Sec 2 Maths: Mensuration of Pyramids, Cones & Spheres practice questions with worked solutions

(a) Volume

$$\begin{aligned}V &= \tfrac43\pi(10)^3\\ &= 4188.8\\ &= 4190 \text{ cm}^3 \text{ (3 s.f.)}\end{aligned}$$

(b) Percentage increase in surface area

The new radius is $1.2r$. Surface area depends on $r^2$, so it is multiplied by $1.2^2$.

$$\begin{aligned}\text{New surface area} &= 4\pi(1.2r)^2 = 1.44\,(4\pi r^2)\\ \text{Percentage increase} &= \frac{1.44 – 1}{1} \times 100\%\\ &= 44\%\end{aligned}$$

(a) Slant height

The slant height runs from $V$ to the midpoint of $BC$. The distance from $O$ to that midpoint is half the side, 12 cm. By Pythagoras’ theorem:

$$\begin{aligned}l &= \sqrt{16^2 + 12^2}\\ &= \sqrt{400} = 20 \text{ cm}\end{aligned}$$

(b) Total surface area

Four congruent triangular faces plus the square base.

$$\begin{aligned}\text{TSA} &= 4\left(\tfrac12 \times 24 \times 20\right) + 24 \times 24\\ &= 960 + 576\\ &= 1536 \text{ cm}^2\end{aligned}$$

(a) Volume

$$\begin{aligned}V &= \tfrac13 \times (12 \times 12) \times 8\\ &= 384 \text{ cm}^3\end{aligned}$$

(b) Slant height $ET$

$OT$ is half the side, 6 cm, and triangle $EOT$ is right-angled at $O$.

$$\begin{aligned}ET &= \sqrt{8^2 + 6^2} = \sqrt{100}\\ &= 10 \text{ cm (shown)}\end{aligned}$$

(c) Total surface area

$$\begin{aligned}\text{TSA} &= 12 \times 12 + 4\left(\tfrac12 \times 12 \times 10\right)\\ &= 144 + 240\\ &= 384 \text{ cm}^2\end{aligned}$$

(a) Volume

The base is a square of side 17 cm.

$$\begin{aligned}V &= \tfrac13 \times 17^2 \times 14\\ &= 1348.7\\ &= 1350 \text{ cm}^3 \text{ (3 s.f.)}\end{aligned}$$

(b) Length $VN$

$N$ is the midpoint of a base edge, so $ON = 17 \div 2 = 8.5$ cm. By Pythagoras’ theorem in triangle $VON$:

$$\begin{aligned}VN^2 &= 14^2 + 8.5^2 = 268.25\\ VN &= 16.378\\ &= 16.4 \text{ cm (3 s.f.)}\end{aligned}$$

(c) Total surface area

$$\begin{aligned}\text{TSA} &= 4 \times \tfrac12(17)(16.378) + 17^2\\ &= 556.85 + 289\\ &= 845.85\\ &= 846 \text{ cm}^2 \text{ (3 s.f.)}\end{aligned}$$

(a) $x = 24$

$$\begin{aligned}\tfrac13 \times x^2 \times 35 &= 6720\\ x^2 &= \frac{6720 \times 3}{35} = 576\\ x &= 24 \quad (x = -24 \text{ rejected, since a length is positive})\end{aligned}$$

(b) Total surface area

Each triangular face has base 24 m and height equal to the slant height, 37 m.

$$\begin{aligned}\text{TSA} &= 24 \times 24 + 4\left(\tfrac12 \times 24 \times 37\right)\\ &= 576 + 1776\\ &= 2352 \text{ m}^2\end{aligned}$$

(a) Slant height $l_1$

$l_1$ is the slant height of the faces on the 8 m edges. From the centre of the base to the midpoint of an 8 m edge is half of 14 m, which is 7 m. By Pythagoras’ theorem:

$$\begin{aligned}l_1 &= \sqrt{20^2 + 7^2} = \sqrt{449}\\ &= 21.190 \text{ m (3 d.p., shown)}\end{aligned}$$

(b) Total surface area

The faces on the 14 m edges have a different slant height $l_2$, using half of 8 m, which is 4 m.

$$\begin{aligned}l_2 &= \sqrt{20^2 + 4^2} = \sqrt{416} = 20.396 \text{ m}\\ \text{Base} &= 14 \times 8 = 112\\ \text{Triangles} &= 2\left(\tfrac12 \times 8 \times 21.190\right) + 2\left(\tfrac12 \times 14 \times 20.396\right)\\ &= 169.52 + 285.54 = 455.06\\ \text{TSA} &= 112 + 455.06 = 567.06\\ &= 567 \text{ m}^2 \text{ (3 s.f.)}\end{aligned}$$

(a) Side of the base

$$\text{side} = (x – 4) \text{ cm}$$

(b) Forming the equation

$$\begin{aligned}(x-4)^2 &= 36\\ x^2 – 8x + 16 &= 36\\ x^2 – 8x – 20 &= 0 \text{ (shown)}\end{aligned}$$

(c) Solving

$$\begin{aligned}(x-10)(x+2) &= 0\\ x = 10 \quad&\text{or}\quad x = -2\end{aligned}$$

(d) Rejecting a solution

$x = -2$ is rejected because $x$ is the height of the box, and a length cannot be negative. (It would also make the side $x – 4 = -6$ cm.)

