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Ecolint Campus des NationsMathématiques
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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 9 · 9.2 Mensuration

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.160 cm³
2.96 cm²
3.90π90\pi cm³
4.24 cm²
5.35 cm²
6.108 cm³
7.144π144\pi cm³
8.36π36\pi cm³
9.108 cm²
10.112π112\pi cm²
Silver
11.600 cm³
12.224 cm²
13.144π144\pi cm²
14.h=V/(πr2)h = V/(\pi r^2)
15.9 cm
16.96 cm³
17.20=25\sqrt{20} = 2\sqrt{5} cm
18.(a) (250/3)π(250/3)\pi (b) 50π50\pi
19.0.4 m
20.(a) 0.25 m³ (b) 250 000 cm³
Gold
21.600 cm³
22.x=6x = 6
23.h=3V/(πr2)=12/πh = 3V/(\pi r^2) = 12/\pi cm
24.18 cm
25.69π69\pi cm²
26.(a) 60π60\pi cm² (b) 96π96\pi cm²
27.r=3,h=4r = 3, h = 4
28.160π+(128/3)π=(608/3)π160\pi + (128/3)\pi = (608/3)\pi
29.12 cm
30.(a) 500 m³ (b) 500 000 L
Platinum
31.126π126\pi cm³
32.(a) 3 cm (b) × 8
33.r=3V/(4π)3r = \sqrt[3]{3V/(4\pi)}
34.(a) 36 036 m² (b) ≈ 179.4 m
35.r=3r = 3
36.160π+32π=192π160\pi + 32\pi = 192\pi
37.≈ 4.9 cm
38.432 cm³
39.√(25 + 9) = √34 ≈ 5.83 m (to midpoint of 8 m side)
40.h=(A−2πr2)/(2πr)h = (A - 2\pi r^2)/(2\pi r)

Pack B — Answers

Bronze
1.180 cm³
2.294 cm²
3.200π200\pi cm³
4.60 cm²
5.40 cm²
6.64 cm³
7.192π192\pi cm³
8.288π288\pi cm³
9.132 cm²
10.60π60\pi cm²
Silver
11.360 cm³
12.144 cm²
13.100π100\pi cm²
14.Same.
15.12 cm
16.120 cm³
17.2 cm
18.(a) 144π144\pi (b) 72π72\pi
19.0.5 m
20.(a) 5 m³ (b) 5 000 000 cm³
Gold
21.360 cm³
22.x=4x = 4
23.h=18/πh = 18/\pi cm
24.27 cm
25.80π80\pi cm²
26.(a) 65π65\pi cm² (b) 90π90\pi cm²
27.Same.
28.54π+18π=72π54\pi + 18\pi = 72\pi
29.10 cm
30.(a) 675 m³ (b) 675 000 L
Platinum
31.224π224\pi cm³
32.Same.
33.Same.
34.(a) 31 200 m² (b) ≈ 152.6 m
35.r=4r = 4
36.72π+18π=90π72\pi + 18\pi = 90\pi
37.≈ 6.2 cm
38.200 cm³
39.√(36 + 4) = √40 ≈ 6.32 m
40.Same.

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) 96π96\pi (b) 83\tfrac{8}{3} cm
Full working
(a) V=(1/3)π(36)(8)=96πV = (1/3)\pi(36)(8) = 96\pi.

(b) π(36)h=96π⇒h=96/36=8/3\pi(36) h = 96\pi \Rightarrow h = 96/36 = 8/3 cm ≈ 2.67 cm.
2Problem 2
Answer
(a) ≈ 2 592 100 m³ (b) ≈ 219.2 m
Full working
(a) V=(1/3)(230)2(147)=(1/3)(52900)(147)=2592100V = (1/3)(230)^2(147) = (1/3)(52900)(147) = 2 592 100 m³.

(b) Half-diagonal of base: 2302/2≈162.6230\sqrt{2}/2 \approx 162.6 m. Slanted edge = 1472+162.62≈219.2\sqrt{147^2 + 162.6^2} \approx 219.2 m.
3Problem 3
Answer
(a) ≈ 50 m (b) ≈ 519 000 m³
Full working
(a) 4πr2=31416⇒r2=2500⇒r=504\pi r^2 = 31416 \Rightarrow r^2 = 2500 \Rightarrow r = 50 m.

(b) Internal r=49.85r = 49.85 m. Internal volume = (4/3)π(49.85)3≈519000(4/3)\pi (49.85)^3 \approx 519 000 m³.
4Problem 4
Answer
≈ 19 cm (2 s.f.)
Full working
Tank vol = 72 000 cm³. Water = 24 000 cm³. Cylinder base =π(400)≈1257= \pi(400) \approx 1257. Depth = 24000/1257≈1924000/1257 \approx 19 cm.
5Problem 5
Answer
(a) r=3V/(4π)3r = \sqrt[3]{3V/(4\pi)} (b) r=3r = 3
Full working
(a) Multiply by 3, divide by 4π4\pi: r3=3V/(4π)r^3 = 3V/(4\pi), r=3V/(4π)3r = \sqrt[3]{3V/(4\pi)}.

