Surface Area of Composite Solids

Find the surface area of a solid made by joining simpler solids: leave out the faces where the parts touch, see when circles cancel, and leave out the base of a solid on the ground.

Before you start

From Surface Area of Prisms: a cuboid $l$ by $b$ by $h$ has total surface area $2(lb + bh + lh)$ , a cube of edge $l$ has $6l^{2}$ , and any prism has $2A + Pl$ (two ends of area $A$ , and the perimeter $P$ of the cross-section times the length $l$ ). An open solid is done face by face.
From Surface Area of Cylinders: a cylinder of radius $r$ and height $h$ has curved surface area $2\pi rh$ and total surface area $2\pi r^{2} + 2\pi rh$ . A half-cylinder of radius $r$ and length $l$ has a curved surface $\pi rl$ , two semicircular ends $\pi r^{2}$ in all, and a flat rectangle $2r \times l$ .
From Volume of Composite Solids: a composite solid is made of simple solids put together. Its volume is the sum of the parts; its surface area needs more care, and that is this lesson. (Solids with a hole drilled through them, pipes, and solids cut into pieces are in the lesson Surface Area of Solids with Holes and Cuts.) Each question says which $\pi$ to use; answers that are not exact are given correct to 3 significant figures, rounding only at the end.

Only the Outside Counts

The surface area of a solid is the area of everything you could touch, paint or wrap on its outside. When two solids are glued together face to face, the place where they touch is inside the new solid: you cannot touch it or paint it, so it is not part of the surface any more.

Look at the two cubes below. Apart, each has 6 faces. Glued together, the right-hand face of the blue cube and the left-hand face of the amber cube are pressed against each other and both are hidden. The new solid has $6 + 6 - 2 = 10$ square faces on its outside.

Surface area of a composite solid

The surface area of a composite solid is the area of its outside only.

Where two parts are joined face to face, the area where they touch is hidden on both parts:

Total surface area $=$ sum of the total surface areas of the parts $-\ 2 \times$ area where they touch.

Or go face by face, adding only the parts of faces that are on the outside. Doing it both ways is a good check.

Example 1: Two cubes joined

Two cubes, each of edge 5 cm, are glued together face to face. Find the total surface area of the solid they make.

Solution

Each face of a cube is a square of area $5 \times 5 = 25\text{ cm}^{2}$ .

The two cubes have $2 \times 6 = 12$ faces. Two of them, one on each cube, are hidden where the cubes meet.

Total surface area $= (12 - 2) \times 25 = 10 \times 25 = 250\text{ cm}^{2}$

Check: the solid is a cuboid 10 cm by 5 cm by 5 cm.

$2(lb + bh + lh) = 2(10 \times 5 + 5 \times 5 + 10 \times 5) = 2(50 + 25 + 50) = 250\text{ cm}^{2}$

The total surface area of the solid is $250\text{ cm}^{2}$ .

Example 2: A cuboid standing on a cuboid

A wooden display stand is made of a cuboid 10 cm wide, 8 cm deep and 12 cm high, standing in the middle of the top of a cuboid 20 cm wide, 16 cm deep and 5 cm high. Find the total surface area of the stand.

Solution

Total surface area of the lower cuboid $= 2(20 \times 16 + 16 \times 5 + 20 \times 5) = 2(320 + 80 + 100) = 1000\text{ cm}^{2}$

Total surface area of the upper cuboid $= 2(10 \times 8 + 8 \times 12 + 10 \times 12) = 2(80 + 96 + 120) = 592\text{ cm}^{2}$

The upper cuboid touches the lower one along its base, a rectangle 10 cm by 8 cm, so area where they touch $= 10 \times 8 = 80\text{ cm}^{2}$ . This hides the upper cuboid's base and $80\text{ cm}^{2}$ of the lower cuboid's top.

Total surface area $= 1000 + 592 - 2 \times 80 = 1432\text{ cm}^{2}$

Check, looking from above: everything that faces up (the uncovered part of the lower top, $320 - 80 = 240$ , and the upper top, 80) adds up to the whole lower top, $320\text{ cm}^{2}$ , and the bottom is $320\text{ cm}^{2}$ too. The sides of the lower cuboid are $2(16 \times 5) + 2(20 \times 5) = 360\text{ cm}^{2}$ and the sides of the upper cuboid are $2(8 \times 12) + 2(10 \times 12) = 432\text{ cm}^{2}$ .

