Surface Area: Practise Online

Either paper 2 marks

The diagram shows a cube. The dashed lines are the edges you cannot see from this side.

Each edge of the cube is \(2\) cm long.

Find the total surface area of the cube.
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\(24\ \text{cm²}\)
Every face of a cube is a square with the same edge, so work out one face and multiply by the number of faces.

One face: \(2 \times 2 = 4\) cm²

A cube has \(6\) faces:

\(6 \times 4 = 24\) cm²
- M1area of one face \(4\) cm² (or a method using 6 faces)
- A1total surface area \(24\) cm²
Either paper 2 marks

Here is a cuboid.

Part of the net of the cuboid has been drawn for you on the centimetre grid. The rectangle with its bottom left corner at \((4, 3)\) is the base of the cuboid.

Two faces are missing. One is joined to the left edge of the base. The other is joined to the top edge of the rectangle above the base.

Complete the net: plot the four corners of each missing face on the grid, including any corner that is already a corner of the net.

You cannot draw on this page. Draw this on paper, then show the answer and check your drawing against the model.
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Face 1: a rectangle 2 cm wide and 4 cm high to the left of the base, with opposite corners at (2, 3) and (4, 7). Face 2: a rectangle 5 cm wide and 4 cm high above the rectangle that is above the base, with opposite corners at (4, 9) and (9, 13).

Each face of the cuboid is a rectangle whose sides are two of the three edges 5 cm, 4 cm and 2 cm. The drawn faces show which pairs of edges are already used.

One missing face goes to the left of the base. It shares the left edge of the base of length 4 cm, and its other side is the remaining edge of 2 cm. So it is a rectangle 2 cm wide and 4 cm high with corners \((2, 3)\) and \((4, 7)\).

The other missing face goes above the rectangle that is above the base. It shares the top edge of the rectangle above the base of length 5 cm, and its other side is 4 cm. So it is a rectangle 5 cm wide and 4 cm high with corners \((4, 9)\) and \((9, 13)\).

Both faces share a whole edge with the net, so they fold onto the edges of the cuboid.
- B1at least six corners plotted, so one missing face is right: 2 cm wide and 4 cm high to the left of the base, corners (2, 3) and (4, 7), or 5 cm wide and 4 cm high above the rectangle that is above the base, corners (4, 9) and (9, 13)
- B1both missing faces right, every corner plotted
Either paper 3 marks

Here is a cuboid.

Calculate the total surface area of the cuboid.
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\(314\ \text{cm²}\)
A cuboid has \(6\) rectangular faces, and they come in matching pairs — front and back, left and right, top and bottom. So work out three of them and double the total.

Front: \(8 \times 9 = 72\) cm²

Side: \(5 \times 9 = 45\) cm²

Top: \(8 \times 5 = 40\) cm²

\(72 + 45 + 40 = 157\) cm²

\(2 \times 157 = 314\) cm²
- M1areas of the three different faces \(72\), \(45\) and \(40\) cm²
- M1sum of the three face areas \(157\) doubled (or all six faces added)
- A1total surface area \(314\) cm²
Either paper 2 marks

Here is cuboid \(A\).

Cuboid \(B\) is made by joining two cuboids like \(A\), so that the faces measuring \(6\) cm by \(7\) cm are touching.

Not drawn accurately.

Volume of \(A\) : volume of \(B\) = \(1 : 2\)

Callum says, "The surface area of B is more than 1.8 times the surface area of A."

Is Callum correct?

Tick one box.

Yes / No / Cannot tell

Give a reason for your answer.
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No. The two faces that are joined are hidden, so \(SA(B) = 2 \times 292 - 2 \times 42 = 500\) cm². This is less than \(1.8 \times 292 = 525.6\) cm², which is \(1.8\) times \(SA(A)\).
Work out the surface area of \(A\).

\(SA(A) = 2(8 \times 6 + 8 \times 7 + 6 \times 7) = 292\) cm²

Two copies of \(A\) would have \(2 \times 292 = 584\) cm² of surface, but where they are joined, two faces of \(42\) cm² are hidden inside the solid and are no longer part of the surface.