(e) Volume

$$\begin{aligned}V &= \tfrac13 \times \text{base area} \times \text{height}\\ &= \tfrac13 \times 36 \times 10\\ &= 120 \text{ cm}^3\end{aligned}$$

(a) Height of the cone

The cone’s radius is the hemisphere’s radius, 5 cm. By Pythagoras’ theorem:

$$\begin{aligned}h &= \sqrt{13^2 – 5^2} = \sqrt{144}\\ &= 12 \text{ cm (shown)}\end{aligned}$$

(b) Volume

$$\begin{aligned}V &= \tfrac13\pi r^2 h + \tfrac23\pi r^3\\ &= \tfrac13\pi(5)^2(12) + \tfrac23\pi(5)^3\\ &= 100\pi + 83.333\pi\\ &= 575.96\\ &= 576 \text{ cm}^3 \text{ (3 s.f.)}\end{aligned}$$

(c) Total surface area

The exposed surfaces are the curved surface of the cone and the curved surface of the hemisphere; the joined circular faces are hidden.

$$\begin{aligned}\text{TSA} &= \pi r l + 2\pi r^2\\ &= \pi(5)(13) + 2\pi(5)^2\\ &= 65\pi + 50\pi = 115\pi\\ &= 361.28\\ &= 361 \text{ cm}^2 \text{ (3 s.f.)}\end{aligned}$$

(a) Volume of the hemisphere

$$\begin{aligned}V &= \tfrac23 \times 3.142 \times 6^3\\ &= 452.45\\ &= 452 \text{ cm}^3 \text{ (3 s.f.)}\end{aligned}$$

(b) Paint needed

The cone touches the hemisphere only at its tip, so every surface is exposed: the hemisphere’s curved surface and flat top ($3\pi r^2$), and the cone’s curved surface and circular base ($\pi r l + \pi r^2$, with $r = 4$ cm).

$$\begin{aligned}\text{Hemisphere} &= 3\pi(6)^2 = 108\pi\\ \text{Cone} &= \pi(4)(7) + \pi(4)^2 = 44\pi\\ \text{Total} &= 152\pi = 152 \times 3.142 = 477.58 \text{ cm}^2\\ \text{Paint} &= \frac{477.58}{400} = 1.194\\ &= 1.19 \text{ litres (3 s.f.)}\end{aligned}$$

(a) Radius of the cylinder

$$\begin{aligned}\pi r^2 &= 150\\ r &= \sqrt{\frac{150}{\pi}} = 6.9099\\ &= 6.91 \text{ cm (3 s.f.)}\end{aligned}$$

(b) Radius of the cone

From the diagram, the cone’s base is 2 cm in from the cylinder’s edge.

$$\begin{aligned}r_{\text{cone}} &= 6.9099 – 2 = 4.9099\\ &= 4.91 \text{ cm (3 s.f.)}\end{aligned}$$

(c) Volume of the solid

$$\begin{aligned}V &= 150 \times 4 + \tfrac13\pi(4.9099)^2(12)\\ &= 600 + 302.94\\ &= 902.94\\ &= 903 \text{ cm}^3 \text{ (3 s.f.)}\end{aligned}$$

The radius of the structure is $10 \div 2 = 5$ cm and each small sphere has radius 1 cm. Melting keeps the volume the same.

$$\begin{aligned}\text{Volume of structure} &= \tfrac23\pi(5)^3 + \pi(5)^2(16)\\ &= 83.333\pi + 400\pi = 483.33\pi\\ &= 1518.4 \text{ cm}^3\\ \text{Volume of one sphere} &= \tfrac43\pi(1)^3 = 4.1888 \text{ cm}^3\\ \text{Number of spheres} &= \frac{1518.4}{4.1888} = 362.5\end{aligned}$$

Only whole spheres count, so 362 spheres can be made. Since $362 \ge 360$, the engineer’s calculation is correct.

(a) $h = 4r$

$$\begin{aligned}\tfrac13\pi r^2 h &= \tfrac43\pi r^3\\ \pi r^2 h &= 4\pi r^3\\ h &= 4r \text{ (dividing by } \pi r^2\text{)}\end{aligned}$$

(b) Value of $r$

By Pythagoras’ theorem, $h^2 + r^2 = l^2$ with $h = 4r$.

$$\begin{aligned}(4r)^2 + r^2 &= 5.4^2\\ 17r^2 &= 29.16\\ r &= \sqrt{\frac{29.16}{17}} = 1.3097\\ &= 1.31 \text{ (3 s.f.)}\end{aligned}$$

(c) Tins of paint

One set is a solid cone (curved surface and base) and a sphere.

$$\begin{aligned}\text{Area of one set} &= \pi r l + \pi r^2 + 4\pi r^2 = \pi r l + 5\pi r^2\\ &= \pi(1.3097)(5.4) + 5\pi(1.3097)^2\\ &= 22.218 + 26.944 = 49.162 \text{ m}^2\\ \text{Area of 10 sets} &= 491.62 \text{ m}^2\\ \text{Tins} &= \frac{491.62}{30} = 16.39\end{aligned}$$

16 tins are not enough, so 17 tins are needed.