$320 + 320 + 360 + 432 = 1432\text{ cm}^{2}$ , as before.

The total surface area of the stand is $1432\text{ cm}^{2}$ .

Two Ways: Separate Parts or One Prism

Some composite solids are also prisms. A step block made of two cuboids has an L-shaped cross-section all along its length, so it is an L-shaped prism and $2A + Pl$ works for it. Doing it both ways checks the answer.

Example 3: A concrete step

A concrete step is 100 cm long. Its end is L-shaped, with the lengths shown beside it. Find its total surface area (a) as two cuboids, (b) as one prism. Give your answer in $\text{cm}^{2}$ and in $\text{m}^{2}$ .

Solution

(a) Cut the step along the level of the lower step's top into a lower cuboid A, 60 cm by 20 cm by 100 cm, and an upper cuboid B, 30 cm by 20 cm by 100 cm.

Total surface area of A $= 2(60 \times 20 + 20 \times 100 + 60 \times 100) = 2(1200 + 2000 + 6000) = 18\,400\text{ cm}^{2}$

Total surface area of B $= 2(30 \times 20 + 20 \times 100 + 30 \times 100) = 2(600 + 2000 + 3000) = 11\,200\text{ cm}^{2}$

B stands on A along a rectangle 30 cm by 100 cm: area where they touch $= 30 \times 100 = 3000\text{ cm}^{2}$ .

Total surface area $= 18\,400 + 11\,200 - 2 \times 3000 = 23\,600\text{ cm}^{2}$

(b) The cross-section is the L-shaped end. Its area is the two rectangles, and its perimeter is the six sides.

$A = 60 \times 20 + 30 \times 20 = 1200 + 600 = 1800\text{ cm}^{2}$

$P = 60 + 20 + 30 + 20 + 30 + 40 = 200$ cm

Total surface area $= 2A + Pl = 2 \times 1800 + 200 \times 100 = 3600 + 20\,000 = 23\,600\text{ cm}^{2}$ , as in (a).

In $\text{m}^{2}$ : $23\,600 \div 10\,000 = 2.36\text{ m}^{2}$

The total surface area of the step is $23\,600\text{ cm}^{2}$ , which is $2.36\text{ m}^{2}$ .

A Cylinder Standing on Another Solid

Stand a cylinder of radius $r$ and height $h$ with its whole base on the top of a cuboid. Two circles of area $\pi r^{2}$ are hidden: the cylinder's base, and a circle cut out of the cuboid's top. But the cylinder's own top, also a circle of area $\pi r^{2}$ , is still on the outside. Face by face:

surface area of the cuboid $- \pi r^{2}$ (the hidden circle on its top) $+\ 2\pi rh$ (the curved surface) $+\ \pi r^{2}$ (the cylinder's top).

The $-\pi r^{2}$ and the $+\pi r^{2}$ cancel: the cylinder's top simply replaces the circle it covers. So all the cylinder adds is its curved surface. Looking down from above, you still see the whole of the cuboid's top, part of it lifted up onto the cylinder.

A cylinder or a cuboid standing on a solid

When a cylinder of radius $r$ and height $h$ stands with its whole base on a flat top face of another solid:

Total surface area $=$ total surface area of the other solid $+\ 2\pi rh$

If any part of the cylinder's base is not on that top face (it hangs over the edge), the circles do not cancel: go face by face.

A cuboid standing with its whole base on a flat top face works in the same way: its base hides a rectangle of that face, and its own top, the same rectangle, replaces it, so the cuboid adds only its four sides. On a cylinder it hides only the rectangle under it (a square, for a cube), not the circle (Practice 8).

Example 4: A cylinder on a cuboid

A solid is made of a cylinder of radius 3 cm and height 10 cm standing in the middle of the top of a cuboid 12 cm by 12 cm by 4 cm. Find its total surface area (a) in terms of $\pi$ , (b) correct to 3 significant figures, using the $\pi$ button on your calculator.

