Exam practice

$2\pi x$ cm

$\pi x^2$ cm²

$2\pi xy$ cm²

$2\pi x(x + y)$ cm²

$\pi x^2 y$ cm³

The radius is $\frac{x}{2}$ cm.
Plane surfaces: $2\pi \left(\frac{x}{2}\right)^2 =$ $\frac{\pi x^2}{2}$ cm².
Volume: $\pi \left(\frac{x}{2}\right)^2 y =$ $\frac{\pi x^2 y}{4}$ cm³.

The curved surface unrolls into a rectangle C long and h wide: Ch cm².

Ah cubic feet (ft³).

$\pi r^2 h$ cubic units

$2\pi rh$ square units

$2\pi r^2$ square units

$2\pi r(r + h)$ square units

TSA $= 2\pi r(r + h) = (2\pi r) \times (r + h) =$ CS square units.

Plane: $2 \times \tfrac{22}{7} \times 49 =$ 308 cm²; curved: $2 \times \tfrac{22}{7} \times 7 \times 5 =$ 220 cm²; volume: $\tfrac{22}{7} \times 49 \times 5 =$ 770 cm³.

Plane: $2 \times \tfrac{22}{7} \times 12.25 =$ 77 m²; curved: $2 \times \tfrac{22}{7} \times 3.5 \times 6 =$ 132 m²; volume: $\tfrac{22}{7} \times 12.25 \times 6 =$ 231 m³.

Plane: $2 \times \tfrac{22}{7} \times 4 \approx$ 25.14 ft²; curved: $2 \times \tfrac{22}{7} \times 2 \times 7 =$ 88 ft²; volume: $\tfrac{22}{7} \times 4 \times 7 =$ 88 ft³.

r = 7, h = 12 (cm). Base $= \tfrac{22}{7} \times 49 =$ 154 cm²; plane surfaces = 2 × 154 = 308 cm²; curved $= 2 \times \tfrac{22}{7} \times 7 \times 12 =$ 528 cm²; total = 308 + 528 = 836 cm²; volume = 154 × 12 = 1848 cm³.

r = 14 ÷ 2 = 7, h = 21 (cm). Base 154 cm²; plane surfaces 308 cm²; curved $= 2 \times \tfrac{22}{7} \times 7 \times 21 =$ 924 cm²; total 1232 cm²; volume = 154 × 21 = 3234 cm³.

r = OA = 7, h = OO′ = 14 (cm). Base 154 cm²; plane surfaces 308 cm²; curved $= 2 \times \tfrac{22}{7} \times 7 \times 14 =$ 616 cm²; total 924 cm²; volume = 154 × 14 = 2156 cm³.

r = 2.8, h = 21 (cm). Base $= \tfrac{22}{7} \times 7.84 =$ 24.64 cm²; plane surfaces 49.28 cm²; curved $= 2 \times \tfrac{22}{7} \times 2.8 \times 21 =$ 369.6 cm²; total = 49.28 + 369.6 = 418.88 cm²; volume = 24.64 × 21 = 517.44 cm³.

Curved surface = C × h = 176 × 30 = 5280 cm².
$2\pi r = 176$, so $r = \frac{176 \times 7}{44} = 28$ cm; the two ends are $2 \times \tfrac{22}{7} \times 28^2 = 4928$ cm².
Total = 5280 + 4928 = 10208 cm².

$2\pi r = 88$, so $r = \frac{88 \times 7}{44} = 14$ cm. Base $= \tfrac{22}{7} \times 196 =$ 616 cm².

h = 24 − 14 = 10 cm. Curved surface = 88 × 10 = 880 cm².

880 + 2 × 616 = 2112 cm² (or C(r + h) = 88 × 24, the same).

616 × 10 = 6160 cm³.

$2\pi r(r + h) = 2992$: $2 \times \tfrac{22}{7} \times r \times 34 = 2992$, so r = 14 cm and h = 20 cm (as in Example 3).
$V = \tfrac{22}{7} \times 196 \times 20 =$ 12320 cm³.

