Further applications of area and volume Test A
Name: ___________________________
Section A Multiple Choice
1
The area of the sector drawn below is
given by:
A
B
C
D
2
B
3
D
The area of the shape is:
A (130 000 + 10 000π) mm²
B (160 000 + 10 000π) mm²
C (160 000 + 5000π) mm²
D (130 000 + 5000π) mm²
275
   32 2
360
85
A
   32 2
360
85
A
   32
360
85
A
 2  32
360
A
Consider the annulus drawn below.
Examine the following figure.
4
The area of a 1-m wide gravel path
around a rectangular lawn 30 m  15 m
is:
A 49 m2
B 94 m2
C 450 m2
D 544 m2
B
B
The area of the annulus can be found
using:
A
B
C
D
A = 2(2.1 – 1.3)
A = (2.12 – 1.32)
A = 2(4.2 – 2.6)
A = (4.22 – 2.62)
Maths Quest General Mathematics HSC Course
Chapter 2 Further applications of area and volume Test A
1
5
Consider the area drawn below.
C
8
The volume of a sphere with a diameter A
of 15 cm is closest to:
A 560 π cm3
B 900 π cm3
C 4500 π cm3
D 36 000 π cm3
9
The circle drawn below has its radius
measured correct to 1 decimal place.
D
Which of the following calculations
correctly approximates the area using
Simpson’s rule?
A
B
C
D
6
7
24
(39  16)
3
48
A
(39  16)
3
24
A
(4  39  16)
3
48
A
(4  39  16)
3
A
The maximum percentage error in
calculating the area of the circle is:
The total surface area of a closed
cylinder of radius 40 cm and height
20 cm is given by:
A 2    202 + 2    20  20
B 2    402 + 2    40  10
C 2    802 + 2    80  20
D 2    402 + 2    40  20
D
The surface area of a sphere with a
diameter of 60 cm can be found using:
A
A
B
C
D
SA = 4    302
SA = 4    602
4
SA =    303
3
4
SA =    603
3
Maths Quest General Mathematics HSC Course
A
B
C
D
10
0.0191%
0.0193%
1.91%
1.93%
The sphere drawn below has its radius
measured to the nearest centimetre.
B
The maximum percentage error in
calculating the surface area of the
sphere is:
A 6.5%
B 6.8%
C 9.7%
D 10.3%
Chapter 2 Further applications of area and volume Test A
2
Further applications of area and volume Test A
Name: ___________________________
Section B Short/Extended answer
1
Find the area of the shaded sector shown to the
nearest cm2.
2
A circular lake in a botanical garden has a
radius of 30 m. The lake has a small island
within it of radius 5 m as shown in the figure
below.
300
  9 2
360
= 212 cm2
Area of sector =
4
(a) Find the area of the island to 1 decimal
place.
(a) A =  r2
=   52
= 78.5 m2
(b) Calculate the area of the lake in square
metres correct to 1 decimal place.
(b) Area of outer circle =  r2
=   302
= 2827.4 m2
Lake area = 2827.4 – 78.5
= 2748.9 m2
Maths Quest General Mathematics HSC Course
2
Chapter 2 Further applications of area and volume Test A
3
3
4
A tree trunk has its cross-section in the shape of
an ellipse as shown in the figure drawn below.
4
(a) Calculate the area of the tree trunk correct
to the nearest square centimetre.
(a) A = ab
=   50  40
= 6283 cm2
(b) A section of the trunk 2 m long is cut off.
Calculate the volume of the trunk in cubic
metres.
(b) V = A  h
= 6283  200
= 1 256 600 cm3
= 1.2566 m2
Find the area of the following figures.
(a)
(b)
6
(a) Combining the ends makes a circle.
A =  r2
=   62
= 113.1 m2
Area of rectangle
A=lb
= 12  14
= 168 m2
Total area = 113.1 + 168
= 281.1 m2
(b) The figure is a trapezium.
1
A = (a + b) h
2
1
= (5 + 17)  10.8
2
= 118.8 m²
Maths Quest General Mathematics HSC Course
Chapter 2 Further applications of area and volume Test A
4
5
Find the area of the shapes below, correct to 1
decimal place.
(a)
(b)
6
(a)
A = area of square + area of semicircle
A = 52 + 1   2.52
2
A = 34.8 m2
(b)
Hexagon contains 6 equilateral triangles.
Height of triangle : a 2  c 2  b 2
h 2  3 2  1.5 2
 9  2.25
 6.75
h  6.75
 2.6 cm
A = 6  ( 1  3  2.6)
2
A = 23.4 cm2
Maths Quest General Mathematics HSC Course
Chapter 2 Further applications of area and volume Test A
5
6
The figure below shows an area bounded by a
road and the beach.
(a) Use one application of Simpson’s rule to
approximate the area.