Solution

Total surface area of the cuboid $= 2(12 \times 12 + 12 \times 4 + 12 \times 4) = 2(144 + 48 + 48) = 480\text{ cm}^{2}$

The cylinder stands wholly on the top face, so the circles cancel and the cylinder adds only its curved surface.

Curved surface area $= 2\pi rh = 2 \times \pi \times 3 \times 10 = 60\pi\text{ cm}^{2}$

(a) Total surface area $= (480 + 60\pi)\text{ cm}^{2}$

Check, face by face: cuboid with a circle missing, $480 - 9\pi$ ; the cylinder's top $9\pi$ and curved surface $60\pi$ . Total $480 - 9\pi + 9\pi + 60\pi = 480 + 60\pi$ .

(b) $480 + 60\pi = 668.495\ldots \approx 668\text{ cm}^{2}$

The total surface area is $(480 + 60\pi)\text{ cm}^{2} \approx 668\text{ cm}^{2}$ .

Example 5: Icing a two-tier cake

Mrs Goh bakes a two-tier cake. The lower tier is a cylinder of diameter 30 cm and height 10 cm, and the upper tier, a cylinder of diameter 20 cm and height 8 cm, stands in the middle of it. She ices the top and the sides of the cake, but not underneath it. Find the area she ices, using the $\pi$ button on your calculator.

Solution

The radii are $30 \div 2 = 15$ cm and $20 \div 2 = 10$ cm.

Tops: the ring of the lower top that shows has area $\pi \times 15^{2} - \pi \times 10^{2} = 225\pi - 100\pi = 125\pi$ , and the upper top is $\pi \times 10^{2} = 100\pi$ . Together $125\pi + 100\pi = 225\pi\text{ cm}^{2}$ : from above you see one circle of radius 15 cm, exactly as the rule says.

Side of the lower tier $= 2\pi rh = 2 \times \pi \times 15 \times 10 = 300\pi\text{ cm}^{2}$

Side of the upper tier $= 2 \times \pi \times 10 \times 8 = 160\pi\text{ cm}^{2}$

The base of the cake on the board and the base of the upper tier are not iced.

Area iced $= 225\pi + 300\pi + 160\pi = 685\pi = 2151.99\ldots \approx 2150\text{ cm}^{2}$

Mrs Goh ices $685\pi \approx 2150\text{ cm}^{2}$ of cake.

A Half-Cylinder on a Cuboid

A half-cylinder lying on a cuboid is joined face to face, as in the first section. If its flat rectangle $2r \times l$ covers the cuboid's top exactly, the flat rectangle and the cuboid's top are both hidden, and what is left of the half-cylinder is its curved surface $\pi rl$ and its two semicircular ends.

The shortcut of the last section does not apply: the top of a half-cylinder is its curved surface, not a copy of the rectangle it covers, so work out the parts that are left on the outside.

Example 6: A letter box

A closed metal letter box is a cuboid 30 cm wide, 35 cm high and 40 cm long with a half-cylinder lying along its top, its flat face covering the top of the cuboid exactly. Find the total surface area of the letter box, using the $\pi$ button on your calculator.

Solution

The flat face of the half-cylinder covers the top exactly, so it is 30 cm wide: the diameter is 30 cm and the radius is $r = 30 \div 2 = 15$ cm. The half-cylinder is 40 cm long.

The cuboid's top and the half-cylinder's flat face are both hidden.

Cuboid without its top: base $30 \times 40 = 1200$ , front and back $2 \times 30 \times 35 = 2100$ , two sides $2 \times 40 \times 35 = 2800$ . Together $1200 + 2100 + 2800 = 6100\text{ cm}^{2}$ .

Curved surface of the half-cylinder $= \pi rl = \pi \times 15 \times 40 = 600\pi\text{ cm}^{2}$

Two semicircular ends $= \pi r^{2} = \pi \times 15^{2} = 225\pi\text{ cm}^{2}$

Total surface area $= 6100 + 600\pi + 225\pi = 6100 + 825\pi = 8691.81\ldots \approx 8690\text{ cm}^{2}$

Check, as one solid with a uniform cross-section (a rectangle with a semicircle on top), in the same way as $2A + Pl$ : the end has area $30 \times 35 + \frac{1}{2}\pi \times 15^{2} = 1050 + 112.5\pi$ and the distance round it is $30 + 35 + 35 + \pi \times 15 = 100 + 15\pi$ . Then $2(1050 + 112.5\pi) + (100 + 15\pi) \times 40 = 2100 + 225\pi + 4000 + 600\pi = 6100 + 825\pi$ , as before.

