Sec 4 E Maths: Mensuration, practice questions & worked solutions

$$\begin{aligned}&\text{Total surface area}\\&=\frac12\times4\pi(6)^2+\pi(6)^2\\&=72\pi+36\pi\\&=108\pi\\&=339\text{ cm}^2\text{ (3 s.f.)}\end{aligned}$$

Answer: $339$ cm$^2$

$$\frac43\pi(2r)^3=2\times\frac13\pi\left(\frac23r\right)^2\times10h$$

$$\frac{32\pi r^3}3=\frac{80\pi r^2h}{27}$$

$$32r\times27=3\times80h$$

$$864r=240h$$

$$\frac hr=\frac{864}{240}$$

$$h:r=18:5$$

Answer: $h:r=18:5$

Perimeter of major sector $=3\times$ Perimeter of minor sector

$$\begin{aligned}x(2\pi-\theta)+2x&=3(x\theta+2x)\\2x\pi-x\theta+2x&=3x\theta+6x\\x(2\pi-\theta+2)&=x(3\theta+6)\\4\theta&=2\pi-6+2\\4\theta&=2\pi-4\\\theta&=\frac{\pi-2}2=0.571\text{ radians (3dp)}\end{aligned}$$

Answer: $\theta=0.571$ radians (3 d.p.)

Let the height of the Great Pyramid be $H$.

Let the height of the replica be $h$

$$\begin{aligned}\left(\frac hH\right)^3&=\frac{\text{Volume of replica}}{\text{Volume of Great Pyramid}}\\\left(\frac hH\right)^3&=\frac{9.22\times10^{-6}}{2.45\times10^4}\\\frac hH&=\sqrt[3]{\frac{9.22\times10^{-6}}{2.45\times10^4}}\\&=7.2197\times10^{-4}\end{aligned}$$

Let $S$ be the surface area of the Great Pyramid and $s$ be the surface area of the replica

$$\begin{aligned}\frac sS&=\left(\frac hH\right)^2\\s&=\left(\frac hH\right)^2\times S\\&=(7.2197\times10^{-4})^2\times8.30\times10^4\\&=4.33\times10^{-2}\text{ m}^2\end{aligned}$$

Answer: $4.33\times10^{-2}$ m$^2$

(a)

$$\sqrt[3]{\frac38}=\frac{40-h}{40}\quad\text{(using cube root)}$$

$$h=40-40\times\sqrt[3]{\frac38}$$

$$h=11.2\text{ (3s.f.)}$$

(b)

$$\frac{A_{\text{smaller cone}}}{A_{\text{larger cone}}}=\left[\left(\frac38\right)^{\frac13}\right]^2\quad\text{(squaring the ratio of lengths)}$$

$$\frac{A_{\text{smaller cone}}}{A_{\text{smaller cone}}+966}=\left(\frac38\right)^{\frac23}$$

$$A_{\text{smaller cone}}=\left(\frac38\right)^{\frac23}\left(A_{\text{smaller cone}}+966\right)$$

$$A_{\text{smaller cone}}=\frac{\left(\dfrac38\right)^{\frac23}(966)}{1-\left(\dfrac38\right)^{\frac23}}=1050\text{ cm}^2\text{ (3s.f.)}$$

Answer: $\text{(a) }h=11.2\text{ cm}\quad\text{(b) }1050\text{ cm}^2$

(a)

$$\begin{aligned}\tan5^\circ&=\frac x{2.1}\\\text{Length of water level}&=0.9+2(2.1\tan5^\circ)\\\text{Area of trapezium}&=\frac12(0.9+0.9+2(2.1\tan5^\circ))\times2.1=2.2758\ \text{m}^2\\\text{Volume of water}&=2.2758\times100=227.5825\approx228\text{m}^3\end{aligned}$$

(b)

$$\begin{aligned}\frac{227.5825}{0.3}\div60&=12.6434h\\&=12\text{ hours }38.60\text{ mins}\\&=12\text{ hours }39\text{ mins}\end{aligned}$$

Answer: (a) $228$ m$^3$ (b) $12$ hours $39$ minutes

(a)

$$\begin{aligned}\text{height}&=\sqrt{17^2-8^2}\\&=15\text{ cm}\\\text{Volume}&=\dfrac13\pi(8)^2(15)+(16\times16\times7.5)\\&=2925.3\\&=2930\text{ cm}^3\text{ (3sf)}\end{aligned}$$

(b)

$$\begin{aligned}\text{Total SA}&=\pi(8)(17)+\left[(16\times16)-\pi(8)^2\right]+\left[(16\times16)+(4\times16\times7.5)\right]\\&=1218.2\\&=1220\text{ cm}^2\text{ (3sf)}\end{aligned}$$

Answer: (a) 2930 cm$^3$ (b) 1220 cm$^2$

Area of shaded region in triangle $ABC$

$$\begin{aligned}&=\left[\frac12(5)^2\sin60^\circ-\frac12\pi(2)^2\right]\text{cm}^2\\&=\left(\frac{25\sqrt3}4-2\pi\right)\text{cm}^2\end{aligned}$$

$$\begin{aligned}\tan1.2&=\frac{EF}{OE}\\EF&=2\tan1.2\\DF&=4\tan1.2\end{aligned}$$

Area of shaded region in triangle $ODF$

$$\begin{aligned}&=\left[\left(\frac12\times2\times4\tan1.2\right)-\frac12(2)^2\,2.4\right]\text{cm}^2\\&=(4\tan1.2-4.8)\text{cm}^2\end{aligned}$$

