Surface Area of Prisms

In these nets the rectangles make one strip, which rolls up into a tube when folded, and one triangle must close each end of the tube: one on the top edge of the strip and one on the bottom edge.

Net A: one triangle on the top edge and one on the bottom edge. It folds into the prism.

Net B: one triangle on the top edge (of the first rectangle) and one on the bottom edge (of the third rectangle). There is still one at each end, and each side of a triangle lies against a rectangle of the same width, so it folds into the prism too. A triangle may be joined to any of the rectangles, by its side of the same length, turned so that its other sides meet the rectangles of the same widths: here each triangle is joined by a 5 cm side, and its 6 cm side meets the 6 cm rectangle when it is folded.

Net C: both triangles are on the top edge. When it is folded, both go to the same end of the tube and the bottom end is left open, so it does not make a closed prism.

First find the lengths not marked, from the folding. Every rectangle in the column is 8 cm long. The second rectangle, 8 cm by 5 cm, is the base. Each side flap is 5 cm long, as marked, and when it is folded up its other sides meet the 3 cm sides of the first and third rectangles, so the flaps are 5 cm by 3 cm. The third rectangle stands opposite the first, so it is 8 cm by 3 cm, and the last rectangle is the lid, lying on the 5 cm tops of the flaps, so it is 8 cm by 5 cm.

So the net has three pairs of identical rectangles: two 8 cm by 5 cm, two 8 cm by 3 cm and two 5 cm by 3 cm.

Total surface area $= 2 \times 8 \times 5 + 2 \times 8 \times 3 + 2 \times 5 \times 3$

$= 80 + 48 + 30$

$= 158\text{ cm}^{2}$

Check with the formula: $2(lb + bh + lh) = 2(8 \times 5 + 5 \times 3 + 8 \times 3) = 2(40 + 15 + 24) = 2 \times 79 = 158\text{ cm}^{2}$.

In a right-angled triangle the two shorter sides are at right angles, so they are the base and the height.

Area of cross-section: $A = \frac{1}{2} \times 15 \times 8 = 60\text{ cm}^{2}$

(a) Volume $= A \times l = 60 \times 5 = 300\text{ cm}^{3}$

(b) Perimeter of cross-section: $P = 8 + 15 + 17 = 40$ cm

Total surface area $= 2A + Pl = 2 \times 60 + 40 \times 5 = 120 + 200 = 320\text{ cm}^{2}$

Volume of a cube $= l^{3}$, so $l^{3} = 343$.

$l = \sqrt[3]{343} = 7$, since $7 \times 7 \times 7 = 343$.

Total surface area $= 6l^{2} = 6 \times 7^{2} = 6 \times 49 = 294\text{ cm}^{2}$

The 8 cm side is at right angles to the parallel sides, so it is the height of the trapezium; it is also one of its sides.

Area of cross-section: $A = \frac{1}{2}(4 + 10) \times 8 = \frac{1}{2} \times 14 \times 8 = 56\text{ cm}^{2}$

Perimeter: $P = 10 + 10 + 4 + 8 = 32$ cm

Total surface area $= 2A + Pl = 2 \times 56 + 32 \times 12 = 112 + 384 = 496\text{ cm}^{2}$

The holder is open at the top, so go face by face: one triangular base and three rectangular sides.

Base: $\frac{1}{2} \times 8 \times 6 = 24\text{ cm}^{2}$ (the 6 cm and 8 cm sides are at right angles)

Sides: $6 \times 12 = 72\text{ cm}^{2}$, $8 \times 12 = 96\text{ cm}^{2}$ and $10 \times 12 = 120\text{ cm}^{2}$

Area of card $= 24 + 72 + 96 + 120 = 312\text{ cm}^{2}$

The trough is open along its top, so go face by face: two trapezium ends, the bottom and two slanting sides.

