radius of \(4\) cm and a total surface area of \(120\pi\text{ cm}^2\). Find its height.
Solution

Substitute what you know, divide every term by \(\pi\), then solve.

\(2\pi \times 4^{2} + 2\pi \times 4 \times h\)\(=\)\(120\pi\)
\(32\pi + 8\pi h\)\(=\)\(120\pi\)
\(32 + 8h\)\(=\)\(120\)
\(8h\)\(=\)\(88\)
\(h\)\(=\)\(\mathbf{11}\text{ cm}\)
32π+8πh=120π,h=11

If the area is a decimal (not a multiple of \(\pi\)), work out \(2\pi r^{2}\) and \(2\pi r\) on the calculator and solve the same way.

Common pitfalls

Count every face once. A rectangular prism has six faces. Listing three and forgetting to double them is the most common error.
Height or sloping side? For the rectangles of a triangular or trapezium prism use the sides of the end shape, not its perpendicular height.
Read for missing faces. An open box, a pool or a shed floor that is not painted has one face less. Glued faces are hidden.
Same units first. Convert centimetres to metres before multiplying, or the area will be in a mix of units.
Check the ends of a cylinder. Closed, open at one end and open at both ends give three different answers. \(2\pi r^{2}\) is not \((2\pi r)^{2}\).

Frequently asked questions

What is the surface area of a prism?

It is the total area of all the faces of the prism, the same as the area of its net. It is measured in square units.

How do you find the surface area of a rectangular prism?

Find the area of the top, the front and the side, add them, then double: \(SA = 2(lw + lh + wh)\).

How do you find the surface area of a triangular prism?

Add the areas of the two triangular ends to the three rectangles. The rectangles together equal the perimeter of the triangle times the length of the prism.

How do you find the surface area of an open box?

Add the areas of the base and the four sides, but leave out the top. That is the closed surface area minus the area of the missing lid.

What is the formula for the surface area of a cylinder?

For a closed cylinder, \(SA = 2\pi r^{2} + 2\pi rh\). The first part is the two circular ends and the second part is the curved surface, which unrolls into a rectangle \(2\pi r\) long and \(h\) wide.

How do you find the surface area of an open cylinder?

Open at one end: \(\pi r^{2} + 2\pi rh\). Open at both ends, like a pipe: \(2\pi rh\) only.