\(SA(B) = 584 - 2 \times 42 = 584 - 84 = 500\) cm²

Check from the dimensions of \(B\): \(2(16 \times 6 + 16 \times 7 + 6 \times 7) = 500\) cm².

The claim involves \(1.8 \times 292 = 525.6\) cm². \(500\) cm² is less than \(525.6\) cm².

So the answer is No. The volumes do double when two cuboids are joined, but the surface area does not, because the joined faces are hidden: \(SA(B)\) is always less than \(2 \times SA(A)\).
- M1joined faces hidden: \(2 \times 292 - 2 \times 42 = 500\) cm² for B, with A at \(292\) cm²
- A1No, since \(500\) cm² is less than \(525.6\) cm²
Either paper 3 marks

Here is a cuboid.

Work out the total surface area of the cuboid.

Give your answer in square metres.
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\(5.6816\ \text{m²}\)
The area is wanted in square metres, so every length must be in m before multiplying. Change each length given in centimetres to metres by dividing by 100.

The three lengths in m are:

\(\text{width} = 0.54\text{ m },\quad \text{depth} = 1.37\text{ m },\quad \text{height} = 1.1\text{ m }\)

A cuboid has \(6\) rectangular faces in three matching pairs. Work out one of each pair.

Front: \(0.54 \times 1.1 = 0.594\) m²

Side: \(1.37 \times 1.1 = 1.507\) m²

Top: \(0.54 \times 1.37 = 0.7398\) m²

\(0.594 + 1.507 + 0.7398 = 2.8408\) m²

\(2 \times 2.8408 = 5.6816\) m²
- M1all three lengths in the same unit: \(0.54\), \(1.37\) and \(1.1\) m
- M1the three different faces \(0.594\), \(1.507\) and \(0.7398\) m², each used twice
- A1total surface area \(5.6816\) m²
Either paper 4 marks

Here is a solid prism that is \(14\) cm long.

The cross-section of the prism is a right-angled triangle with sides \(3\) cm, \(4\) cm and \(5\) cm. The right angle is between the \(3\) cm side and the \(4\) cm side.

Find the total surface area of the prism.
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\(180\ \text{cm²}\)
The prism has \(5\) faces: the two triangular ends, and three rectangles between them. Write them all down before adding anything.

The two triangles. The right angle is between the \(3\) cm and \(4\) cm sides, so those two are the base and the height:

\(\frac{1}{2} \times 3 \times 4 = 6\) cm², and there are two of them: \(12\) cm²

The three rectangles. Each is \(14\) cm long, and their widths are the three sides of the triangle:

\(3 \times 14 = 42\) cm²

\(4 \times 14 = 56\) cm²

\(5 \times 14 = 70\) cm²

Those three together are \((3 + 4 + 5) \times 14 = 12 \times 14 = 168\) cm² — the perimeter of the cross-section times the length, which is worth knowing as a shortcut.

Total: \(12 + 168 = 180\) cm²
- M1area of one triangle \(6\) cm², so both ends \(12\) cm²
- M1areas of the three rectangles \(42\), \(56\) and \(70\) cm²
- M1adds the two triangles and the three rectangles, \(12 + 168\)
- A1total surface area \(180\) cm²
Either paper 3 marks

The diagram shows a solid cube placed on a horizontal table.

The force exerted by the cube on the table is \(196\) newtons. The pressure on the table due to the cube is \(4\) newtons/cm².

\(\text{pressure} = \text{force} \div \text{area}\)

Show that the total surface area of the cube is less than \(310\) cm².
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\(\text{base area} = 196 \div 4 = 49\) cm², so \(\text{total surface area} = 6 \times 49 = 294\) cm², and \(294 < 310\).
Only one face of the cube touches the table, so the pressure acts on the base.