The total surface area of the letter box is $(6100 + 825\pi)\text{ cm}^{2} \approx 8690\text{ cm}^{2}$ .

Painting a Solid on the Ground

A solid standing on the ground cannot be painted underneath, so the face on the ground is left out. Read each question for what is covered: a solid against a wall is not painted at the back, an open box has no top, a cake is not iced underneath.

Example 7: Painting a pedestal

A concrete pedestal stands on the ground. It is a cuboid 1.2 m by 1.2 m by 0.5 m high with a cylinder of radius 0.4 m and height 1.5 m standing in the middle of its top. All of it except the bottom is to be painted. One tin of paint covers $2\text{ m}^{2}$ and costs $14.50. Using the $\pi$ button on your calculator, find (a) the area to be painted, (b) the number of tins needed and their cost.

Solution

(a) Cuboid without its bottom: top $1.2 \times 1.2 = 1.44$ , four sides $4 \times 1.2 \times 0.5 = 2.4$ . Together $1.44 + 2.4 = 3.84\text{ m}^{2}$ .

The cylinder stands wholly on the top, so the circles cancel: the cylinder adds its curved surface only.

Curved surface area $= 2\pi rh = 2 \times \pi \times 0.4 \times 1.5 = 1.2\pi\text{ m}^{2}$

Area to be painted $= 3.84 + 1.2\pi = 7.6099\ldots \approx 7.61\text{ m}^{2}$

(b) Number of tins $= 7.6099\ldots \div 2 = 3.80\ldots$ . Three tins cover only $6\text{ m}^{2}$ , so round up: 4 tins.

Cost $= 4 \times 14.50 = 58$ , so the cost is $58.00.

(a) About $7.61\text{ m}^{2}$ is painted. (b) 4 tins are needed, costing $58.00.

Common mistakes

Adding the parts' surface areas and stopping. Where two parts touch, that area is hidden: take it away.
Taking the joined area away only once. It is hidden on both parts, so subtract it twice: $1000 + 592 - 2 \times 80$ , not $1000 + 592 - 80$ .
Taking away the circle under a cylinder and forgetting its top. A cylinder standing wholly on a top face adds only its curved surface, $2\pi rh$ .
Cancelling a circle that is not hidden. When a cuboid stands on a cylinder, its base hides only the rectangle under it (a square, for a cube): take away that rectangle twice, not the circle.
Counting the flat face of a half-cylinder. When it covers a cuboid's top, the flat face and the top are both hidden.
Painting the bottom, or the back against a wall. A solid standing on the ground is not painted underneath, and a face against a wall is not painted.
Rounding a number of tins down. Tins are rounded up, or the last part is not painted.

Practice questions

Try each question before you open its solution. Answers that are not exact are correct to 3 significant figures.

Practice 1: A cube on a cube

A cube of edge 3 cm is glued in the middle of the top of a cube of edge 6 cm. Find the total surface area of the solid.

ShowHide solution

Large cube: $6 \times 6^{2} = 216\text{ cm}^{2}$ . Small cube: $6 \times 3^{2} = 54\text{ cm}^{2}$ .

Area where they touch $= 3 \times 3 = 9\text{ cm}^{2}$ , hidden on both cubes.

Total surface area $= 216 + 54 - 2 \times 9 = 252\text{ cm}^{2}$

Check, face by face: large cube $5 \times 36 + (36 - 9) = 207$ ; small cube, five faces $5 \times 9 = 45$ ; $207 + 45 = 252\text{ cm}^{2}$ .

The total surface area is $252\text{ cm}^{2}$ .

Practice 2: Three cubes in a row

Three cubes of edge 4 cm are glued face to face in a row. Find the total surface area of the solid, and check your answer by treating it as a cuboid.

