Year 8 Measurement Worksheets 5b4ao
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Year 8 Measurement Worksheets
C6.1
Circumference using = 3.14 To find the circumference of a circle you can use either of two formulae: 1. If you know the diameter, use C = D (where D is the diameter).
2.
If you know the radius, use C = 2 r (where r is the radius).
Find the circumference of each of the following circles, using = 3.14. (a) (b)
(a)
1
diameter = 10 cm use C = D = 3.14 10 = 31.4 cm
(b)
radius = 8 cm use C = 2r = 2 3.14 8 = 50.24 cm
Find the circumference of each of the following circles, using = 3.14. (a)
(b)
Radius = …… Use C = ………
…………………… = …… Use C = ………
= …… …… ……
= …… ……
= …… cm
= …… m
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
C6.1 (continued)
(c)
(d)
…………………… = ……
…………………… = ……
Use C = ………
Use C = ………
= …… ……
= …… …… ……
= …… m
= …… m
(e)
(f)
…………………… = ……
…………………… = ……
Use C = ………
Use C = ………
= …… …… ……
= …… ……
= …… m
= …… cm
(g)
(h)
…………………… = ……
…………………… = ……
Use C = ………
Use C = ………
= …… …… ……
= …… ……
= …… m
= …… cm
(i)
(j)
…………………… = …… Use C = ………
…………………… = …… Use C = ………
= …… ……
= …… …… ……
= …… mm
= …… m
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
C6.2
Perimeter using the circumference of a circle Find the perimeter of the shape on the right.
First, label the key points. Perimeter = straight edge (OB) + straight edge (OA) + quarter circle (AB) Straight edge (OB) = Straight edge (OA) = 6 cm Quarter circle (AB) has radius = 6 cm Quarter circle (AB) =
1 4
2r =
1 4
2 3.14 6 = 9.42 cm
Perimeter = 6 + 6 + 9.42 = 21.42 cm
1
Find the perimeter of each of the following shapes. (a)
Perimeter = straight edge (……) + half circle (……) Straight edge (AB) = …… Half circle (BA) has ………………… 10 cm 1 1 Half circle (BA) = ……… = 3.14 …… = ……… 2 2 Perimeter = ……… + ……… = ……… cm (b)
Perimeter = straight edge (……) + straight edge (……) + straight edge (……) + half circle (……) Straight edge (……) = Straight edge (……) = …… Straight edge (……) = …… Half circle (……) has diameter …… 1 1 Half circle (……) = …… = …… …… = …… 2 2 Perimeter = …… + …… + …… + …… = …… cm Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
C6.3
Area of a parallelogram To show that the area of a parallelogram is the same as that of the corresponding rectangle, you will need to: 1. 2. 3. 4.
Cut out shapes and given below. Paste rectangle into your book. Cut parallelogram into two pieces along the dotted line. Place the two pieces formed in part 3 on top of rectangle so that they exactly cover it without any overlapping. Paste them in place.
You have now shown that: Area of a parallelogram = Area of a rectangle A = bh
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
C6.4
Area of a trapezium The rule for the area of a trapezium is: 1 A = (a + b)h 2 where b is the base, h is the height and a is the side parallel to the base.
