Module 2 Topic Test

Measurement

Maths Standard Year 11 · All 22 lessons · MC checkpoint plus separate short-answer practice

L1 — Formulas & Units L2 — Area L3 — Pythagoras L4 — Intro Trig L5 — Perimeter & Arc L6 — Sectors & Annuli L7 — SA Prisms L8 — SA Pyramids L9 — Vol Prisms L10 — Vol Pyramids L11 — Rates L12 — Scale L13 — Errors L14 — Trig Sides L15 — Trig Angles L16 — Elevation & Depression L17 — Bearings L18 — Energy & Mass L19 — Trapezoidal Rule L20 — Timetables L21 — Time Zones L22 — Lat & Long
25 MC 7 SA ~50 min
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Complete the 25 multiple choice questions to unlock a sharper next move. The short-answer section below is separate practice.

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Part A — Multiple Choice (1 mark each, 25 marks total)
1 Which of the following is equivalent to $3.2\text{ km}^2$ expressed in $\text{m}^2$? L1
A $3\,200\text{ m}^2$
B $320\,000\text{ m}^2$
C $3\,200\,000\text{ m}^2$
D $32\,000\text{ m}^2$
C — $3\,200\,000\text{ m}^2$. $1\text{ km} = 1000\text{ m}$, so $1\text{ km}^2 = 1000^2 = 1\,000\,000\text{ m}^2$. Therefore $3.2 \times 1\,000\,000 = 3\,200\,000\text{ m}^2$.
2 A rhombus has diagonals of $10\text{ cm}$ and $8\text{ cm}$. What is its area? L2
A $80\text{ cm}^2$
B $40\text{ cm}^2$
C $20\text{ cm}^2$
D $160\text{ cm}^2$
B — $40\text{ cm}^2$. $A = \tfrac{1}{2} \times d_1 \times d_2 = \tfrac{1}{2} \times 10 \times 8 = 40\text{ cm}^2$.
3 The diagonal of a square measures $10\text{ cm}$. What is the side length of the square, to 2 decimal places? L3
A $5.00\text{ cm}$
B $14.14\text{ cm}$
C $7.07\text{ cm}$
D $6.32\text{ cm}$
C — $7.07\text{ cm}$. The diagonal $d = s\sqrt{2}$, so $s = \dfrac{d}{\sqrt{2}} = \dfrac{10}{\sqrt{2}} = 5\sqrt{2} \approx 7.07\text{ cm}$.
4 In a right-angled triangle, which ratio defines the cosine of an angle? L4
A $\dfrac{\text{opposite}}{\text{hypotenuse}}$
B $\dfrac{\text{adjacent}}{\text{opposite}}$
C $\dfrac{\text{adjacent}}{\text{hypotenuse}}$
D $\dfrac{\text{opposite}}{\text{adjacent}}$
C — $\dfrac{\text{adjacent}}{\text{hypotenuse}}$. SOH-CAH-TOA: Cos = Adjacent / Hypotenuse.
5 A circle has circumference $40\pi\text{ cm}$. What is its radius? L5
A $20\text{ cm}$
B $40\text{ cm}$
C $10\text{ cm}$
D $80\text{ cm}$
A — $20\text{ cm}$. $C = 2\pi r \Rightarrow 40\pi = 2\pi r \Rightarrow r = 20\text{ cm}$.
6 A sector has area $18\pi\text{ cm}^2$ and radius $6\text{ cm}$. What is the central angle of the sector? L6
A $90°$
B $120°$
C $180°$
D $270°$
C — $180°$. $A = \dfrac{\theta}{360°} \times \pi r^2 \Rightarrow 18\pi = \dfrac{\theta}{360°} \times 36\pi \Rightarrow \theta = \dfrac{18 \times 360°}{36} = 180°$.