The curved surface is $2\pi rh = \pi dh$. Find dh, then two numbers with that product and sum 28.

$\pi dh = 462$: $dh = \frac{462 \times 7}{22} = 147$. Two numbers with sum 28 and product 147 are 7 and 21 (7 + 21 = 28, 7 × 21 = 147). So there are two cylinders:

• d = 7, h = 21: r = 3.5, ends $= 2 \times \tfrac{22}{7} \times 12.25 = 77$; TSA = 462 + 77 = 539 cm².
• d = 21, h = 7: r = 10.5, ends $= 2 \times \tfrac{22}{7} \times 110.25 = 693$; TSA = 462 + 693 = 1155 cm².

Both fit the question: a tall thin one and a short wide one.

$r = \frac{1}{4} \times 56 = 14$ cm, $h = \frac{3}{4} \times 56 = 42$ cm.
Curved $= 2 \times \tfrac{22}{7} \times 14 \times 42 =$ 3696 cm²; total = 3696 + 2 × 616 = 4928 cm².

Let r = k, h = 3k. $2\pi \times k \times 3k = 924$, so $6 \times \tfrac{22}{7} \times k^2 = 924$, $k^2 = 49$, k = 7: r = 7 cm, h = 21 cm.
Base $= \tfrac{22}{7} \times 49 =$ 154 cm²; volume = 154 × 21 = 3234 cm³.

$2\pi rh = 3\pi r^2$, so 2h = 3r, h = 1.5r. Then r + 1.5r = 25: r = 10 cm, h = 15 cm.
$V = \tfrac{22}{7} \times 100 \times 15 \approx$ 4714.29 cm³.

From $C = 2\pi r$, $r = \frac{C}{2\pi}$, so $V = \pi r^2 h = \frac{C^2 h}{4\pi}$.
First pot: $V = \frac{21^2 \times 14}{4 \times \frac{22}{7}} = \frac{6174 \times 7}{88} \approx 491.11$ cm³.
Second pot: $V = \frac{14^2 \times 21}{4 \times \frac{22}{7}} = \frac{4116 \times 7}{88} \approx 327.41$ cm³.
The first pot (circumference 21 cm) holds about 163.70 cm³ more, about 0.1637 liters. The wider pot wins even though it is shorter, for the reason in Example 6.

One holder is the curved surface and one base:
$2 \times \tfrac{22}{7} \times 3 \times 10.5 + \tfrac{22}{7} \times 3^2 = 198 + 28.29 = 226.29$ cm².
For 35 students: $35 \times \left(198 + \tfrac{198}{7}\right) = 6930 + 990 =$ 7920 cm² (a little more in practice, for the overlaps).

r = 3.5 cm. One glass: $\tfrac{22}{7} \times 3.5^2 \times 6 = 231$ cm³.
For 250: 231 × 250 = 57750 cm³ = 57.75 liters.

The mat goes once round the bin: its length is the bin’s circumference.

$\pi r^2 h = 1.4$ with h = 1.1 m: $r^2 = \frac{1.4 \times 7}{22 \times 1.1} = \frac{49}{121}$, so $r = \frac{7}{11}$ m.
Length $= 2\pi r = 2 \times \tfrac{22}{7} \times \tfrac{7}{11} =$ 4 m, with a little more for the overlap where the ends are tied.

$4\pi x^2$ square units

$\frac{4}{3}\pi x^3$ cubic units

$2\pi x$ units

$\pi x^2$ square units

Four great circles: 4a square units.

$4\pi r^2 = 4x$, so $\pi r^2 = x$. The hemisphere’s total surface is $3\pi r^2 =$ 3x square units.

4y square units.