Find the area of each of the following trapeziums. (a) (b)
(a)
a = 6 b = 10 h = 5
(b)
1 A = (a + b) h 2 1 = (6 + 10) 5 2 1 = 16 5 2 = 40 cm2
1
a = 9 b = 5 h = 4
1 (a + b) h 2 1 = (9 + 5) 4 2 1 = 14 4 2 = 28 m2
A =
Find the area of each of the following trapeziums. (a)
(b)
a = …… b = …… h = …… 1 A = (a + b) h 2 1 = (…… + ……) …… 2 1 = …… …… 2 = ……… mm2
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
a = …… b = …… h = …… 1 A = (…… + ……) …… 2 1 = (…… + ……) …… 2 1 = …… …… 2 = ……… cm2
C6.4 (continued)
(c)
(d)
a = …… b = …… h = ……
a = …… b = …… h = ……
A =
1 (…… + ……) …… 2
A =
1 (…… + ……) …… 2
=
1 (…… + ……) …… 2
=
1 (…… + ……) …… 2
=
1 …… …… 2
=
1 …… …… 2
= ……… m2 (e)
= ……… m2 (f)
a = …… b = …… h = ……
a = …… b = …… h = ……
A =
1 (…… + ……) …… 2
A =
1 (…… + ……) …… 2
=
1 (…… + ……) …… 2
=
1 (…… + ……) …… 2
=
1 …… …… 2
=
1 …… …… 2
= ……… m2
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
= ……… mm2
C6.5
Perimeter, circumference and area Bicycles were invented as a mode of transport some 200 years ago. A variety of bicycles exist, including recreational bikes, collapsible bikes, racing bikes, mountain bikes and motorised bikes. The design of a bicycle incorporates a number of mathematical components, including ratio, circumference, diameter and mass. The correct height of the bicycle seat is very important for rider safety and comfort, and in of the physical positioning of the body. The most suitable seat height has been found to be a fixed percentage of the length of the rider’s leg. What is this fixed percentage? Find out the answer by first solving the following, showing all your working in the spaces provided. Then write the question’s code letter above its solution in the boxes at the end of the next page. For each of the three shapes below, find both the perimeter and area. N
Perimeter (m)
C
Area (m2)
P
Perimeter (cm)
I
Area (cm2)
A
Perimeter (mm)
T
Area (mm2)
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
C6.5 (continued)
For each of the three circles below, calculate the circumference and area. Use the button on your calculator and round your answer to two decimal places. O
Circumference (m)
D
Area (m2)
R
Circumference (cm)
E
Area (cm2)
H
Circumference (mm)
U
Area (mm2)
What is this fixed percentage?
78.54
24
907.92
56.55
254.47
24
490.87
54
24
490.87
24
98
24
907.92
46
907.92
106.81
20
907.92
24
150
106.81
907.92
490.87
C6.6 Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
Finding the area of part of a shape To find the shaded area of a shape you must use the rule: Shaded area = total area – unshaded area
Find the area of the shaded shape on the right.
Total area (circle) = r 2 = 3.14 82 = 200.96 cm2 Unshaded area (rectangle) = l w = 10 5 = 50 cm2 Shaded area = 200.96 – 50 = 150.96 cm2
1
Find the shaded area in each of the following examples. (a)
Total area (………………) = l2 = ……2 = ……… Unshaded area (………………) = ……… = 3.14 …… = ……… Shaded area = total area – unshaded area = ……… – ……… = ……… cm2
(b)
Total area (………………) = =
...... …… ...... ...... …… …… ......
= ……… 1 Unshaded area (………………) = (…… + ……) …… 2 1 = (…… + ……) 2 …… 1 = (……) …… 2 = ……… Shaded area = ……………… area – ……………… area = ……… – ……… = ……… cm2
C6.7 Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
Composite areas, volume, capacity, and speed In mathematics, the study of the motion of a body, without reference to the cause of the motion, is called ‘kinematics’. Within this study quantities such as position, displacement, distance, speed, velocity and acceleration are considered. All of these quantities depend on time. You will have met a formula for average speed, given as: distance travelled average speed = time taken The unit of speed is commonly metre per second (m/s) or kilometre per hour (km/h). Human beings have achieved all sorts of speed records. We even have a speed record for travel on the moon! What top speed was reached by the astronauts’ moon buggy in 1972? Find out the answer by first solving the following, showing all your working in the spaces provided. Then write the question’s code letter above its solution in the boxes at the end of the next page. Find the area of each composite shape below. State your answer correct to two decimal places. E R
U
M
Find the shaded area of each shape below. State your answer correct to two decimal places. H A
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
C6.7 (continued)
For each of the three solids below find both the volume and capacity in the units given. S
Volume (m3)
L
Capacity (L)
O
Volume (cm3)
X
Capacity (L)
T
Volume (cm3)
I
Capacity (L)
N Calculate the average speed of a car that travels 490 km in 5¼ hours.
K Stephanie rides her bike for 20 minutes at an average speed of 0.75 km/min. How far did she travel in the 20 minutes?