7 A closed cylinder has radius $2\text{ cm}$ and height $7\text{ cm}$. What is its total surface area, to 2 decimal places? L7
A $56.55\text{ cm}^2$
B $87.96\text{ cm}^2$
C $113.10\text{ cm}^2$
D $226.19\text{ cm}^2$
C — $113.10\text{ cm}^2$. $SA = 2\pi r^2 + 2\pi r h = 2\pi(4) + 2\pi(2)(7) = 8\pi + 28\pi = 36\pi \approx 113.10\text{ cm}^2$.
8 A square pyramid has a base of $6\text{ cm} \times 6\text{ cm}$ and slant height $5\text{ cm}$. What is its total surface area? L8
A $60\text{ cm}^2$
B $96\text{ cm}^2$
C $156\text{ cm}^2$
D $120\text{ cm}^2$
B — $96\text{ cm}^2$. Base $= 36\text{ cm}^2$. Four triangular faces: $4 \times \tfrac{1}{2} \times 6 \times 5 = 60\text{ cm}^2$. Total $= 36 + 60 = 96\text{ cm}^2$.
9 A cylindrical tank has radius $3\text{ m}$ and height $5\text{ m}$. How many litres of water can it hold? (Recall: $1\text{ m}^3 = 1000\text{ L}$, answer to nearest whole litre.) L9
A $45\,000\text{ L}$
B $141\,372\text{ L}$
C $471\,239\text{ L}$
D $282\,743\text{ L}$
B — $141\,372\text{ L}$. $V = \pi r^2 h = \pi(9)(5) = 45\pi \approx 141.372\text{ m}^3 = 141\,372\text{ L}$.
10 A hemisphere has radius $6\text{ cm}$. What is its volume, to 2 decimal places? L10
A $904.78\text{ cm}^3$
B $226.19\text{ cm}^3$
C $452.39\text{ cm}^3$
D $113.10\text{ cm}^3$
C — $452.39\text{ cm}^3$. $V = \tfrac{1}{2} \times \tfrac{4}{3}\pi r^3 = \tfrac{2}{3}\pi(216) = 144\pi \approx 452.39\text{ cm}^3$.
11 Water flows into a tank at $15\text{ L/min}$. How many hours does it take to fill a $27\,000\text{ L}$ tank? L11
A $18\text{ hours}$
B $30\text{ hours}$
C $45\text{ hours}$
D $1800\text{ hours}$
B — $30\text{ hours}$. Time $= \dfrac{27\,000}{15} = 1800\text{ min} = \dfrac{1800}{60} = 30\text{ hours}$.
12 On a map with scale $1:25\,000$, a road appears as $3.2\text{ cm}$. What is the actual road length in kilometres? L12
A $8\text{ km}$
B $0.08\text{ km}$
C $80\text{ km}$
D $0.8\text{ km}$
D — $0.8\text{ km}$. $3.2 \times 25\,000 = 80\,000\text{ cm} = 800\text{ m} = 0.8\text{ km}$.
13 A length is recorded as $7.6\text{ m}$ with an absolute error of $0.05\text{ m}$. Which interval gives the limits of accuracy? L13
A $7.5\text{ m} \leq x < 7.7\text{ m}$
B $7.55\text{ m} \leq x < 7.65\text{ m}$
C $7.1\text{ m} \leq x < 8.1\text{ m}$
D $7.56\text{ m} \leq x < 7.64\text{ m}$
B — $7.55\text{ m} \leq x < 7.65\text{ m}$. Lower bound $= 7.6 - 0.05 = 7.55\text{ m}$; Upper bound $= 7.6 + 0.05 = 7.65\text{ m}$.
14 A ramp is inclined at $15°$ to the horizontal and has a horizontal run of $12\text{ m}$. How high does the ramp rise, to 2 decimal places? L14
A $11.59\text{ m}$
B $3.21\text{ m}$
C $46.39\text{ m}$
D $12.42\text{ m}$
B — $3.21\text{ m}$. $\tan 15° = \dfrac{h}{12} \Rightarrow h = 12 \times \tan 15° \approx 12 \times 0.2679 \approx 3.21\text{ m}$.
