Surface area and volume of prisms and cylinders

$248$ cm $^2$ . Method: $2(60 + 40 + 24)$ .
$294$ cm $^2$ . Method: $6 \times 49$ .
$171.65$ cm $^2$ (approx). Method: two triangles $2 \times \tfrac{1}{2}(5)(4.33) = 21.65$ ; three rectangles $3 \times 5 \times 10 = 150$ ; total $\approx 171.65$ .
$78\pi \approx 245.0$ cm $^2$ . Method: $2\pi(9) + 2\pi(3)(10) = 18\pi + 60\pi$ .
$720\pi$ cm $^3$ . Method: $\pi(36)(20)$ .
(a) $64\pi \approx 201.1$ cm $^2$ . (b) $96\pi \approx 301.6$ cm $^2$ . (c) $128\pi \approx 402.1$ cm $^3$ .
$2.5$ L. Method: $2500 \div 1000$ .
$\approx 703.7$ cm $^2$ . Method: $r = 7$ ; $2\pi(49) + 2\pi(7)(9) = 98\pi + 126\pi = 224\pi$ .

$r \approx 5.2$ cm. Method: $\pi r^2 \times 12 = 1000$ ; $r^2 = \tfrac{1000}{12\pi} \approx 26.53$ ; $r \approx 5.2$ .
$h = 5$ cm. Method: $\pi(25)(20) = \pi(100)h$ ; $h = \tfrac{500}{100} = 5$ .
$h = 20$ cm. Method: $2\pi(100) + 2\pi(10)h = 600\pi$ ; $200\pi + 20\pi h = 600\pi$ ; $20\pi h = 400\pi$ .
$\approx 1034.5$ cm $^3$ . Method: prism $= 1600$ ; cylinder hole $= \pi(9)(20) = 180\pi \approx 565.5$ ; $1600 - 565.5$ .
Cube SA $= 600$ cm $^2$ . Cylinder SA $= 2\pi(25) + 2\pi(5)(10) = 50\pi + 100\pi = 150\pi \approx 471.2$ cm $^2$ . The cube has greater surface area.
$\approx 445.8$ cm $^2$ . Method: two trapezium ends $= 2 \times \tfrac{1}{2}(6+10)(4) = 64$ ; four rectangles: $6 \times 15 = 90$ , $10 \times 15 = 150$ , $4 \times 15 = 60$ , $5.66 \times 15 \approx 84.9$ ; total $\approx 64 + 384.9 \approx 448.9$ . (Accept minor rounding differences.)

The first term $2\pi r^2$ is the area of the two circular ends (top and bottom). The second term $2\pi r h$ is the curved lateral surface — the rectangle you get when you unroll the cylinder, whose width is the circumference $2\pi r$ and whose height is $h$ .
The height must double. Since $V = \pi r^2 h$ and $r$ is fixed, doubling $V$ requires doubling $h$ .
Volume increases by a factor of $4$ . Since $V = \pi r^2 h$ , replacing $r$ with $2r$ gives $\pi(2r)^2 h = 4\pi r^2 h$ . The $r^2$ term means radius has a squared effect on volume.
$\approx 226.2$ cm $^2$ . Method: cylinder curved SA $= 2\pi(3)(10) = 60\pi$ ; cylinder base circle $= \pi(9) = 9\pi$ (only the bottom; the top is covered by the hemisphere); hemisphere curved SA $= 2\pi(9) = 18\pi$ ; total $= (60 + 9 + 18)\pi = 87\pi \approx 273.3$ cm $^2$ . (Accept equivalent working.)
$\approx 392.7$ L. Method: $r = 0.05$ m; $V = \pi(0.05)^2(50) = \pi(0.0025)(50) = 0.125\pi \approx 0.3927$ m $^3$ ; $\times 1000 = 392.7$ L.

Side $\approx 9.3$ cm. Method: cylinder $V = \pi(25)(10) = 250\pi \approx 785.4$ cm $^3$ ; cube side $= \sqrt[3]{785.4} \approx 9.3$ .

Express $h$ : from $2\pi r^2 + 2\pi r h = 200\pi$ , divide through by $2\pi$ : $r^2 + r h = 100$ , so $h = \dfrac{100 - r^2}{r} = \dfrac{100}{r} - r$ .

Substitute into $V$ : $V = \pi r^2 h = \pi r^2 \left(\dfrac{100}{r} - r\right) = 100\pi r - \pi r^3$ .

Find the maximum by table of values (trial and improvement). Evaluate $V = \pi(100r - r^3)$ at whole-number $r$ :

$r$	$100r - r^3$	$V \approx$
3	$300 - 27 = 273$	$857.7$
4	$400 - 64 = 336$	$1055.6$
5	$500 - 125 = 375$	$1178.1$
6	$600 - 216 = 384$	$\mathbf{1206.4}$
7	$700 - 343 = 357$	$1121.5$
8	$800 - 512 = 288$	$904.8$

The maximum is near $r = 6$ . Refine with decimals:

$r$	$100r - r^3$	$V \approx$
5.7	$570 - 185.193 = 384.807$	$1208.9$
5.8	$580 - 195.112 = 384.888$	$\mathbf{1208.9}$
5.9	$590 - 205.379 = 384.621$	$1208.0$

The radius that maximises the volume is $r \approx 5.8$ cm. (Then $h = \dfrac{100}{5.8} - 5.8 \approx 11.5$ cm, and $V \approx 1209$ cm $^3$ .)

(a) Volume: prism $= 480$ ; half-cylinder $= \tfrac{1}{2}\pi(16)(10) = 80\pi \approx 251.3$ ; total $\approx 731.3$ cm $^3$ . (b) SA: prism base $= 10 \times 8 = 80$ ; two prism ends $= 2(8 \times 6) = 96$ ; two prism long sides $= 2(10 \times 6) = 120$ ; prism top has rectangle $10 \times 8$ minus half-cylinder footprint (the diameter strip is part of the prism top, but the half-cylinder sits on it): exposed top $= 80 - 8 \times 10 = 0$ (the half-cylinder covers the entire top, so no exposed top); half-cylinder curved $= \tfrac{1}{2}(2\pi)(4)(10) = 40\pi \approx 125.7$ ; two half-circle ends $= 2 \times \tfrac{1}{2}\pi(16) = 16\pi \approx 50.3$ ; total SA $\approx 80 + 96 + 120 + 125.7 + 50.3 = 472.0$ cm $^2$ . (Accept reasonable variations depending on which faces are considered exposed.)
$V = \tfrac{1}{2}\pi r^2 L = \tfrac{1}{2}\pi(0.5)^2(1.2) = 0.15\pi \approx 0.4712$ m $^3 \approx 471.2$ L.

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