$\therefore$ area of shaded region of the logo

$$\begin{aligned}&=\left(\frac{25\sqrt3}4-2\pi+4\tan1.2-4.8\right)\text{cm}^2\\&=10.030739\text{ cm}^2\\&\simeq10.0\text{ cm}^2\end{aligned}$$

Answer: $10.0$ cm$^2$

(a)

Using similarity,

$$\frac h{h-10}=\frac{24}{18}$$

$$18h=24h-240$$

$$h=40$$

$$\text{Volume of frustum}=\frac13\pi(12)^2\times40-\frac13\pi(9)^2\times30$$

$$=1110\pi\text{ cm}^3$$

$$=1.11\pi\text{ litres}$$

(b)

$$\frac34\pi r^2(20)=1110\pi$$

$$r^2=74$$

$$r=\sqrt{74}\approx8.6023$$

$$\text{Therefore, area not in contact of water}=2\pi(\sqrt{74})\left[\frac14(20)\right]$$

$$\approx270.250$$

$$=270\text{ cm}^2\text{ (to 3 sf)}$$

Answer: $\text{(a) }1.11\pi\text{ litres}\quad\text{(b) }270\text{ cm}^2$

(a)

Slant height of cone,

$$\sqrt{9^2+5^2}=10.296$$

$$\text{Curved SA}=\pi(5)(10.296)$$

$$=162\text{ cm}^2$$

(b(i))

The ratio of area is not proportional to the ratio of heights. There is a larger surface area at the top compared to the bottom. Hence less than two-third of the inner surface area of the cup is in contact with the water.

(b(ii))

$$\frac{V_{\mathrm{Water}}}{V_{\mathrm{Cup}}}=\left(\frac{h_{\mathrm{Water}}}{h_{\mathrm{Cup}}}\right)^3=\left(\frac69\right)^3=\frac8{27}$$

Volume of water in cup,

$$\frac8{27}\times\left(\frac13\right)\pi(5^2)(9)=69.813$$

Let the radius of the bottle be $r$.

$$\pi r^2(0.4)=69.813$$

$$r=7.45$$

Answer: (a) $162$ cm$^2$   (b)(i) the water touches a similar cone with $\tfrac69$ of the height, so only $\left(\tfrac69\right)^2=\tfrac49$ of the inner surface, which is less than two thirds   (b)(ii) $7.45$ cm

(a)

$$\begin{aligned}\text{Vol of hemisphere}&=\frac23\times\pi\times(3x)^3\\&=18\pi x^3\\\text{Vol of cone}&=\frac13\times\pi\times\left(\frac32x\right)^2\times y\\&=\frac13\times\pi\times\frac94x^2\times y\\&=\frac{3\pi}{4}x^2y\end{aligned}$$

$$\begin{aligned}18\pi x^3&=3\times\frac{3\pi}{4}x^2y\\18x^3&=\frac94x^2y\\y&=\frac{18x^3\times4}{9x^2}\end{aligned}$$

$y=8x$ (shown)

(b)

$$\begin{aligned}l&=\sqrt{y^2+\left(\frac32x\right)^2}\\&=\sqrt{(8x)^2+\left(\frac32x\right)^2}\\&=\sqrt{64x^2+\frac94x^2}\\&=\sqrt{\frac{265x^2}{4}}\\&=8.1394x\ /\ \frac{\sqrt{265}}{2}x\end{aligned}$$

$$\begin{aligned}\text{CSA of cone}&=\pi\left(\frac32x\right)l\\&=\pi\left(\frac32x\right)\times8.1394x\\&=38.356x^2\\\text{CSA of cylinder}&=2\pi\left(\frac32x\right)\times3x\\&=28.274x^2\\\text{CSA of hemisphere}&=2\pi(3x)^2\\&=56.549x^2\\\text{rim area}&=\pi(3x)^2-\pi\left(\frac32x\right)^2\\&=21.206x^2\end{aligned}$$

$$\begin{aligned}\text{Total SA}&=450\\38.356x^2+28.274x^2+56.549x^2+21.206x^2&=450\\144.385x^2&=450\\x^2&=3.1167\\x&=1.7654\\\text{Total ht, }h&=y+3x+3x\\h&=8x+3x+3x\\h&=14x\\h&=14(1.7654)\\h&=24.7156\\h&=24.7\ \text{(3sf)}\end{aligned}$$

Answer: (b) $h=24.7$ cm

(a)

$$\begin{aligned}(6-r)^2+2.5^2&=r^2\\36-12r+r^2+6.25&=r^2\\12r&=42.25\\r&=3.5\text{ m (shown)}\end{aligned}$$

(b)(i)

$$\begin{aligned}\tfrac12\angle EOF&=\cos^{-1}\left(\frac{6-3.5-0.3}{3.5}\right)\\\tfrac12\angle EOF&=\cos^{-1}\left(\frac{2.2}{3.5}\right)\\\angle EOF&=1.78216\text{ rad}\\\angle EOF&=1.78\text{ rad (3sf)}\end{aligned}$$

(b)(ii)

$$\begin{aligned}&3.5(2\pi-1.78216)\\&=15.8\text{ m}\end{aligned}$$

(b)(iii)

$$\begin{aligned}&\tfrac12(3.5^2)(2\pi-1.78216)+\tfrac12(3.5^2)\sin1.78216\\&=33.55747\\&=33.6\text{ m}^2\text{ (3sf)}\end{aligned}$$

(b)(iv)

$$\begin{aligned}&33.55747\times1400\\&=47\,000\text{ m}^3\text{ (3sf)}\end{aligned}$$

Answer: (b)(i) $1.78$ rad (ii) $15.8$ m (iii) $33.6$ m$^2$ (iv) $47\,000$ m$^3$