15 In a right-angled triangle, the side opposite $\theta$ is $11\text{ cm}$ and the hypotenuse is $14\text{ cm}$. Find $\theta$ to the nearest minute. L15
A $38°13'$
B $51°47'$
C $52°13'$
D $38°47'$
B — $51°47'$. $\sin\theta = \dfrac{11}{14} \Rightarrow \theta = \sin^{-1}(0.7857) \approx 51.788° = 51° + 0.788 \times 60' \approx 51°47'$.
16 An observer $80\text{ m}$ from the base of a cliff looks up at an angle of elevation of $32°$. How high is the cliff, to 2 decimal places? L16
A $67.82\text{ m}$
B $42.38\text{ m}$
C $50.00\text{ m}$
D $100.48\text{ m}$
C — $50.00\text{ m}$. $\tan 32° = \dfrac{h}{80} \Rightarrow h = 80 \times \tan 32° \approx 80 \times 0.6249 \approx 50.00\text{ m}$.
17 From the top of a $45\text{ m}$ lighthouse, the angle of depression to a boat is $18°$. How far is the boat from the base, to 1 decimal place? L16
A $14.6\text{ m}$
B $47.3\text{ m}$
C $138.5\text{ m}$
D $146.4\text{ m}$
C — $138.5\text{ m}$. The angle of depression equals the angle of elevation from the boat. $\tan 18° = \dfrac{45}{d} \Rightarrow d = \dfrac{45}{\tan 18°} \approx \dfrac{45}{0.3249} \approx 138.5\text{ m}$.
18 A ship sails on a bearing of $070°$. What is the back bearing (the bearing back to the starting point)? L17
A $070°$
B $110°$
C $250°$
D $290°$
C — $250°$. Since the bearing is less than $180°$, back bearing $= 070° + 180° = 250°$.
19 A bushwalker travels N$35°$E for $8\text{ km}$. How far north has she travelled, to 2 decimal places? L17
A $4.59\text{ km}$
B $9.76\text{ km}$
C $6.55\text{ km}$
D $13.95\text{ km}$
C — $6.55\text{ km}$. N-component $= 8\cos 35° \approx 8 \times 0.8192 \approx 6.55\text{ km}$ north.
20 Point C is on a bearing of $040°$ from point A. What is the bearing from C back to A? L17
A $040°$
B $140°$
C $220°$
D $310°$
C — $220°$. $040° < 180°$, so back bearing $= 040° + 180° = 220°$.
21 A 2.5 kW air conditioner runs for 4 hours per day. How much energy does it use in one day? L18
A $2.5\text{ kWh}$
B $6.5\text{ kWh}$
C $10\text{ kWh}$
D $0.625\text{ kWh}$
C — $10\text{ kWh}$. $E = P \times t = 2.5\text{ kW} \times 4\text{ h} = 10\text{ kWh}$.
22 The trapezoidal rule is applied to a shape with $h = 3\text{ m}$, first width $d_f = 4\text{ m}$, middle width $d_m = 7\text{ m}$, and last width $d_l = 4\text{ m}$. What is the estimated area? L19
A $24\text{ m}^2$
B $33\text{ m}^2$
C $36\text{ m}^2$
D $22\text{ m}^2$
B — $33\text{ m}^2$. $A \approx \dfrac{h}{2}(d_f + 2d_m + d_l) = \dfrac{3}{2}(4 + 14 + 4) = \dfrac{3}{2} \times 22 = 33\text{ m}^2$.
23 A train departs at $11{:}48\text{ pm}$ and arrives at $2{:}15\text{ am}$ the following morning. What is the journey time? L20
A $2\text{ h } 15\text{ min}$
B $2\text{ h } 27\text{ min}$
C $3\text{ h } 33\text{ min}$
D $14\text{ h } 27\text{ min}$
B — $2\text{ h } 27\text{ min}$. Count up: 11:48 pm → midnight = 12 min; midnight → 2:15 am = 2 h 15 min. Total $= 12 + 135 = 147\text{ min} = 2\text{ h } 27\text{ min}$.
