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WORKSHEET General 2 MathematicsTopic Areas:Measurement
MM2/4 – Perimeter, Area and Volume Prelim – Perimeter, Area and Volume (MM2) Areas and Volumes Simpson's Rule/Measurement Error
Teacher: PETER HARGRAVESSource: HSC exam questionsExam Equivalent Time: 79.5 minutesWorked Solutions: IncludedNote: Each question has designated marks. Use this information as both a guide to the question's difficultyand as a timing indicator, whereby each mark should equate to 1.5 minutes of working (examination) time.
Questions
1. Measurement, 2UG 2011 HSC 1 MC
Which of the solids shown is a prism?
2. Measurement, 2UG 2012 HSC 6 MC
What is the volume of this rectangular prism in cubic centimetres?
(A)
(B)
(C)
(D)
3. Measurement, 2UG 2008 HSC 2 MC
What is the surface area of the open box?
(A) cm²
(B) cm
(C) cm
(D) cm
6 cm³
600 cm³
60 000 cm³
6 000 000 cm³
10
30 2
52 2
62 2
4. Measurement, 2UG 2010 HSC 17 MC
During a flood hectares of land was covered by water to a depth of .
How many kilolitres of water covered the land? ( hectare )
(A)
(B)
(C)
(D)
5. Measurement, 2UG 2011 HSC 13 MC
The diagram represents a field.
What is the area of the field, using two applications of Simpson’s rule?
(A)
(B)
(C)
(D)
1.5 17 cm
1 = 10 000 m²
2.55 kL
2550 kL
255 000 kL
2 550 000 kL
99 m²
126 m²
198 m²
396 m²
6. Measurement, 2UG 2010 HSC 3 MC
A field diagram has been drawn from an offset survey.
What is the distance from to correct to the nearest metre?
(A)
(B)
(C)
(D)
G H
11
13
16
20
7. Measurement, 2UG 2013 HSC 12 MC
A square pyramid fits exactly on top of a cube to form a solid.
What is the volume of the solid?
(A)
(B)
(C)
(D)
513 cm³
999 cm³
1242 cm³
1539 cm³
8. Measurement, 2UG 2013 HSC 16 MC
The shaded region shows a quadrant with a rectangle removed.
What is the area of the shaded region, to the nearest cm ?
(A)
(B)
(C)
(D)
2
38 cm²
52 cm²
61 cm²
70 cm²
9. Measurement, 2UG 2008 HSC 11 MC
The diagram shows the floor of a shower. The drain in the floor is a circle with a diameter of cm.
What is the area of the shower floor, excluding the drain?
(A)
(B)
(C)
(D)
10. Measurement, 2UG 2014 HSC 10 MC
The top of the Sydney Harbour Bridge is measured to be above sea level.
What is the percentage error in this measurement?
(A)
(B)
(C)
(D)
10
9686 cm²
9921 cm²
9969 cm²
10 000 cm²
138.4 m
0.036%
0.050%
0.072%
0.289%
11. Measurement, 2UG 2014 HSC 12 MC
A path metres wide surrounds a circular lawn of radius metres.
What is the approximate area of the path?
(A)
(B)
(C)
(D)
12. Measurement, 2UG 2009 HSC 19 MC
Two identical spheres fit exactly inside a cylindrical container, as shown.
The diameter of each sphere is .
What is the volume of the cylindrical container, to the nearest cubic centimetre?
(A)
(B)
(C)
(D)
1.5 3
7.1 m²
21.2 m²
35.3 m²
56.5 m²
12 cm
1357 cm³
2714 cm³
5429 cm³
10 857 cm³
13. Measurement, 2UG 2013 HSC 19 MC
A logo is designed using half of an annulus.
What is the area of the logo, to the nearest cm²?
(A)
(B)
(C)
(D)
25 cm²
33 cm²
132 cm²
143 cm²
14. Measurement, 2UG 2013 HSC 25 MC
A net is made using four rectangles and two trapeziums. It is folded to form a solid.
What is the volume of the solid, in cm ?
(A)
(B)
(C)
(D)
3
360 cm³
434 cm³
440 cm³
576 cm³
Find the area of the terrace. (2 marks)
Tiles are sold in boxes. Each box holds one square metre of tiles and costs .When buying the tiles, more tiles are needed, due to cutting and wastage.
Find the total cost of the boxes of tiles required for the terrace. (1 mark)
15. Measurement, 2UG 2014 HSC 25 MC
A grain silo is made up of a cylinder with a hemisphere (half a sphere) on top. The outside ofthe silo is to be painted.
What is the area to be painted?
(A)
(B)
(C)
(D)
16. Measurement, 2UG 2009 HSC 23c
The diagram shows the shape and dimensions of a terrace which is to be tiled.