$\pi d$ cm

$\pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$ cm²

$\frac{4}{3}\pi \left(\frac{d}{2}\right)^3 = \frac{\pi d^3}{6}$ cm³

$4\pi \left(\frac{d}{2}\right)^2 = \pi d^2$ cm²

$4 \times \tfrac{22}{7} \times 10.5^2 =$ 1386 cm². (Its circumference is $2 \times \tfrac{22}{7} \times 10.5 = 66$ cm, right in the middle of the 65–67 cm the international rules allow.)

r = 40 m: $4 \times \tfrac{22}{7} \times 40^2 \approx$ 20114.29 m², about two hectares of glass, nearly three football pitches. (In fact the bottom of the sphere sits on its base, so a little less is glass.)

r = 7 cm. Surface $= 4 \times \tfrac{22}{7} \times 49 =$ 616 cm²; volume $= \tfrac{4}{3} \times \tfrac{22}{7} \times 343 \approx$ 1437.33 cm³.

r = 14 ÷ 2 = 7 cm, the same sphere as in a.: 616 cm² and about 1437.33 cm³.

A solid hemisphere, r = 28 cm. Total surface $= 3 \times \tfrac{22}{7} \times 784 =$ 7392 cm²; volume $= \tfrac{2}{3} \times \tfrac{22}{7} \times 21952 \approx$ 45994.67 cm³.

$2\pi r = 88$, so r = 14 cm. Total surface $= 3 \times \tfrac{22}{7} \times 196 =$ 1848 cm²; volume $= \tfrac{2}{3} \times \tfrac{22}{7} \times 2744 \approx$ 5749.33 cm³.

r = 17.5 cm. Surface $= 4 \times \tfrac{22}{7} \times 17.5^2 =$ 3850 cm²; volume $= \tfrac{4}{3} \times \tfrac{22}{7} \times 17.5^3 \approx$ 22458.33 cm³.

$2\pi r = 44$, so r = 7 cm. TSA $= 3 \times \tfrac{22}{7} \times 49 =$ 462 cm².

$4 \times \tfrac{22}{7} \times r^2 = 2464$: $r^2 = 196$, r = 14 cm. The diameter is 28 cm.

$\tfrac{4}{3} \times \tfrac{22}{7} \times r^3 = 38808$: $r^3 = \frac{38808 \times 21}{88} = 9261$, so $r = \sqrt[3]{9261} =$ 21 m.

$3\pi r^2 = 243\pi$: $r^2 = 81$, r = 9 cm. $V = \tfrac{2}{3}\pi \times 729 = 486\pi \approx$ 1527.43 cm³.

$\tfrac{4}{3}\pi r^3 = 2304\pi$: $r^3 = 1728$, r = 12 cm. Surface $= 4\pi \times 144 =$ $576\pi \approx 1810.29$ cm².

If the Moon’s radius is r, the Earth’s is 4r: $\frac{4\pi r^2}{4\pi (4r)^2} = \frac{1}{16}$. Moon : Earth = 1 : 16.
(With the measured diameters, 3,475 km and 12,742 km, the Moon’s is 0.27 of the Earth’s, a little more than a quarter, and the ratio is nearer 1 : 13.4.)

$\frac{4\pi r^2}{\frac{4}{3}\pi r^3} = \frac{3}{r} = \frac{1}{3}$, so r = 9 cm. The diameter is 18 cm, and the volume $= \tfrac{4}{3} \times \tfrac{22}{7} \times 729 \approx$ 3054.86 cm³.

$\frac{4}{3}\pi (2r)^3 = 8 \times \frac{4}{3}\pi r^3$: the new volume is 8 times the old, so it has grown by 7 times the old volume.

$\frac{4\pi \times 7^2}{4\pi \times 14^2} = \frac{49}{196} = \frac{1}{4}$: 1 : 4. The rubber is stretched over four times the area, so it is about a quarter as thick.

$r^3 = 2^3 + 12^3 + 16^3 = 8 + 1728 + 4096 = 5832$, so $r = \sqrt[3]{5832} = 18$ cm. The diameter is 36 cm.