24 When it is $8{:}00\text{ pm}$ in Sydney (AEST, UTC$+10$), what time is it in Dubai (UTC$+4$)? L21
A $2{:}00\text{ am}$
B $10{:}00\text{ pm}$
C $2{:}00\text{ pm}$
D $6{:}00\text{ pm}$
C — $2{:}00\text{ pm}$. Sydney is 6 hours ahead of Dubai (UTC$+10$ vs UTC$+4$). So Dubai time $= 8{:}00\text{ pm} - 6\text{ h} = 2{:}00\text{ pm}$.
25 City A is at longitude $150°\text{E}$ and City B is at longitude $30°\text{E}$. Using the rule that Earth rotates $15°$ per hour, what is the time difference between the two cities? L22
A $6\text{ hours}$
B $4\text{ hours}$
C $8\text{ hours}$
D $12\text{ hours}$
C — $8\text{ hours}$. $\Delta\lambda = 150° - 30° = 120°$. Time difference $= 120° \div 15°/\text{h} = 8\text{ hours}$. City A is 8 hours ahead of City B.
Part B — Short Answer (show all working)
1 L5 & L6
A circular swimming pool has diameter $8\text{ m}$.
(a) Find the circumference of the pool, to 2 decimal places.
(b) Find the area of the pool, to 2 decimal places.
(c) A circular fence is built $1\text{ m}$ outside the pool. Find the area of the annular gap between pool edge and fence, to 2 decimal places.
(a) $r = 4\text{ m}$. $C = 2\pi r = 2\pi(4) = 8\pi \approx 25.13\text{ m}$
(b) $A = \pi r^2 = \pi(16) = 16\pi \approx 50.27\text{ m}^2$
(c) Outer radius $= 4 + 1 = 5\text{ m}$. Annulus area $= \pi(5^2 - 4^2) = \pi(25 - 16) = 9\pi \approx 28.27\text{ m}^2$
2 L7 & L9
A cylindrical water storage tank has radius $2.5\text{ m}$ and height $8\text{ m}$. The tank is closed at top and bottom.
(a) Find the total surface area of the tank, to 2 decimal places.
(b) Find the volume of the tank in $\text{m}^3$, to 2 decimal places.
(c) How many litres of water does the tank hold?
(a) $SA = 2\pi r^2 + 2\pi r h = 2\pi(6.25) + 2\pi(2.5)(8) = 12.5\pi + 40\pi = 52.5\pi \approx 164.93\text{ m}^2$
(b) $V = \pi r^2 h = \pi(6.25)(8) = 50\pi \approx 157.08\text{ m}^3$
(c) $157.08\text{ m}^3 \times 1000 = 157\,080\text{ L}$ (or $50\,000\pi \approx 157\,080\text{ L}$)
3 L11 & L13
A car's speedometer reads $65\text{ km/h}$ with an absolute error of $2\text{ km/h}$.
(a) Write the car's speed as an interval (upper and lower bounds).
(b) Find the percentage error of the speedometer reading, to 2 decimal places.
(c) How far would the car travel in $2.5$ hours at exactly $65\text{ km/h}$?
(a) $63\text{ km/h} \leq v < 67\text{ km/h}$
(b) $\%\text{ error} = \dfrac{2}{65} \times 100 \approx 3.08\%$
(c) $D = S \times T = 65 \times 2.5 = 162.5\text{ km}$
4 L14 & L16
From point A on level ground, the angle of elevation to the top T of a building is $42°$. From point B, which is $20\text{ m}$ closer to the building than A, the angle of elevation to T is $58°$. Find the height of the building, to 1 decimal place.
(a) Let $d$ = horizontal distance from B to the building. Write two equations for $h$ (the height) in terms of $d$.
(b) Solve for $d$, then find $h$.