(i)
(ii)
8143 m²
11 762 m²
12 667 m²
23 524 m²
$5510%
Calculate the percentage error in the measurement of the longer side. (1 mark)
Between what lower and upper limits does the actual area of the top of the choppingboard lie? (2 marks)
17. Measurement, 2UG 2013 HSC 27d
A rectangular wooden chopping board is advertised as being by , with eachside measured to the nearest centimetre.
(i)
(ii)
18. Measurement, 2UG 2008 HSC 23b
The capacity of a bottle is measured as litres correct to the nearest millilitres.
What is the percentage error for this measurement? (1 mark)
17 cm 25 cm
1.25 10
19. Measurement, 2UG 2014 HSC 28d
An aerial diagram of a swimming pool is shown.
The swimming pool is a standard length of metres but is not in the shape of a rectangle.
(i) Given , determine the scale of the diagram such that
(1 mark)
(ii) If the length of a carpark next to the pool measured (not shown), how long would itbe in real life? (1 mark)
(iii) In the diagram of the swimming pool, the five widths are measured to be:
The average depth of the pool is
Calculate the approximate volume of the swimming pool, in cubic metres. In your calculations,use TWO applications of Simpson’s Rule. (3 marks)
50
AB = 8 cm
1 cm = x m
5 cm
CD = 21.88 m
EF = 25.63 m
GH = 31.88 m
IJ = 36.25 m
KL = 21.88 m
1.2 m
Use Simpson’s Rule to find the approximate area of the lake’s surface. (3 marks)
The lake is deep. Bozo the clown thinks he can empty the lake using a fourlitre bucket.How many times would he have to fill his bucket from the lake in order to empty thelake? (Note that . (2 marks)
20. Measurement, 2UG 2009 HSC 25c
There is a lake inside the rectangular grass picnic area , as shown in the diagram.
(i)
(ii)
ABCD
60 cm
1 m³ = 1000 L)
What are the dimensions of the rectangular box? (4 marks)
What is the volume of the remaining triangular prism of cheese? Answer to thenearest cubic centimetre. (2 marks)
21. Measurement, 2UG 2008 HSC 25c
Pieces of cheese are cut from cylindrical blocks with dimensions as shown.
Twelve pieces are packed in a rectangular box. There are three rows with four pieces ofcheese in each row. The curved surface is face down with the pieces touching as shown.
(i)
To save packing space, the curved section is removed.
(ii)
The area of the crosssection must be m . The tunnel is m wide.
Find an expression for the area of the crosssection using TWO applications ofSimpson’s rule. (2 marks)
If the value of a increases by m, by how much will change? (2 marks)
22. Measurement, 2UG 2014 HSC 27c
The base of a water tank is in the shape of a rectangle with a semicircle at each end, asshown.
The tank is mm long, mm wide, and has a height of mm.
What is the capacity of the tank, to the nearest litre? (4 marks)
23. Measurement, 2UG 2008 HSC 28b
A tunnel is excavated with a crosssection as shown.
(i)
(ii)
1400 560 810
600 2 80
2 b
Cans are packed in boxes that are rectangular prisms with dimensions .
What is the maximum number of cans that can be packed into one of these boxes? (1 mark)
The shaded label on the can shown wraps all the way around the can with nooverlap.What area of paper is needed to make the labels for all the cans in this box whenthe box is full? (2 marks)
The company is considering producing larger cans. Monica says if you double thediameter of the can this will double the volume. Is Monica correct? Justify youranswer with suitable calculations. (2 marks)
The company wants to produce a can with a volume of , using the leastamount of metal.
Monica is given the job of determining the dimensions of the can to be produced.She considers the following graphs.
24. Measurement, 2UG 2010 HSC 28b
Moivre’s manufacturing company produces cans of Magic Beans. The can has a diameter of and a height of .
(i)
(ii)
(iii)
(iv)
10 cm 10 cm
30 cm × 40 cm × 60 cm
1570 cm³
What radius and height should Monica recommend that the company use tominimise the amount of metal required to produce these cans? Justify your choiceof dimensions with reference to the graphs and/or suitable calculations. (2 marks)
Copyright © 200914 The State of New South Wales (Board of Studies, Teaching and Educational Standards NSW)
♦♦ Mean mark 34%NOTE: The unit conversion 1 m³= 1,000 L is contained in theFormulae and Data sheet givenout in the exam.