What top speed was reached by the astronauts’ moon buggy in 1972?
150.8
2.464
0.26
2464
47.24
47.24
93.33
15
2.464
150 800
260
950.56
47.24
2464
105.68
93.33
385
260
139.4
408
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
408
47.24
150.8
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
C6.1
Circumference using = 3.14 To find the circumference of a circle you can use either of two formulae: 1. If you know the diameter, use C = D (where D is the diameter).
2.
If you know the radius, use C = 2 r (where r is the radius).
Find the circumference of each of the following circles, using = 3.14. (a) (b)
(a)
1
diameter = 10 cm use C = D = 3.14 10 = 31.4 cm
(b)
radius = 8 cm use C = 2r = 2 3.14 8 = 50.24 cm
Find the circumference of each of the following circles, using = 3.14. (a)
(b)
Radius = …… Use C = ………
…………………… = …… Use C = ………
= …… …… ……
= …… ……
= …… cm
= …… m
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
C6.1 (continued)
(c)
(d)
…………………… = ……
…………………… = ……
Use C = ………
Use C = ………
= …… ……
= …… …… ……
= …… m
= …… m
(e)
(f)
…………………… = ……
…………………… = ……
Use C = ………
Use C = ………
= …… …… ……
= …… ……
= …… m
= …… cm
(g)
(h)
…………………… = ……
…………………… = ……
Use C = ………
Use C = ………
= …… …… ……
= …… ……
= …… m
= …… cm
(i)
(j)
…………………… = …… Use C = ………
…………………… = …… Use C = ………
= …… ……
= …… …… ……
= …… mm
= …… m
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
C6.2
Perimeter using the circumference of a circle Find the perimeter of the shape on the right.
First, label the key points. Perimeter = straight edge (OB) + straight edge (OA) + quarter circle (AB) Straight edge (OB) = Straight edge (OA) = 6 cm Quarter circle (AB) has radius = 6 cm Quarter circle (AB) =
1 4
2r =
1 4
2 3.14 6 = 9.42 cm
Perimeter = 6 + 6 + 9.42 = 21.42 cm
1
Find the perimeter of each of the following shapes. (a)
Perimeter = straight edge (……) + half circle (……) Straight edge (AB) = …… Half circle (BA) has ………………… 10 cm 1 1 Half circle (BA) = ……… = 3.14 …… = ……… 2 2 Perimeter = ……… + ……… = ……… cm (b)
Perimeter = straight edge (……) + straight edge (……) + straight edge (……) + half circle (……) Straight edge (……) = Straight edge (……) = …… Straight edge (……) = …… Half circle (……) has diameter …… 1 1 Half circle (……) = …… = …… …… = …… 2 2 Perimeter = …… + …… + …… + …… = …… cm Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
C6.3
Area of a parallelogram To show that the area of a parallelogram is the same as that of the corresponding rectangle, you will need to: 1. 2. 3. 4.
Cut out shapes and given below. Paste rectangle into your book. Cut parallelogram into two pieces along the dotted line. Place the two pieces formed in part 3 on top of rectangle so that they exactly cover it without any overlapping. Paste them in place.
You have now shown that: Area of a parallelogram = Area of a rectangle A = bh
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
C6.4
Area of a trapezium The rule for the area of a trapezium is: 1 A = (a + b)h 2 where b is the base, h is the height and a is the side parallel to the base.