\(\text{area} = \text{force} \div \text{pressure}\)

\(\text{base area} = 196 \div 4 = 49\) cm²

A cube has \(6\) identical square faces:

\(\text{total surface area} = 6 \times 49 = 294\) cm²

Compare with the bound:

\(294 < 310\)

So the total surface area of the cube is less than \(310\) cm².
- M1base area \(196 \div 4 = 49\) cm²
- M1total surface area \(6 \times 49 = 294\) cm²
- A1compares \(294\) with \(310\) and concludes \(294 < 310\)
Either paper 4 marks

A solid cube has a total surface area of \(486\) cm².

Find the volume of the solid.
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\(729\ \text{cm³}\)
The edge is not given, so work back to it from the surface area. A cube has \(6\) identical square faces.

One face: \(486 \div 6 = 81\) cm²

The face is a square, so the edge is its square root:

\(\sqrt{ 81 } = 9\) cm

Now cube the edge to find the volume:

\(9^3 = 729\) cm³
- M1divide the surface area by 6 to get one face, \(81\) cm²
- M1square root of the face area gives the edge, \(9\) cm
- M1cube the edge, \(9^3\)
- A1volume \(729\) cm³
Either paper 5 marks

The diagram shows a cuboid. The height of the cuboid is \(h\) cm.

(a)
Assume that the total surface area of the cuboid is \(108\) cm².

Work out the volume of the cuboid.

(b)
In fact, the total surface area of the cuboid is larger than \(108\) cm².

What does this mean about the volume of the cuboid?

Tick one box and give a reason for your answer.

- It is bigger than the answer to part (a) - It is smaller than the answer to part (a) - It is the same as the answer to part (a) - It could be any of the above
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(a) \(72\ \text{cm³}\)
(b) It is bigger than the answer to part (a), because 6 cm and 4 cm stay the same, so the volume changes only with h, which is larger
The top and bottom are each \(6\) cm by \(4\) cm, so together they have area

\(2 \times 6 \times 4 = 48\) cm²

The four upright faces form a band of length \(2 \times (6 + 4) = 20\) cm and height \(h\) cm. The total surface area is

\(48 + 20h = 108\)

\(20h = 60\)

\(h = 60 \div 20 = 3\)

Volume \(= 6 \times 4 \times 3 = 72\) cm³
Two edges of the cuboid are fixed, so the surface area is \(48 + 20h\) and the only thing that can change is \(h\).

The area is larger than \(108\) cm², so \(h\) is larger than \(3\) cm.

The volume is \(6 \times 4 \times h\), so it changes in the same direction as \(h\).

The volume is bigger than the answer to part (a).
- (a) M1surface area as \(2 \times 6 \times 4 + 20h = 108\)
- (a) M1solves to \(h = 3\)
- (a) A1volume \(72\) cm³
- (b) A1ticks: it is bigger than the answer to part (a)
- (b) B1reason: the base edges \(6\) cm and \(4\) cm are fixed, so the volume \(6 \times 4 \times h\) changes with \(h\), which is larger
Either paper 3 marks

The diagram shows a cube and a cuboid.

The cube is placed on the top face of the cuboid to make a solid. The whole of the base of the cube is on the top face of the cuboid.

The diagrams are not drawn to scale.

Calculate the total surface area of the solid.
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\(202\ \text{cm²}\)
The base of the cube is hidden against the cuboid. It covers a square of the cuboid's top face the same size, so that square is hidden too.

Cuboid, all six faces:

\(5 \times 7 + 4 \times 7 + 5 \times 4 = 83\) cm²

\(2 \times 83 = 166\) cm²

The top of the cube is \(3 \times 3 = 9\) cm², which is the same size as the square it covers on the cuboid, so the two cancel. That leaves only the four sides of the cube to add:

\(4 \times 9 = 36\) cm²

\(166 + 36 = 202\) cm²
- M1surface area of the cuboid \(166\) cm²
- M1the cube adds \(36\) cm² (its top, less the covered patch of the cuboid, and the four sides)
- A1total \(202\) cm²