ShowHide solution

Each face has area $4 \times 4 = 16\text{ cm}^{2}$ , and the three cubes have $3 \times 6 = 18$ faces.

There are two joins, and each hides 2 faces: $2 \times 2 = 4$ faces are hidden.

Total surface area $= (18 - 4) \times 16 = 14 \times 16 = 224\text{ cm}^{2}$

Check: a cuboid 12 cm by 4 cm by 4 cm has $2(12 \times 4 + 4 \times 4 + 12 \times 4) = 2(48 + 16 + 48) = 224\text{ cm}^{2}$ .

The total surface area is $224\text{ cm}^{2}$ .

Practice 3: A cuboid on the corner of a cuboid

A cuboid 8 cm by 8 cm by 3 cm high stands on the back left-hand corner of a cuboid 14 cm wide, 10 cm deep and 4 cm high, as shown. Find the total surface area of the solid.

ShowHide solution

Lower cuboid: $2(14 \times 10 + 10 \times 4 + 14 \times 4) = 2(140 + 40 + 56) = 472\text{ cm}^{2}$

Upper cuboid: $2(8 \times 8 + 8 \times 3 + 8 \times 3) = 2(64 + 24 + 24) = 224\text{ cm}^{2}$

Area where they touch $= 8 \times 8 = 64\text{ cm}^{2}$

Total surface area $= 472 + 224 - 2 \times 64 = 568\text{ cm}^{2}$

Check from above: tops together $= 140$ , bottom $140$ , lower sides $2(10 \times 4) + 2(14 \times 4) = 192$ , upper sides $4 \times 8 \times 3 = 96$ . $140 + 140 + 192 + 96 = 568\text{ cm}^{2}$ . (The upper cuboid's back and left faces are level with the lower cuboid's, but they are still on the outside.)

The total surface area is $568\text{ cm}^{2}$ .

Practice 4: A trophy

A trophy is a solid cylinder of radius 2.5 cm and height 8 cm standing in the middle of the top of a cuboid base 9 cm by 9 cm by 3 cm. Find its total surface area (a) in terms of $\pi$ , (b) taking $\pi = 3.142$ .

ShowHide solution

Cuboid: $2(9 \times 9 + 9 \times 3 + 9 \times 3) = 2(81 + 27 + 27) = 270\text{ cm}^{2}$

The cylinder stands wholly on the top, so it adds its curved surface: $2\pi rh = 2 \times \pi \times 2.5 \times 8 = 40\pi\text{ cm}^{2}$

(a) Total surface area $= (270 + 40\pi)\text{ cm}^{2}$

(b) $270 + 40 \times 3.142 = 270 + 125.68 = 395.68 \approx 396\text{ cm}^{2}$

(a) $(270 + 40\pi)\text{ cm}^{2}$ (b) $\approx 396\text{ cm}^{2}$

Practice 5: Finding the edge

Two identical cubes are glued face to face to make a cuboid with a total surface area of $360\text{ cm}^{2}$ . Find (a) the length of an edge of each cube, (b) the volume of the cuboid.

ShowHide solution

(a) Let the edge be $x$ cm. Each face has area $x^{2}\text{ cm}^{2}$ , and $12 - 2 = 10$ faces are on the outside.

$10x^{2} = 360$

$x^{2} = 36$

$x = 6$ (a length is positive)

(b) Volume $= 2 \times 6^{3} = 2 \times 216 = 432\text{ cm}^{3}$

(a) Each edge is 6 cm. (b) The volume is $432\text{ cm}^{3}$ .

Practice 6: A trophy in three parts

A trophy has three parts, each standing in the middle of the top of the one below: a cuboid 20 cm by 20 cm by 4 cm, a cuboid 12 cm by 12 cm by 6 cm, and a cylinder of radius 3 cm and height 15 cm. Find its total surface area (a) in terms of $\pi$ , (b) using the $\pi$ button on your calculator.