(a) From A (distance $d + 20$ from building): $h = (d+20)\tan 42°$
From B (distance $d$ from building): $h = d\tan 58°$
(b) Setting equal: $(d+20)\tan 42° = d\tan 58°$
$d \cdot 0.9004 + 20 \times 0.9004 = d \times 1.6003$
$18.008 = d(1.6003 - 0.9004) = 0.6999d$
$d \approx 25.73\text{ m}$
$h = 25.73 \times \tan 58° \approx 25.73 \times 1.6003 \approx 41.2\text{ m}$
5 L17
A ship leaves port P and sails $60\text{ km}$ on a bearing of $030°$ to point Q. It then sails $40\text{ km}$ on a bearing of $120°$ to point R.
(a) Find the eastward and northward displacements for the leg P → Q.
(b) Find the eastward and northward displacements for the leg Q → R.
(c) Find the straight-line distance PR, to 1 decimal place.
(a) P → Q (bearing 030°):
East: $60\sin 30° = 60 \times 0.5 = 30\text{ km}$
North: $60\cos 30° = 60 \times 0.8660 = 51.96\text{ km}$
(b) Q → R (bearing 120°):
East: $40\sin 120° = 40\sin 60° = 40 \times 0.8660 = 34.64\text{ km}$
North: $40\cos 120° = 40 \times (-0.5) = -20\text{ km}$ (i.e. $20\text{ km}$ south)
(c) Total east $= 30 + 34.64 = 64.64\text{ km}$; Total north $= 51.96 - 20 = 31.96\text{ km}$
$PR = \sqrt{64.64^2 + 31.96^2} = \sqrt{4178.3 + 1021.4} = \sqrt{5199.7} \approx 72.1\text{ km}$
6 L18 & L19
A farmer uses a 3.5 kW water pump to irrigate a paddock. The pump runs for 4 hours each day for 7 days.
(a) Calculate the total energy used by the pump over the 7 days, in kWh.
(b) Electricity costs $0.25 per kWh. How much does the irrigation cost for the week?
(c) The paddock has an irregular shape. Five cross-sectional widths are measured 8 m apart: 6 m, 10 m, 14 m, 12 m, 8 m. Use the trapezoidal rule to estimate the paddock's area.
(a) Total time $= 4 \times 7 = 28\text{ h}$. $E = P \times t = 3.5 \times 28 = 98\text{ kWh}$
(b) Cost $= 98 \times \$0.25 = \$24.50$
(c) 5 measurements → 4 strips, $h = 8\text{ m}$.
$A \approx \dfrac{8}{2}(6 + 2 \times 10 + 2 \times 14 + 2 \times 12 + 8) = 4(6 + 20 + 28 + 24 + 8) = 4 \times 86 = 344\text{ m}^2$
7 L20, L21 & L22
City A is at longitude $120°\text{E}$ (UTC$+8$) and City B is at longitude $165°\text{E}$ (UTC$+11$). A train timetable for City B shows: depart 07:42, arrive Central 10:18, arrive Eastport 12:05.
(a) How long is the train journey from the departure station to Eastport?
(b) A video call is scheduled for 09:00 in City A (UTC$+8$). What time is it in City B (UTC$+11$) at this moment?
(c) Using the longitude rule ($15°$ per hour), calculate the theoretical time difference between City A and City B. Does it agree with the UTC offset difference?
(a) Depart 07:42, arrive Eastport 12:05.
$12{:}05 - 07{:}42 = 4\text{ h } 23\text{ min}$
(b) City B is UTC$+11$, City A is UTC$+8$ — City B is 3 hours ahead.
$09{:}00 + 3\text{ h} = 12{:}00$ noon in City B.
(c) $\Delta\lambda = 165° - 120° = 45°$. Time difference $= 45 \div 15 = 3\text{ hours}$.
UTC offset difference: $11 - 8 = 3\text{ hours}$. Yes — they agree. City B is 3 hours ahead of City A.
Module 2 Complete

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