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 Spheres 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

24 MC 7 SA ~50 min

Beyond the syllabus. This is the legacy 22-lesson Measurement topic test, not an Applications of measurement (MST-11-05) paper, and four of its lessons are trigonometry that belongs to Year 12: L14 and L15 finding sides and angles, L16 angle of elevation and depression, and L17 Bearings. You will not be assessed on any of it this year, so skip those questions rather than losing time on them. Everything else here is examinable and is the reason to sit this quiz: units and formulas, perimeter and arc length, area and composite shapes, surface area, volume, rates, scale, measurement error, energy and mass, the trapezoidal rule, and the time and location questions at the end.

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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 $7.07\text{ cm}$

B $14.14\text{ cm}$

C $5.00\text{ cm}$

D $6.32\text{ cm}$

A, $7.07\text{ cm}$. 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 $\dfrac{\text{opposite}}{\text{adjacent}}$

D $\dfrac{\text{adjacent}}{\text{hypotenuse}}$

D, $\dfrac{\text{adjacent}}{\text{hypotenuse}}$. SOH-CAH-TOA: C os = A djacent / H ypotenuse.

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 $270°$

D $180°$

D, $180°$. $A = \dfrac{\theta}{360°} \times \pi r^2 \Rightarrow 18\pi = \dfrac{\theta}{360°} \times 36\pi \Rightarrow \theta = \dfrac{18 \times 360°}{36} = 180°$.

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 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 $282\,743\text{ L}$

C $471\,239\text{ L}$

D $141\,372\text{ L}$

D, $141\,372\text{ L}$. $V = \pi r^2 h = \pi(9)(5) = 45\pi \approx 141.372\text{ m}^3 = 141\,372\text{ L}$.

9 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, $452.39\text{ cm}^3$. $V = \tfrac{1}{2} \times \tfrac{4}{3}\pi r^3 = \tfrac{2}{3}\pi(216) = 144\pi \approx 452.39\text{ cm}^3$.

10 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 $= \dfrac{27\,000}{15} = 1800\text{ min} = \dfrac{1800}{60} = 30\text{ hours}$.

11 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, $0.8\text{ km}$. $3.2 \times 25\,000 = 80\,000\text{ cm} = 800\text{ m} = 0.8\text{ km}$.

12 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 $7.55\text{ m} \leq x < 7.65\text{ m}$

B $7.5\text{ m} \leq x < 7.7\text{ m}$

C $7.1\text{ m} \leq x < 8.1\text{ m}$

D $7.56\text{ m} \leq x < 7.64\text{ m}$

A, $7.55\text{ m} \leq x < 7.65\text{ m}$. Lower bound $= 7.6 - 0.05 = 7.55\text{ m}$; Upper bound $= 7.6 + 0.05 = 7.65\text{ m}$.

13 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, $3.21\text{ m}$. $\tan 15° = \dfrac{h}{12} \Rightarrow h = 12 \times \tan 15° \approx 12 \times 0.2679 \approx 3.21\text{ m}$.

14 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, $51°47'$. $\sin\theta = \dfrac{11}{14} \Rightarrow \theta = \sin^{-1}(0.7857) \approx 51.788° = 51° + 0.788 \times 60' \approx 51°47'$.

15 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, $50.00\text{ m}$. $\tan 32° = \dfrac{h}{80} \Rightarrow h = 80 \times \tan 32° \approx 80 \times 0.6249 \approx 50.00\text{ m}$.

16 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 $138.5\text{ m}$

B $47.3\text{ m}$

C $14.6\text{ m}$

D $146.4\text{ m}$

A, $138.5\text{ m}$. The angle of depression equals the angle of elevation from the boat. $\tan 18° = \dfrac{45}{d} \Rightarrow d = \dfrac{45}{\tan 18°} \approx \dfrac{45}{0.3249} \approx 138.5\text{ m}$.

17 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 $290°$

D $250°$

D, $250°$. Since the bearing is less than $180°$, back bearing $= 070° + 180° = 250°$.

18 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.

19 Point C is on a bearing of $040°$ from point A. What is the bearing from C back to A? L17

A $040°$

B $220°$

C $140°$

D $310°$

B, $220°$. $040° < 180°$, so back bearing $= 040° + 180° = 220°$.

20 A 2.5 kW air conditioner runs for 4 hours per day. How much energy does it use in one day? L18

A $10\text{ kWh}$

B $6.5\text{ kWh}$

C $2.5\text{ kWh}$

D $0.625\text{ kWh}$

A, $10\text{ kWh}$. $E = P \times t = 2.5\text{ kW} \times 4\text{ h} = 10\text{ kWh}$.

21 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, $33\text{ m}^2$. $A \approx \dfrac{h}{2}(d_f + 2d_m + d_l) = \dfrac{3}{2}(4 + 14 + 4) = \dfrac{3}{2} \times 22 = 33\text{ m}^2$.

22 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 $14\text{ h } 27\text{ min}$

C $3\text{ h } 33\text{ min}$

D $2\text{ h } 27\text{ min}$

D, $2\text{ h } 27\text{ min}$. Count up: 11:48 pm → midnight = 12 min; midnight → 2:15 am = 2 h 15 min. Total $= 12 + 135 = 147\text{ min} = 2\text{ h } 27\text{ min}$.

23 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 $= 8{:}00\text{ pm} - 6\text{ h} = 2{:}00\text{ pm}$.

24 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 $12\text{ hours}$

D $8\text{ hours}$

D, $8\text{ hours}$. $\Delta\lambda = 150° - 30° = 120°$. 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.

(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.

(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.

(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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