$3\pi r^2 = 243\pi$, so r = 9 cm. Rim $= 2 \times \tfrac{22}{7} \times 9 \approx$ 56.57 cm; volume $= \tfrac{2}{3} \times \tfrac{22}{7} \times 729 \approx$ 1527.43 cm³.

$s = \frac{20 + 21 + 13}{2} = 27$; $A = \sqrt{27 \times 7 \times 6 \times 14} = \sqrt{15876} =$ 126 cm².

$s = \frac{12 + 15 + 18}{2} = 22.5$; $A = \sqrt{22.5 \times 10.5 \times 7.5 \times 4.5} = \sqrt{7973.4375} \approx$ 89.29 cm².

$9^2 + 12^2 = 81 + 144 = 225 = 15^2$, so each tile is right-angled: $A = \frac{1}{2} \times 9 \times 12 = 54$ cm². (Heron gives the same: s = 18, $\sqrt{18 \times 9 \times 6 \times 3} = \sqrt{2916} = 54$.)
28 tiles: 28 × 54 = 1512 cm². Cost = 1512 × 10 = Rs 15,120.

$3 \times 2\pi rh = 2 \times 2\pi r(r + h)$; divide by $2\pi r$: 3h = 2r + 2h, so h = 2r.
CSA $= 2\pi r \times 2r = 4\pi r^2 = 616$: $r^2 = 49$, r = 7 cm.
Circumference $= 2 \times \tfrac{22}{7} \times 7 =$ 44 cm; height = 2 × 7 = 14 cm.

The difference is the two ends: $2\pi r^2 = 308$, $r^2 = 49$, r = 7 cm.
Diameter 14 cm; circumference $= 2 \times \tfrac{22}{7} \times 7 =$ 44 cm.

$2\pi r^2 = 1232$: $r^2 = 196$, r = h = 14 cm. $V = \tfrac{22}{7} \times 196 \times 14 =$ 8624 cm³.

$4\pi r^2 = 616$: $r^2 = 49$, r = 7 cm. The diameter is 14 cm.

r = 28 cm. TSA $= 3 \times \tfrac{22}{7} \times 28^2 =$ 7392 cm².

Area painted = 30060 ÷ 180 = 167 m².
Four walls = 2(12 + 10) × h = 44h; door and windows = 3 + 4 × 1.5 = 9 m².
44h − 9 = 167, so 44h = 176: the room is 4 m high.

The front is five squares of 4 cm: A = 5 × 16 = 80 cm²; its perimeter is 12 sides of 4 cm, P = 48 cm.
TSA = 2A + Ph = 160 + 48 × 9 = 592 cm²; V = Ah = 80 × 9 = 720 cm³.

The end is right-angled at B: $AB = \sqrt{5^2 - 4^2} = 3$ cm, so $A = \frac{1}{2} \times 3 \times 4 = 6$ cm² and P = 3 + 4 + 5 = 12 cm.
TSA = 2 × 6 + 12 × 10 = 132 cm²; V = 6 × 10 = 60 cm³.

r = 10.5 cm. The surface is one flat end, the curved side of the cylinder and the hemisphere’s curve:
$\pi r^2 + 2\pi r \times 40 + 2\pi r^2 = 3\pi r^2 + 80\pi r = 1039.5 + 2640 = 3679.5$ cm².
Cost = 3679.5 × 180 = Rs 6,62,310.
What it means: Rs 180 for every square centimeter is far more than paint really costs. Painters charge by the square meter (10,000 cm²): at Rs 180 per m², as in Question 8, this solid would cost about Rs 66.

The hemisphere’s radius is 54 − 40 = 14 cm, the cylinder’s too.
Surface $= 2\pi r^2 + 2\pi r \times 40 + \pi r^2 = 3\pi r^2 + 80\pi r = 1848 + 3520 = 5368$ cm².
Cost = 5368 × 180 = Rs 9,66,240.