Exam practice

V = A × h = 35 × 10 = 350 cm³.

LSA = P × h = 36 × 8 = 288 cm².

TSA = 2A + LSA = 2 × 40 + 175 = 255 cm².

Take the front, 10 × 8, as the cross-section; the length is 6 cm.
A = 10 × 8 = 80 cm²; P = 2(10 + 8) = 36 cm.
LSA = 36 × 6 = 216 cm²; TSA = 2 × 80 + 216 = 376 cm²; V = 80 × 6 = 480 cm³.

A cube with edge 5 cm. $A = 5^2 =$ 25 cm²; P = 20 cm.
LSA = 20 × 5 = 100 cm²; TSA = 6 × 25 = 150 cm²; $V = 5^3 =$ 125 cm³.

The front is a column 4 × 9 with an arm 4 × 3: A = 36 + 12 = 48 cm².
Going round: P = 4 + 3 + 4 + 3 + 4 + 3 + 4 + 9 = 34 cm. (A quicker way: every side going right is matched by sides going left, and every side going up by sides going down, so P = 2(8 + 9) = 34.)
LSA = 34 × 6 = 204 cm²; TSA = 2 × 48 + 204 = 300 cm²; V = 48 × 6 = 288 cm³.

The plus sign is a column 4 × 11 (4 + 3 + 4 tall) with two 3 × 3 arms: A = 44 + 9 + 9 = 62 cm².
P = 2(10 + 11) = 42 cm (width 3 + 4 + 3, height 4 + 3 + 4, as in c.).
LSA = 42 × 8 = 336 cm²; TSA = 2 × 62 + 336 = 460 cm²; V = 62 × 8 = 496 cm³.

The end is a rectangle 4 × 6 = 24 cm² with an isosceles triangle on top, sides 3, 3 and base 4:
$$\frac{4}{4}\sqrt{4 \times 3^2 - 4^2} = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5} \approx 4.47$$
$A = 24 + 2\sqrt{5} \approx$ 28.47 cm²; P = 4 + 6 + 3 + 3 + 6 = 22 cm.
LSA = 22 × 10 = 220 cm²; TSA = 2 × 28.47 + 220 = 276.94 cm²; $V = 28.47 \times 10 \approx$ 284.72 cm³.

The end is a staircase: rows 30, 20 and 10 cm wide, each 10 cm high, so A = 300 + 200 + 100 = 600 cm².
P = 2(30 + 30) = 120 cm (the steps’ sides add up to the full width and height).
LSA = 120 × 30 = 3,600 cm²; TSA = 2 × 600 + 3600 = 4,800 cm²; V = 600 × 30 = 18,000 cm³ (18 liters).

Standing on its 6 cm × 5 cm base (height 8 cm): A = 6 × 5 = 30 cm².
Some books take the 6 cm × 8 cm face as the cross-section (so the prism is 5 cm long): then A = 6 × 8 = 48 cm². Either is right if you say which face is the base.

2h(l + b) = 2 × 8 × (6 + 5) = 176 cm².
A cuboid can stand on any face, and the “lateral” faces change with it: lying on its 6 cm × 8 cm face (so 5 cm is its height), its cross-section is 48 cm² and its lateral surface 2 × 5 × (6 + 8) = 140 cm². Say which face is the base when you answer.

6 × 5 × 8 = 240 cm³, whichever face it stands on.

V = 2 × 1.5 × 1 = 3 m³ = 3 × 1000 = 3,000 liters.

The big cube is $20^3 = 8000$ cm³, so each small cube is 8000 ÷ 8 = 1000 cm³ and its edge is $\sqrt[3]{1000} =$ 10 cm.
(One cut through the middle each way, three cuts, makes the 8 cubes: each edge is halved.)

The end is right-angled at B: $A = \frac{1}{2} \times 3 \times 4 =$ 6 cm². The third side is $\sqrt{3^2 + 4^2} = 5$ cm, so P = 3 + 4 + 5 = 12 cm.
LSA = 12 × 12 = 144 cm²; TSA = 2 × 6 + 144 = 156 cm²; V = 6 × 12 = 72 cm³.

The end is equilateral with side 6 cm: $A = \frac{\sqrt{3}}{4} \times 6^2 = 9\sqrt{3} \approx$ 15.59 cm²; P = 18 cm.
LSA = 18 × 15 = 270 cm²; TSA $= 2 \times 9\sqrt{3} + 270 = 18\sqrt{3} + 270 \approx$ 301.18 cm²; $V = 9\sqrt{3} \times 15 = 135\sqrt{3} \approx$ 233.83 cm³.

