Volume and surface area

Volume $= 8 \times 5 \times 4 = 160\text{ cm}^3$

Volume = length × width × height

$= 15 \times 10 \times 4 = 600$

Answer: $600\text{ cm}^3$

(a) Base area $= 10 \times 4 = 40$ cm², so height $= 200 \div 40 = 5$ cm

(b) Faces: $6 \times 5 = 30$, $6 \times 3 = 18$, $5 \times 3 = 15$

Each appears twice: $2(30 + 18 + 15) = 126$ cm²

Crate in cm: $300 \times 100 \times 150$

$300 \div 20 = 15$, $100 \div 10 = 10$, $150 \div 75 = 2$

Boxes: $15 \times 10 \times 2 = 300$

(Check with volumes: $4\,500\,000 \text{ cm}^3 \div 15\,000 \text{ cm}^3 = 300$ boxes)

A cube has 6 equal faces: $150 \div 6 = 25$ cm²

Edge length: $\sqrt{25} = 5$ cm

Volume: $5 \times 5 \times 5 = 125$ cm³

Change the packet to cm: 8 cm by 6 cm by 2 cm.

Along each edge: $72 \div 8 = 9$, $48 \div 6 = 8$, $24 \div 2 = 12$

$9 \times 8 \times 12 = 864$

(Check: $82944 \div 96 = 864$.)

Cube: $6^3 = 216$ cm³

Pyramid: $\frac13 \times 8 \times 8 \times h = \frac{64}{3}h$

$\frac{64}{3}h = 216 \Rightarrow h = \dfrac{216 \times 3}{64} = 10.125$ cm

Volume $= \pi \times 15^2 \times 43 = 30\,394.9\ldots\text{ cm}^3$

$1000\text{ cm}^3 = 1$ litre, so this is $30.3949\ldots$ litres.

Time $= 30.3949 \div 0.47 = 64.67\ldots$, so 65 minutes

The plan is a circle 6 cm across, so the radius is 3 cm.

The front elevation is 5 cm tall, so the height is 5 cm.

$V = \pi r^2 h = \pi \times 3^2 \times 5 = 45\pi$ cm³

Cuboid faces: $2(7\times5) + 2(6\times5) + 2(7\times6) = 70 + 60 + 84 = 214$

Cube faces: $6 \times 16 = 96$

The join hides one $4 \times 4$ face on each solid: $214 + 96 - 2 \times 16 = 278\text{ cm}^2$

One tank: $\pi \times 40^2 \times 160 \approx 804\,248\text{ cm}^3$

Four tanks: $\approx 3\,216\,991\text{ cm}^3 \approx 3217$ litres

Fertiliser is 1 part in 101: $3217 \div 101 \approx 31.85$ litres

$31.85 < 32$, so yes, she has enough.

Cross-section area: $750 \div 25 = 30$ cm²

Triangle area: $\frac{1}{2} \times 5 \times h = 30$

$2.5h = 30$, so $h = 12$ cm

Cube surface area: $6x^2$

Sphere surface area: $4\pi \times 3^2 = 36\pi$

$6x^2 = 36\pi$

$x^2 = 6\pi$

$x = \sqrt{6\pi}$, so $k = 6$, an integer.

Cuboid surface area: $2(6\times8 + 6\times18 + 8\times18) = 2(48+108+144) = 600\text{ cm}^2$

Each cube face: $600 \div 6 = 100$, so the side is $\sqrt{100} = 10$ cm

Cube volume $= 10^3 = 1000\text{ cm}^3$; cuboid volume $= 6 \times 8 \times 18 = 864\text{ cm}^3$

$1000 \ne 864$, so Janet is not correct.

Volume $= 30 \times 19 \times 6 = 3420\text{ cm}^3$

Water $= \frac{2}{3} \times 3420 = 2280\text{ cm}^3 = 2280\text{ ml}$

$2280 \div 275 = 8.29...$

Only 8 cups can be completely filled.

Radius $= 0.8$ m

Top and bottom: $2 \times \pi \times 0.8^2 = 4.021...$

Curved surface: $2 \times \pi \times 0.8 \times 1.8 = 9.047...$

One tank: $13.069...$ m²; three tanks: $39.21$ m²

Paint covers $7 \times 5 = 35$ m² $< 39.21$, so No, he does not have enough.