ShowHide solution

Lower cuboid: $2(20 \times 20 + 20 \times 4 + 20 \times 4) = 2(400 + 80 + 80) = 1120\text{ cm}^{2}$

Middle cuboid: $2(12 \times 12 + 12 \times 6 + 12 \times 6) = 2(144 + 72 + 72) = 576\text{ cm}^{2}$

The middle cuboid touches the lower one along $12 \times 12 = 144\text{ cm}^{2}$ , hidden on both: take away $2 \times 144 = 288\text{ cm}^{2}$ .

The cylinder stands wholly on the middle cuboid's top, so it adds its curved surface: $2\pi rh = 2 \times \pi \times 3 \times 15 = 90\pi\text{ cm}^{2}$

(a) Total surface area $= 1120 + 576 - 288 + 90\pi = (1408 + 90\pi)\text{ cm}^{2}$

Check from above: all that faces up adds up to the lower top, 400, and the bottom is 400; the sides are $4 \times 20 \times 4 = 320$ and $4 \times 12 \times 6 = 288$ . $400 + 400 + 320 + 288 = 1408$ , and the curved surface $90\pi$ .

(b) $1408 + 90\pi = 1690.74\ldots \approx 1690\text{ cm}^{2}$

(a) $(1408 + 90\pi)\text{ cm}^{2}$ (b) $\approx 1690\text{ cm}^{2}$

Practice 7: Concrete steps against a wall

A block of three concrete steps is 1 m long. Each step is 15 cm high and 25 cm deep, as its end shows. (a) Find the total surface area of the block in $\text{m}^{2}$ . (b) The block stands on the ground with its 45 cm high back against a wall. Find the area of concrete that can be seen, in $\text{m}^{2}$ .

ShowHide solution

(a) The block is a prism. Its end is three rectangles stacked, 75 cm, 50 cm and 25 cm wide, each 15 cm high.

$A = 75 \times 15 + 50 \times 15 + 25 \times 15 = 1125 + 750 + 375 = 2250\text{ cm}^{2}$

Going round the end: the bottom 75, three risers of 15 and three treads of 25, and the back 45. $P = 75 + 3 \times 15 + 3 \times 25 + 45 = 240$ cm.

The length is $1\text{ m} = 100$ cm. Total surface area $= 2A + Pl = 2 \times 2250 + 240 \times 100 = 4500 + 24\,000 = 28\,500\text{ cm}^{2}$

$28\,500 \div 10\,000 = 2.85\text{ m}^{2}$

(b) Leave out the bottom, $75 \times 100 = 7500\text{ cm}^{2}$ , and the back, $45 \times 100 = 4500\text{ cm}^{2}$ .

Area seen $= 28\,500 - 7500 - 4500 = 16\,500\text{ cm}^{2} = 1.65\text{ m}^{2}$

Check: three treads $3 \times 25 \times 100 = 7500$ , three risers $3 \times 15 \times 100 = 4500$ , two ends $2 \times 2250 = 4500$ ; $7500 + 4500 + 4500 = 16\,500\text{ cm}^{2}$ .

(a) $2.85\text{ m}^{2}$ (b) $1.65\text{ m}^{2}$

Practice 8: A cube on a cylinder

A cube of edge 8 cm stands in the middle of the top of a solid cylinder of diameter 20 cm and height 6 cm. (a) What shape is hidden on the top of the cylinder? (b) Find the total surface area of the solid, using the $\pi$ button on your calculator.

ShowHide solution

(a) The cube's base, a square 8 cm by 8 cm, hides a square of the cylinder's top. The rest of the circle is still on the outside, so this time no circle cancels: only the square is hidden, on both solids.

(b) The radius is $20 \div 2 = 10$ cm.

Cylinder: $2\pi r^{2} + 2\pi rh = 2 \times \pi \times 10^{2} + 2 \times \pi \times 10 \times 6 = 200\pi + 120\pi = 320\pi\text{ cm}^{2}$

Cube: $6 \times 8^{2} = 384\text{ cm}^{2}$ . Area where they touch $= 8 \times 8 = 64\text{ cm}^{2}$ .

Total surface area $= 320\pi + 384 - 2 \times 64 = 320\pi + 256 = 1261.31\ldots \approx 1260\text{ cm}^{2}$

Check, face by face: the cylinder's top with a square missing, $100\pi - 64$ ; its bottom $100\pi$ and curved surface $120\pi$ ; the cube's top and four sides, $5 \times 64 = 320$ . Total $320\pi - 64 + 320 = 320\pi + 256$ . The cube's top replaces the square it covers, just as a cylinder's top replaces its circle.