Worked Solutions
1. Measurement, 2UG 2011 HSC 1 MC
2. Measurement, 2UG 2012 HSC 6 MC
3. Measurement, 2UG 2008 HSC 2 MC
4. Measurement, 2UG 2010 HSC 17 MC
⇒ D
V = l × b × h
= 30 × 5 × 400
= 60 000 cm³⇒ C
Surface Area
= 2 × (3 × 5) + 2 × (2 × 3) + (5 × 2)
= 30 + 12 + 10
= 52 cm²
⇒ C
Area covered = 1.5 × 10 000
= 15 000 m²
Volume = Ah
= 15 000 × 0.17
= 2550 m³
Since 1 m³ = 1000 L = 1 kL
Volume = 2550 kL⇒ B
5. Measurement, 2UG 2011 HSC 13 MC
6. Measurement, 2UG 2010 HSC 3 MC
7. Measurement, 2UG 2013 HSC 12 MC
A = ( + 4 + ) ... applied twiceh
3y0 y1 y2
= (5 + 4 × 7 + 12) + (12 + 4 × 8 + 10)33
33
= 45 + 54
= 99 m²⇒ A
Using Pythagoras
GH2 = +122 (16 − 11)2
= 144 + 25
= 169
∴ GH = (GH > 0)169− −−√
= 13m⇒ B
Volume = Vol (cube) + Vol (pyramid)
= + Ahl3 13
= (9 × 9 × 9) + ( × 9 × 9 × 10)13
= 999 cm³⇒ B
COMMENT: Students shouldsee that answers are all in cm²,and therefore should convert allmeasurements into cm forcalculations.
♦ Mean mark 48%
8. Measurement, 2UG 2013 HSC 16 MC
9. Measurement, 2UG 2008 HSC 11 MC
10. Measurement, 2UG 2014 HSC 10 MC
11. Measurement, 2UG 2014 HSC 12 MC
Shaded area = Area segment − Area rectangle
= π − (6 × 2)14
r2
= π × − 1214
92
= 51.617... cm²⇒ B
Area = Square - Circle
= (100 × 100) − π ⋅ 52
= 10 000 − 78.5398...
= 9921.46... cm²⇒ B
Absolute error = ± 0.05 m
% error = × 1000.05
138.4= 0.036 %
⇒ A
Area of annulus
= π( − )R2 r2
= π( − )4.52 32
= π(11.25)
= 35.3 m² (1 d.p.)
⇒ C
♦♦ Mean mark 35%COMMENT: Note that in theArea formula is different to the used in the Volume formula.
12. Measurement, 2UG 2009 HSC 19 MC
13. Measurement, 2UG 2013 HSC 19 MC
14. Measurement, 2UG 2013 HSC 25 MC
Since diameter sphere = 12cm
⇒ Radius of cylinder = 6cm
Height of cylinder = 2 × diameter of sphere
= 2 × 12
= 24cm
Volume cylinder = π hr2
= π × × 2462
= 2714.336... cm³
⇒ B
Area = × π( − )12
R2 r2
= × π( − )12
52 22
= π212
= 33 cm² (nearest cm²)⇒ B
Volume = Ah where A is the area of a trapezium
hh
A = h(a + b)12
= × 8(11 + 5)12
= 64 cm²
∴ V = Ah = 64 × 9 = 576 cm³⇒ D
♦ Mean mark 40%
15. Measurement, 2UG 2014 HSC 25 MC
Total Area = Area of cylinder + ½ sphere
Area of cylinder = 2πrh
= 2π × 24 × 30
= 4523.9
Area of ½ sphere = × 4π12
r2
= × 4π ×12
242
= 3619.1
∴ Total area = 4523.9 × 3619.1
= 8143 m²⇒ A
16. Measurement, 2UG 2009 HSC 23c
(i)
(ii)
Area = Area of big square – Area of 2 cut-out squares
= (2.7 + 1.8) × (2.7 + 1.8) − 2 × (1.8 × 1.8)
= 20.25 − 6.48
= 13.77 m²
Tiles required = (13.77 + 10 % ) × 13.77
= 15.147 m²∴ 16 boxes are needed
Cost of boxes = 16 × $55
= $880
♦♦ Mean mark 23%MARKER'S COMMENT: Beaware that measurementsaccurate to the nearest cm havean absolute error forcalculation purposes of 0.5 cm.
♦ Mean mark 35%
17. Measurement, 2UG 2013 HSC 27d
(i)
(ii)
18. Measurement, 2UG 2008 HSC 23b
Longer side = 25 cm
Absolute error = ± 0.5 cm
% error = ±0.525
× 100
= ± 2 %
Area = l × b
Area (upper) = 25.5 × 17.5
= 446.25 cm²
Area (lower) = 24.5 × 16.5
= 404.25 cm²
∴ Area is between 404.25 cm² and 446.25 cm².
Absolute error = 5 mL
∴ % error = × 1005
1250= 0.4 %
♦ Mean mark 50%. Be carefulnot to give away easy marks!