(a) Volume of the sculpture

Hemisphere radius 1 m; cone radius 0.6 m. The cone’s height comes from Pythagoras’ theorem.

$$\begin{aligned}\text{Hemisphere} &= \tfrac23\pi(1)^3 = 0.66667\pi\\ h_{\text{cone}} &= \sqrt{1^2 – 0.6^2} = 0.8 \text{ m}\\ \text{Cone} &= \tfrac13\pi(0.6)^2(0.8) = 0.096\pi\\ \text{Sculpture} &= 0.66667\pi – 0.096\pi = 0.57067\pi\\ &= 1.7928 = 1.79 \text{ m}^3 \text{ (3 s.f., shown)}\end{aligned}$$

(b) Volume of sand

Sand is 5 parts out of $3 + 5 + 7 = 15$.

$$\begin{aligned}\text{Sand} &= \tfrac{5}{15} \times 1.7928 = 0.5976\\ &= 0.598 \text{ m}^3 \text{ (3 s.f.)}\end{aligned}$$

(c) Mass of sand

$$\begin{aligned}\text{Mass} &= 0.5976 \times 1530 = 914.33\\ &= 914 \text{ kg (3 s.f.)}\end{aligned}$$

(d) Are 36 bags enough?

$$\begin{aligned}36 \times 25 &= 900 \text{ kg}\\ \frac{914.33}{25} &= 36.57\end{aligned}$$

No. 36 bags hold only 900 kg, less than the 914 kg needed, so 37 bags are required.

(a) Height $h$

The removed small cone is similar to the full cone, with radii in the ratio $6:9$.

$$\begin{aligned}\text{Height of small cone} &= \tfrac{6}{9} \times 12 = 8 \text{ m}\\ h &= 12 – 8 = 4 \text{ m (shown)}\end{aligned}$$

(b) Volume of water

$$\begin{aligned}V &= \tfrac13\pi(9)^2(12) – \tfrac13\pi(6)^2(8)\\ &= 324\pi – 96\pi\\ &= 228\pi \text{ m}^3 \;(= 716 \text{ m}^3 \text{ to 3 s.f.})\end{aligned}$$

Note: the school’s answer key prints this as $716\pi$ m$^3$; the working gives $228\pi$ m$^3$, which is 716 m$^3$, so the $\pi$ in the key is a slip.

(c) Total surface area

Slant heights by Pythagoras’ theorem, then curved surface of the frustum plus the top and base circles (the container is closed).

$$\begin{aligned}l_{\text{large}} &= \sqrt{9^2 + 12^2} = 15 \text{ m}\\ l_{\text{small}} &= \sqrt{6^2 + 8^2} = 10 \text{ m}\\ \text{TSA} &= \pi(9)(15) – \pi(6)(10) + \pi(6)^2 + \pi(9)^2\\ &= 135\pi – 60\pi + 36\pi + 81\pi\\ &= 192\pi \text{ m}^2\end{aligned}$$

(a) Height of cone $OAB$

Let the height of cone $OAB$ be $h$ cm, so cone $OQR$ has height $h + 10$. The two cones are similar, so radius and height are in the same ratio.

$$\begin{aligned}\frac{6}{12} &= \frac{h}{h + 10}\\ 12h &= 6h + 60\\ h &= 10 \text{ cm (shown)}\end{aligned}$$

(b) Total surface area of the frustum

Large cone: radius 12, height 20. Small cone: radius 6, height 10. Slant heights by Pythagoras’ theorem.

$$\begin{aligned}l_{\text{large}} &= \sqrt{12^2 + 20^2} = \sqrt{544}\\ l_{\text{small}} &= \sqrt{6^2 + 10^2} = \sqrt{136}\\ \text{TSA} &= \pi(12)\sqrt{544} – \pi(6)\sqrt{136} + \pi(12)^2 + \pi(6)^2\\ &= 879.29 – 219.82 + 452.39 + 113.10\\ &= 1224.96\\ &= 1220 \text{ cm}^2 \text{ (3 s.f.)}\end{aligned}$$

(c) Clay required

Volume of the frustum minus the hemisphere of radius 6 cm.

$$\begin{aligned}V &= \tfrac13\pi(12)^2(20) – \tfrac13\pi(6)^2(10) – \tfrac23\pi(6)^3\\ &= 960\pi – 120\pi – 144\pi\\ &= 696\pi = 2186.5\\ &= 2190 \text{ cm}^3 \text{ (3 s.f.)}\end{aligned}$$