(a) From the front you see a triangle: base 6 cm, height 4 cm, apex above the middle of the base.

(b) Each sloping face: $\frac{1}{2} \times 6 \times 5 = 15\text{ cm}^2$

Four faces: $4 \times 15 = 60$. Base: $6 \times 6 = 36$

Total $= 60 + 36 = 96\text{ cm}^2$

(a) From the front you see a triangle: base 6 cm, height 4 cm, apex above the middle of the base.

(b) Each sloping face is a triangle with base 6 and slant height 5: $\frac{1}{2} \times 6 \times 5 = 15\text{ cm}^2$

Base $= 6 \times 6 = 36\text{ cm}^2$

Total $= 4 \times 15 + 36 = 96\text{ cm}^2$

Base of triangle: $\sqrt{8.4^2 - 7.2^2} = \sqrt{18.72} = 4.327$ cm

Triangle area: $\frac{1}{2} \times 4.327 \times 7.2 = 15.58$ cm²

Volume: $15.58 \times 18 = 280.4 \approx 280$ cm³

Volume $= 100 \times 60 \times 40 = 240000$ cm³

$1$ litre $= 1000$ cm³, so that is $240$ litres.

Bags: $240 \div 8 = 30$

Cost: $30 \times £2.50 = £75$, which is more than £70, so Sally is wrong.

Volume $= \pi r^2 h = \pi \times 100^2 \times 30 = 942\,478$ cm³

Time $= 942\,478 \div 250 = 3769.9$ seconds

$3769.9 \div 60 = 62.8$, so about $63$ minutes.

The triangle is isosceles, so split it into two right-angled triangles with base $5$ cm.

Height $h = 5\tan 40^\circ \approx 4.195$ cm

Cross-section $= 10 \times 12 + \frac{1}{2} \times 10 \times 4.195 \approx 140.98$ cm²

Volume $= 140.98 \times 20 \approx 2819.5 \approx 2820$ cm³

Radius $= 6 \div 2 = 3$ cm.

Cone: $\frac{1}{3} \pi \times 3^2 \times 10 = 30\pi$

Hemisphere: $\frac{1}{2} \times \frac{4}{3} \pi \times 3^3 = 18\pi$

Total $= 48\pi$, so $k = 48$.

Radius $= 8.4 \div 2 = 4.2$ cm

Volume $= \frac{1}{2} \times \frac{4}{3}\pi \times 4.2^3 = 155.17...$

$= 155\text{ cm}^3$ (3 s.f.)

Face areas 15, 27, 45 need edges with pair products 15, 27, 45: the edges are 3, 5 and 9 cm.

Volume of cuboid $= 3 \times 5 \times 9 = 135$ cm$^3$

Volume of one cube $= 2.5^3 = 15.625$ cm$^3$

$135 \div 15.625 = 8.64$, so at most 8 whole cubes.

Let height $= h$, so $r = \frac{3}{4}h$.

$l = \sqrt{h^2 + \frac{9}{16}h^2} = \frac{5}{4}h$

Total area $= \pi r l + \pi r^2 = \pi\left(\frac{15}{16}h^2 + \frac{9}{16}h^2\right) = \frac{24}{16}\pi h^2$

$\frac{3}{2}h^2 = 54 \Rightarrow h^2 = 36 \Rightarrow h = 6$ cm

Small cone: height 10 is $\frac{2}{3}$ of 15, so its radius is $\frac{2}{3} \times 12 = 8$ cm.

Large cone $= \frac{1}{3}\pi(12^2)(15) = 720\pi$; small cone $= \frac{1}{3}\pi(8^2)(10) = \frac{640}{3}\pi$

Frustum $= 720\pi - \frac{640}{3}\pi = \frac{1520}{3}\pi$

Hemisphere $= \frac{2}{3}\pi(12^3) = 1152\pi$

$S = 1152\pi - \frac{1520}{3}\pi = 2027.37... \approx 2030\text{ cm}^3$

(a) Radius $= 3.5$ cm. Hemisphere: $\frac{2}{3}\pi(3.5)^3 = 28.583...\pi$

Cone volume $= 120\pi - 28.583...\pi = 91.416...\pi$

$\frac{1}{3}\pi(3.5)^2 h = 91.416...\pi$ gives $h = 91.416... \div 4.0833... = 22.387...$

$y = h + 3.5 = 25.887...$, so $y = 25.9$

(b) With a wider base the same volume needs less height, so $y$ would be smaller.