(b) r3=3(36π)/(4π)=27⇒r=3r^3 = 3(36\pi)/(4\pi) = 27 \Rightarrow r = 3.
6Problem 6
Answer
(a) 4 cm (b) x=20x = 20 cm
Full working
(a) Height equals the cut depth = 4 cm.

(b) V=x2×4=1600⇒x=20V = x^2 \times 4 = 1600 \Rightarrow x = 20 cm.
7Problem 7
Answer
(a) 3 cm (b) ≈ 280π cm³ — see working
Full working
(a) Similar triangles: at height 8, distance to apex = 4 above the cut. Radius shrinks linearly from 9 (at base) to 0 (at apex). At height 8: radius = 9×(12−8)/12=9×1/3=39 \times (12 - 8)/12 = 9 \times 1/3 = 3.

(b) Big cone vol = (1/3)π(81)(12)=324π(1/3)\pi(81)(12) = 324\pi. Small (top sliced off) cone vol = (1/3)π(9)(4)=12π(1/3)\pi(9)(4) = 12\pi. Frustum = 324π−12π=312π324\pi - 12\pi = 312\pi.

Hmm — let me recheck: the slice is at height 8 cm above the base. The small cone above the slice has height 12−8=412 - 8 = 4 cm and radius 3. Vol =(1/3)π(9)(4)=12π= (1/3)\pi(9)(4) = 12\pi. Frustum = 324π−12π=312π324\pi - 12\pi = 312\pi cm³.
8Problem 8
Answer
(a) 250π+(250/3)π=(1000/3)π250\pi + (250/3)\pi = (1000/3)\pi (b) Cyl side 100π100\pi + base 25π25\pi + hemi 50π=175π50\pi = 175\pi
Full working
(a) Cyl vol = π(25)(10)=250π\pi(25)(10) = 250\pi. Hemi vol = (2/3)π(125)=(250/3)π(2/3)\pi(125) = (250/3)\pi. Total =(1000/3)π= (1000/3)\pi cm³.

(b) Curved cyl side = 2π(5)(10)=100π2\pi(5)(10) = 100\pi. Base circle = 25π25\pi. Hemi surface = 2π(25)=50π2\pi(25) = 50\pi. Total = 175π175\pi.
9Problem 9
Answer
(a) 4 cm (b) 4104\sqrt{10} cm (c) ≈ 71.6°
Full working
(a) G is centre of square base; M is midpoint of QR (a side). GM = half-side = 4.

(b) VM=122+42=160=410VM = \sqrt{12^2 + 4^2} = \sqrt{160} = 4\sqrt{10} cm.

(c) Angle VMG: tan⁡=12/4=3\tan = 12/4 = 3. θ=tan⁡−1(3)≈71.6°\theta = \tan^{-1}(3) \approx 71.6°.
10Problem 10
Answer
(a) (160/3)π(160/3)\pi cm³ ≈ 167.6 (b) ≈ 3.35 s (c) No — glass holds 200π200\pi
Full working
(a) V=(1/3)π(16)(10)=(160/3)π≈167.6V = (1/3)\pi(16)(10) = (160/3)\pi \approx 167.6 cm³.

(b) 167.6/50≈3.35167.6/50 \approx 3.35 s.

(c) Glass cap =π(25)(8)=200π≈628= \pi(25)(8) = 200\pi \approx 628 cm³. Funnel is much smaller — no overflow.
11Problem 11
Answer
(a) 1728+144π≈2180.41728 + 144\pi \approx 2180.4 cm³ (b) Cube SA (one face replaced) + hemi curved + circle ≈ see working
Full working
(a) Cube V=123=1728V = 12^3 = 1728. Hemi V=(2/3)π(216)=144π≈452V = (2/3)\pi(216) = 144\pi \approx 452. Total ≈2180\approx 2180 cm³.

(b) Cube has 6 faces of 144144. Top face replaced by hemisphere curved 2π(36)=72π2\pi(36) = 72\pi. Plus the circular bottom of the hemisphere (visible if the cube is hollow), but assumed not exposed. SA = 5 faces of cube + curved hemi = 5(144)+72π=720+72π≈9465(144) + 72\pi = 720 + 72\pi \approx 946 cm².
12Problem 12
Answer
(a) 8 (b) 400 (c) 36
Full working
(a) Volume scale factor = 23=82^3 = 8.

(b) 50×8=40050 \times 8 = 400 cm³.

(c) SA scale factor = 22=42^2 = 4. Smaller = 144/4=36144/4 = 36 cm².