(a) A square 8 cm by 8 cm. (b) The total surface area is $(320\pi + 256)\text{ cm}^{2} \approx 1260\text{ cm}^{2}$ .

Practice 9: A concrete block against a wall

A concrete block at an HDB void deck is made of a cuboid 120 cm long, 40 cm deep and 50 cm high, with a cuboid 60 cm long, 40 cm deep and 30 cm high standing in the middle of its top; their fronts and backs are level. The block stands on the ground with its back against a wall. Every surface that can be seen is painted. Find the area painted, in $\text{m}^{2}$ .

ShowHide solution

Not painted: the bottom (on the ground), both backs (against the wall), and the two faces where the cuboids touch.

Lower cuboid: front $120 \times 50 = 6000$ , two ends $2 \times 40 \times 50 = 4000$ , the top that shows $120 \times 40 - 60 \times 40 = 4800 - 2400 = 2400$ .

Upper cuboid: front $60 \times 30 = 1800$ , two ends $2 \times 40 \times 30 = 2400$ , top $60 \times 40 = 2400$ .

Area painted $= 6000 + 4000 + 2400 + 1800 + 2400 + 2400 = 19\,000\text{ cm}^{2}$

Check from above: the tops together are the whole lower top, $120 \times 40 = 4800$ ; fronts $6000 + 1800 = 7800$ ; ends $4000 + 2400 = 6400$ . $4800 + 7800 + 6400 = 19\,000\text{ cm}^{2}$ .

$19\,000 \div 10\,000 = 1.9\text{ m}^{2}$

The area painted is $19\,000\text{ cm}^{2}$ , which is $1.9\text{ m}^{2}$ .

Practice 10: Covering a play structure

A play structure at a community club playground is a cuboid 1.4 m wide, 0.6 m high and 2.4 m long with a half-cylinder lying along its top, its flat face covering the top exactly. It stands on the ground. All of it except the bottom is covered with rubber that costs $35 per $\text{m}^{2}$ . Taking $\pi = \frac{22}{7}$ , find the area covered and the cost.

ShowHide solution

The half-cylinder's flat face covers the top exactly, so its diameter is 1.4 m and its radius is $r = 1.4 \div 2 = 0.7$ m; its length is 2.4 m.

Hidden: the cuboid's top and the half-cylinder's flat face. Not covered: the bottom.

Cuboid: front and back $2 \times 1.4 \times 0.6 = 1.68$ , two long sides $2 \times 2.4 \times 0.6 = 2.88$ . Together $4.56\text{ m}^{2}$ .

Curved surface $\pi rl = \frac{22}{7} \times 0.7 \times 2.4 = 5.28\text{ m}^{2}$

Two semicircular ends $\pi r^{2} = \frac{22}{7} \times 0.7^{2} = 1.54\text{ m}^{2}$

Area covered $= 4.56 + 5.28 + 1.54 = 11.38\text{ m}^{2}$

Cost $= 11.38 \times 35 = 398.30$ , so the cost is $398.30.

The area covered is $11.38\text{ m}^{2}$ and the rubber costs $398.30.

Key points

The surface area of a composite solid is the area of its outside only.
Where two parts touch, the area of contact is hidden on both: subtract $2 \times$ the area where they touch.
A cuboid or an upright cylinder standing with its whole base on a flat top face hides a patch of that face, and its own top, the same shape, replaces the patch: a cylinder adds only its curved surface, $2\pi rh$ , and a cuboid only its four sides.
If a solid's base is not wholly on the face (it hangs over the edge), go face by face. A cuboid standing on a cylinder hides only the rectangle under it (a square, for a cube), not a circle.
A half-cylinder whose flat face covers a cuboid's top adds $\pi rl$ and its two semicircular ends; the flat face and the top are both hidden.
A solid standing on the ground is not painted underneath, nor where it stands against a wall; tins of paint are rounded up.
Check a composite solid a second way: face by face, as one prism, or by looking from above.