19. Measurement, 2UG 2014 HSC 28d
(i)
(ii)
(iii)
8 cm = 50 m
1 cm =508
= 6.25 m∴ x = 6.25 m
Using scale from (i)
5 cm = 5 × 6.25
= 31.25 m∴ The carpark would be 31.25 m long
h = = 12.5 m504
A ≈ [ + 4( ) + ] ... applied twiceh
3y0 y1 y2
≈ [21.88 + 4(25.63) + 31.88]12.5
3
+ [31.88 + 4(36.25) + 21.88]12.5
3
≈ [156.28] + [198.76]12.5
312.5
3
≈ 1, 479.33 m²
V = Ah
≈ 1479.33 × 1.2
≈ 1775.2
≈ 1775 m³
♦ Mean mark 44%STRATEGY: Most students whodid calculations in and made errors. Keepingcalculations in metres is mucheasier here.
20. Measurement, 2UG 2009 HSC 25c
(i)
(ii)
21. Measurement, 2UG 2008 HSC 25c
(i)
Area of lake = Area of rectangle − Area of grass
Area of rectangle = 24 × 55
= 1320 m²
Area of grass ≈ [ + 4 + ] ... applied twiceh
3y0 y1 y2
≈ [20 + 4 × 5 + 10] + [35 + 4 × 22 + 30]123
123
≈ 4[50] + 4[153]
≈ 200 + 612
≈ 812 m²
∴ Area of lake ≈ 1320 − 812
≈ 508 m²
cm² cm³
V = Ah
= 508 × 0.6
= 304.8 m³
= 304 800 L (since 1 m³ = 1000 L)
# Times to fill bucket =304 800
4= 76 200
∴ Bozo would have to fill his bucket
76 200 times to empty the lake.
Box height = 15 cm
(radius of the arc)
Box width = 3 × 7
= 21 cm
Box length = 4x
(ii)
Using cosine rule
c2 = + − 2ab cos Ca2 b2
⇒ x2 = + − 2 × 15 × 15 ×152 152 cos 40∘
= 450 − 344.7199...
= 105.2800...
x = 10.2606...
⇒ Box length = 4 × 10.2606...
= 41.04...
∴ Dimensions are 41 cm × 21 cm × 15 cm
Volume = Ah
h = 7 cm
Need to find A
Using A = ab sin C12
⇒ A = × 15 × 15 ×12
sin 40∘
= 72.3136...
∴ V = 72.3136... × 7
= 506.195...
= 506 cm³ (nearest whole)
♦ Mean mark 41%STRATEGY: Another examplewhere adjusting measurementsto metres makes the finalconversion to litres simple.
22. Measurement, 2UG 2014 HSC 27c
V = Ah
Finding Area of base
Semi-circles have radius 280mm = 0.28 m
∴ Area of 2 semicircles
= 2 × × π12
r2
= π × (0.28)2
= 0.2463... m²
Area of rectangle
= l × b
= (1.4 − 2 × 0.28) × 0.56
= 0.4704 m²
∴ Volume = Ah
= (0.2463... + 0.4704) × 0.810
= 0.58077... m³
= 580.77... L (using 1m³ = 1000 L)
= 581 L (nearest L)
♦♦ Mean mark 27%
♦ Mean mark 38%MARKER'S COMMENT: Manystudents didn't account for theclearance of 0.5 cm at the topand bottom of each can.
♦ Mean mark 44%MARKER'S COMMENT: Manystudents performed calculationsin this part without concluding if
23. Measurement, 2UG 2008 HSC 28b
(i)
(ii)
24. Measurement, 2UG 2010 HSC 28b
(i)
(ii)
(iii)
A ≈ [ + 4 + ] + [ + 4 + ]h
3y0 y1 y2
h
3y0 y1 y2
≈ [0 + 4a + b] + [b + 4a + 0]h
3h
3
≈ (4a + b)2h
3
Given A = 600 m²
If 80 m wide ⇒ h = 20
Using A = (4a + b)2h
3
600 = (4a + b)(2 × 20)
3
4a + b =600 × 3
40b = 45 − 4a
∴ If a ↑ by 2 m
⇒ b ↓ by 8 m
Maximum # Cans = 4 × 3 × 6
= 72 cans
Label Area (1 can) = 2πrh
= 2 × π × 5 × 9
= 90π
= 282.7433... cm²
∴ Label Area (72 cans) = 72 × 282.7433...
= 20 357.52...
= 20 358 cm² (nearest cm²)
Original volume = π hr2
= π × × 1052
= 785.398... cm³
Monica is correct or not. Readthe question carefully.
♦♦ Mean mark 26%
(iv)
Copyright © 2015 M2 Mathematics Pty Ltd (SmarterMaths.com.au)
= 785.398... cm³
If the diameter doubles, radius = 10cm
New volume = π × × 10102
= 3141.592... cm³
∴ Monica is incorrect because the volume
doesn't double. It increases by a factor of 4.
Minimum metal used when surface area is a minimum.
From graph, minimum surface area when r = 6.3 cm
When r = 6.3 cm, h = 12.6 cm (from graph)
∴ She should recommend radius 6.3 cm and height 12.6 cm