$\frac{1}{4} \times \frac{4}{3}\pi r^3 = 576\pi$, so $r^3 = 1728$ and $r = 12$

Curved part: $\frac{1}{4} \times 4\pi \times 12^2 = 144\pi$

Two flat semicircles: $2 \times \frac{1}{2}\pi \times 12^2 = 144\pi$

Total $= 288\pi = 904.8 \approx 905\text{ cm}^2$ (3 s.f.)

Hemisphere volume: $\frac{2}{3}\pi r^3 = \frac{250}{3}\pi$, so $r^3 = 125$ and $r = 5$.

Curved surface: $\frac{1}{2} \times 4\pi \times 5^2 = 50\pi$

Flat face: $\pi \times 5^2 = 25\pi$

Total: $50\pi + 25\pi = 75\pi$ cm²

Volume: $20wh = 300$, so $wh = 15$

Surface area: $2(20w + 20h + wh) = 370 \Rightarrow 40w + 40h + 30 = 370 \Rightarrow w + h = 8.5$

Substitute $w = \frac{15}{h}$: $\frac{15}{h} + h = 8.5 \Rightarrow 2h^2 - 17h + 30 = 0$

$(2h - 5)(h - 6) = 0$, so $h = 2.5$ or $6$

The width is greater than the height, so $h = 2.5$ cm (and $w = 6$ cm)

Small cone radius: $\dfrac{6}{10} \times 3 = 1.8$ cm

Curved area of frustum $= \pi \times 3 \times 10 - \pi \times 1.8 \times 6 = 19.2\pi = 60.3...$

Bottom circle: $\pi \times 3^2 = 9\pi$; top circle: $\pi \times 1.8^2 = 3.24\pi$

Total $= (19.2 + 9 + 3.24)\pi = 31.44\pi = 98.77...$, so $98.8$ cm²

(a) $\frac{4}{3}\pi r^3 = \frac{1}{3}\pi r^2 h \Rightarrow 4r = h$, so $r : h = 1 : 4$

(b) $4\pi r^2 = \pi r^2 + \pi r l \Rightarrow 3r = l$

$l^2 = r^2 + h^2 \Rightarrow 9r^2 = r^2 + h^2 \Rightarrow h^2 = 8r^2$

$h = \sqrt{8}\,r$, so $r : h = 1 : \sqrt{8}$

Whole cone: $r = 10$, $l = 25$, so curved surface area $= \pi \times 10 \times 25 = 250\pi$.

The grey part is a smaller similar cone with slant height $25 - 10 = 15$.

Its radius is $10 \times \frac{15}{25} = 6$, so its area is $\pi \times 6 \times 15 = 90\pi$.

Unpainted area $= 250\pi - 90\pi = 160\pi$ cm².

Frustum $= \frac{1}{3}\pi(3.6)^2(6.4) - \frac{1}{3}\pi(1.8)^2(3.2) = 76.00$ cm³

Hemisphere $= \frac{2}{3}\pi(3.6)^3 = 97.72$ cm³

Mass $= 76.00 \times 2.4 + 97.72 \times 4.8 = 182.40 + 469.04 = 651.44$ g

Density $= \frac{651.44}{173.72} = 3.75$ g/cm³

Radius $= 12x$, height $= 16x$, so slant height $l = \sqrt{(12x)^2 + (16x)^2} = 20x$

Curved surface area: $\pi \times 12x \times 20x = 240x^2\pi = 2160\pi$, so $x^2 = 9$ and $x = 3$

So $r = 36$ and $h = 48$

Volume $= \frac{1}{3}\pi \times 36^2 \times 48 = 20\,736\pi$, so $V = 20\,736$