(b) Approximate the area using two
applications of Simpson’s rule.
(c) Explain which of the two approximations
will be the more accurate.
Maths Quest General Mathematics HSC Course
8
(a) h = 60  2
= 30
df = 0, dm = 40, dl = 50
h
A  ( d f  4d m  d l )
3
30
=
(0 + 4  40 + 50)
3
= 10  210
= 2100 m2
(b) h = 60  4
= 15
For the left half of the figure:
df = 0, dm = 70, dl = 40
h
A  ( d f  4d m  d l )
3
15
= (0 + 4  70 + 40)
3
= 5  320
= 1600 m2
For the right half of the figure:
df = 40, dm = 60, dl = 50
h
A  ( d f  4d m  d l )
3
15
=
(40 + 4  60 + 50)
3
= 5  330
= 1650 m2
Total area = 1600 + 1650
= 3250 m2
(c) It will be more accurate using Simpson’s
rule twice because the more offsets will
account for more changes in the distance
between the beach and the road.
Chapter 2 Further applications of area and volume Test A
6
7
Calculate the surface area of:
(a) a closed cylinder with height 15 cm and
diameter 9 cm
(a)
SA = 2r2 + 2rh
SA = 2    4.52 + 2    4.52  15
SA = 2035.75 cm2
(b)
(b)
SA = 4r2
SA = 4    16.52
SA = 3421.2 cm2
a sphere with diameter 33 cm.
Give answers correct to 1 decimal place.
8
A cylindrical bucket has a diameter of 30 cm
and a height of 40 cm as shown below.
SA = r2 + 2rh
=   152 + 2    15  40
= 4 477 cm2
6
4
V = r2h
=   152  40
= 28 274 cm3
Calculate the surface area and volume of the
bucket correct to the nearest whole number.
9
A hemispherical bowl with a diameter of 40 cm
4
Volume of sphere = πr3
will be used to hold a pre-mixed fruit drink for
3
a party. If you want to fill it to the brim, how
4
=  π  203
many litres of pre-mixed drink can you pour
3
in? Give your answer correct to 1 decimal
= 33 510 cm3
place.
Volume of hemisphere = 33 510  2
= 16 755 cm3
Capacity = 16 755 mL
 16.8 L
Maths Quest General Mathematics HSC Course
Chapter 2 Further applications of area and volume Test A
2
7
10
Examine the diagram below.
4
For the object shown:
(a) what is the volume contained by the solid
and framed sections (to 1 decimal place)?
(a)
V = π r² h
= π  3²  6
 169.7 m³
(b) what is the volume of the solid part only?
(b)
V=
4
π r³
3
4
π 3³
3
 113.1 m³
=
Maths Quest General Mathematics HSC Course
Chapter 2 Further applications of area and volume Test A
8
11
The figure below shows a rectangular prism
that has had all its measurements taken correct
to the nearest centimetre.
8
(a) Calculate the volume of the prism.
(a)
V = lbh
= 28  18  12
= 6048 cm3
(b) Calculate the maximum possible
percentage error in the volume correct to 1
decimal place.
(b)
Smallest possible dimensions:
27.5  17.5  11.5
V = lbh
= 27.5  17.5  11.5
= 5534.375 cm3
Error = 6048 – 5534.375
= 513.625 cm3
Largest possible dimensions:
28.5  18.5  12.5
V = lbh
= 28.5  18.5  12.5
= 6590.625 cm3
Error = 6590.625 – 6048
= 542.625 cm3
Max error = 542.625 cm3
542.625
Max percent error =
 100%
6048
= 9.0%
Maths Quest General Mathematics HSC Course
Chapter 2 Further applications of area and volume Test A
9
12
A closed cylinder has a radius of 4.3 cm and a
height of 9.7 cm measured correct to 1 decimal
place
8
(a) Find the surface area of the cylinder correct (a)
to 1 decimal place.
SA = 2r2 + 2rh
SA = 2    4.32 + 2    4.3  9.7
SA = 378.2 cm2
(b) Find the smallest possible surface area of
the cylinder.
(b)
Smallest dimensions: r = 4.25, h = 9.65
SA = 2r2 + 2rh
SA = 2    4.252 + 2    4.25  9.65
SA = 371.2 cm2
(c) Find the largest possible surface area of the (c)
cylinder.
Largest dimensions: r = 4.35, h = 9.75
SA = 2r2 + 2rh
SA = 2    4.352 + 2    4.35  9.75
SA = 385.4 cm2
(d) Find the maximum possible percentage
error in calculating the surface area of the
cylinder.
Max error = 385.4 – 378.2
Max error = 7.2
7 .2
Percentage error =
 100%
378.2
Percentage error = 1.9%
Maths Quest General Mathematics HSC Course
(d)
Chapter 2 Further applications of area and volume Test A
10