Isosceles end, a = 6, b = 8: $A = \frac{8}{4}\sqrt{4 \times 36 - 64} = 2\sqrt{80} = 8\sqrt{5} \approx$ 17.89 cm²; P = 6 + 6 + 8 = 20 cm.
LSA = 20 × 13 = 260 cm²; TSA $= 2 \times 17.89 + 260 \approx$ 295.78 cm²; $V = 8\sqrt{5} \times 13 = 104\sqrt{5} \approx$ 232.55 cm³.

Scalene end, sides 8, 10, 12: s = 15, $A = \sqrt{15 \times 7 \times 5 \times 3} = \sqrt{1575} = 15\sqrt{7} \approx$ 39.69 cm²; P = 30 cm.
LSA = 30 × 16 = 480 cm²; TSA $= 2 \times 39.686 + 480 \approx$ 559.37 cm²; $V = 39.686 \times 16 \approx$ 634.98 cm³.
Keep the area unrounded (or use $15\sqrt{7}$) until the end: doubling 39.69 gives 559.38, a hundredth off.

$A = \frac{1}{2} \times 8 \times 9 = 36$ cm². 36 × h = 864, so h = 24 cm.

The hypotenuse is $\sqrt{4^2 + 3^2} = 5$ cm, so the three rectangles are
4 × 30 = 120 cm², 3 × 30 = 90 cm² and 5 × 30 = 150 cm², together 360 cm² (= P × h = 12 × 30).

Perpendicular $= \sqrt{13^2 - 12^2} = \sqrt{25} = 5$ cm. $A = \frac{1}{2} \times 12 \times 5 = 30$ cm²; P = 12 + 5 + 13 = 30 cm.
2 × 30 + 30 × h = 660, so 30h = 600 and h = 20 cm.

$s = \frac{4 + 13 + 15}{2} = 16$; $A = \sqrt{16 \times 12 \times 3 \times 1} = \sqrt{576} = 24$ cm².
24 × l = 480, so l = 20 cm.

4 × 20 = 80 cm², 13 × 20 = 260 cm² and 15 × 20 = 300 cm²: 640 cm² together.

2 × 24 + 640 = 688 cm².

54 × h = 2160, so h = 40 cm.

V = 126 × 40 = 5,040 cm³.

Bin A. Floor: $s = \frac{15 + 13 + 4}{2} = 16$; $A = \sqrt{16 \times 1 \times 3 \times 12} = \sqrt{576} = 24$ ft². Volume = 24 × 8 = 192 ft³. Sheet = 2 × 24 + (15 + 13 + 4) × 8 = 48 + 256 = 304 ft².
Bin B. Volume = 6 × 4 × 8 = 192 ft³. Sheet = 2(24 + 32 + 48) = 208 ft².
Both hold the same, 192 ft³ of rice. Bin B needs 96 ft² less sheet (304 − 208), almost a third less. The long, thin triangle has a lot of wall for its floor.

The roof’s triangle has sides 3, 3 and base 4.8 m; its height is $\sqrt{3^2 - 2.4^2} = \sqrt{3.24} = 1.8$ m, so its area is $\frac{1}{2} \times 4.8 \times 1.8 = 4.32$ m².
One end = 4.8 × 3 + 4.32 = 18.72 m².
Two side walls = 2 × 3 × 10 = 60 m²; two roof slopes = 2 × 3 × 10 = 60 m².
The floor is carpet and one end is open, so the cloth is one end, the walls and the roof: 18.72 + 60 + 60 = 138.72 m².
(If both ends were made of cloth, with a door cut in one, it would be 2 × 18.72 + 120 = 157.44 m².)

The floor: 4.8 × 10 = 48 m².

$48 \div 7 \approx 6.86$: 6 cabins (a seventh would need 49 m²). In practice there must also be room to walk between them.

Tank A: V = 8 × 6 × 8 = 384 ft³; TSA = 2(48 + 48 + 64) = 320 ft².

Tank B: V = 16 × 4 × 6 = 384 ft³; TSA = 2(64 + 24 + 96) = 368 ft².

They hold the same water, 384 ft³ (about 10,900 liters), but tank A needs 48 ft² less material. Tank A is the better design. It is closer to a cube: Back to the tank again.