Two ends: $2 \times \frac{1}{2}(30 + 60) \times 20 = 2 \times 900 = 1800\text{ cm}^{2}$

Bottom: $30 \times 100 = 3000\text{ cm}^{2}$

Two slanting sides: $2 \times 25 \times 100 = 5000\text{ cm}^{2}$

Area of metal $= 1800 + 3000 + 5000 = 9800\text{ cm}^{2}$

In $\text{m}^{2}$: $9800 \div 10\,000 = 0.98\text{ m}^{2}$

The sides not marked: the lower tread is $60 - 30 = 30$ cm and the upper riser is $30 - 15 = 15$ cm.

Area of each end: $60 \times 15 + 30 \times 15 = 900 + 450 = 1350\text{ cm}^{2}$, so the two ends: $2 \times 1350 = 2700\text{ cm}^{2}$

The rectangles, each 80 cm long, except the bottom: front riser $15 \times 80 = 1200$; lower tread $30 \times 80 = 2400$; upper riser $15 \times 80 = 1200$; top $30 \times 80 = 2400$; back $30 \times 80 = 2400$ (all in $\text{cm}^{2}$).

Area painted $= 2700 + 1200 + 2400 + 1200 + 2400 + 2400 = 12\,300\text{ cm}^{2}$

In $\text{m}^{2}$: $12\,300 \div 10\,000 = 1.23\text{ m}^{2}$

Total surface area $= 2(lb + bh + lh)$. Put in the numbers given:

$2(12 \times 5 + 5h + 12h) = 256$

$2(60 + 17h) = 256$

$120 + 34h = 256$

$34h = 136$ (subtract 120 from both sides)

$h = 4$ (divide both sides by 34)

Check: $2(60 + 5 \times 4 + 12 \times 4) = 2(60 + 20 + 48) = 2 \times 128 = 256$.

$2.4\text{ L} = 2.4 \times 1000 = 2400\text{ cm}^{3}$, since $1\text{ L} = 1000\text{ cm}^{3}$.

Volume $= l \times b \times h$, so $20 \times 10 \times h = 2400$

$200h = 2400$, so $h = 12$ (divide both sides by 200)

Total surface area $= 2(lb + bh + lh) = 2(20 \times 10 + 10 \times 12 + 20 \times 12)$

$= 2(200 + 120 + 240) = 2 \times 560 = 1120\text{ cm}^{2}$

(a) Face by face, leaving out the base:

Top: $4 \times 3 = 12\text{ m}^{2}$

Front and back: $2 \times 4 \times 2.5 = 20\text{ m}^{2}$

Two ends: $2 \times 3 \times 2.5 = 15\text{ m}^{2}$

Area painted $= 12 + 20 + 15 = 47\text{ m}^{2}$

(b) $47 \div 10 = 4.7$. Four tins cover only $40\text{ m}^{2}$, so 5 tins are needed (round up).

(c) $5 \times 23.90 = 119.50$, so the paint costs $119.50.

(a) $4 \times 6 \times 6 = 144\text{ cm}^{2}$

(b) The box is $40 - 12 = 28$ cm long, $30 - 12 = 18$ cm wide and 6 cm high. Face by face:

Base: $28 \times 18 = 504\text{ cm}^{2}$; two long sides: $2 \times 28 \times 6 = 336\text{ cm}^{2}$; two short sides: $2 \times 18 \times 6 = 216\text{ cm}^{2}$

Total surface area $= 504 + 336 + 216 = 1056\text{ cm}^{2}$

Check: $40 \times 30 - 144 = 1200 - 144 = 1056\text{ cm}^{2}$.

(c) Volume $= 28 \times 18 \times 6 = 3024\text{ cm}^{3}$, and $3024 \div 1000 = 3.024$ L.

Area of cross-section: $A = \frac{1}{2}(6 + 12) \times 4 = 36\text{ cm}^{2}$

Perimeter: $P = 12 + 5 + 6 + 5 = 28$ cm

(a) $2A + Pl = 632$, so $2 \times 36 + 28l = 632$

$72 + 28l = 632$

$28l = 560$ (subtract 72 from both sides)

$l = 20$ (divide both sides by 28)

Check: $72 + 28 \times 20 = 72 + 560 = 632$.

(b) Volume $= A \times l = 36 \times 20 = 720\text{ cm}^{3}$