Find the area of each of the following trapeziums. (a) (b)
(a)
a = 6 b = 10 h = 5
(b)
1 A = (a + b) h 2 1 = (6 + 10) 5 2 1 = 16 5 2 = 40 cm2
1
a = 9 b = 5 h = 4
1 (a + b) h 2 1 = (9 + 5) 4 2 1 = 14 4 2 = 28 m2
A =
Find the area of each of the following trapeziums. (a)
(b)
a = …… b = …… h = …… 1 A = (a + b) h 2 1 = (…… + ……) …… 2 1 = …… …… 2 = ……… mm2
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
a = …… b = …… h = …… 1 A = (…… + ……) …… 2 1 = (…… + ……) …… 2 1 = …… …… 2 = ……… cm2
C6.4 (continued)
(c)
(d)
a = …… b = …… h = ……
a = …… b = …… h = ……
A =
1 (…… + ……) …… 2
A =
1 (…… + ……) …… 2
=
1 (…… + ……) …… 2
=
1 (…… + ……) …… 2
=
1 …… …… 2
=
1 …… …… 2
= ……… m2 (e)
= ……… m2 (f)
a = …… b = …… h = ……
a = …… b = …… h = ……
A =
1 (…… + ……) …… 2
A =
1 (…… + ……) …… 2
=
1 (…… + ……) …… 2
=
1 (…… + ……) …… 2
=
1 …… …… 2
=
1 …… …… 2
= ……… m2
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
= ……… mm2
C6.5
Perimeter, circumference and area Bicycles were invented as a mode of transport some 200 years ago. A variety of bicycles exist, including recreational bikes, collapsible bikes, racing bikes, mountain bikes and motorised bikes. The design of a bicycle incorporates a number of mathematical components, including ratio, circumference, diameter and mass. The correct height of the bicycle seat is very important for rider safety and comfort, and in of the physical positioning of the body. The most suitable seat height has been found to be a fixed percentage of the length of the rider’s leg. What is this fixed percentage? Find out the answer by first solving the following, showing all your working in the spaces provided. Then write the question’s code letter above its solution in the boxes at the end of the next page. For each of the three shapes below, find both the perimeter and area. N
Perimeter (m)
C
Area (m2)
P
Perimeter (cm)
I
Area (cm2)
A
Perimeter (mm)
T
Area (mm2)
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
C6.5 (continued)
For each of the three circles below, calculate the circumference and area. Use the button on your calculator and round your answer to two decimal places. O
Circumference (m)
D
Area (m2)
R
Circumference (cm)
E
Area (cm2)
H
Circumference (mm)
U
Area (mm2)
What is this fixed percentage?
78.54
24
907.92
56.55
254.47
24
490.87
54
24
490.87
24
98
24
907.92
46
907.92
106.81
20
907.92
24
150
106.81
907.92
490.87
C6.6 Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
Finding the area of part of a shape To find the shaded area of a shape you must use the rule: Shaded area = total area – unshaded area
Find the area of the shaded shape on the right.
Total area (circle) = r 2 = 3.14 82 = 200.96 cm2 Unshaded area (rectangle) = l w = 10 5 = 50 cm2 Shaded area = 200.96 – 50 = 150.96 cm2
1
Find the shaded area in each of the following examples. (a)
Total area (………………) = l2 = ……2 = ……… Unshaded area (………………) = ……… = 3.14 …… = ……… Shaded area = total area – unshaded area = ……… – ……… = ……… cm2
(b)
Total area (………………) = =
...... …… ...... ...... …… …… ......
= ……… 1 Unshaded area (………………) = (…… + ……) …… 2 1 = (…… + ……) 2 …… 1 = (……) …… 2 = ……… Shaded area = ……………… area – ……………… area = ……… – ……… = ……… cm2
C6.7 Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
Composite areas, volume, capacity, and speed In mathematics, the study of the motion of a body, without reference to the cause of the motion, is called ‘kinematics’. Within this study quantities such as position, displacement, distance, speed, velocity and acceleration are considered. All of these quantities depend on time. You will have met a formula for average speed, given as: distance travelled average speed = time taken The unit of speed is commonly metre per second (m/s) or kilometre per hour (km/h). Human beings have achieved all sorts of speed records. We even have a speed record for travel on the moon! What top speed was reached by the astronauts’ moon buggy in 1972? Find out the answer by first solving the following, showing all your working in the spaces provided. Then write the question’s code letter above its solution in the boxes at the end of the next page. Find the area of each composite shape below. State your answer correct to two decimal places. E R
U
M
Find the shaded area of each shape below. State your answer correct to two decimal places. H A
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
C6.7 (continued)
For each of the three solids below find both the volume and capacity in the units given. S
Volume (m3)
L
Capacity (L)
O
Volume (cm3)
X
Capacity (L)
T
Volume (cm3)
I
Capacity (L)
N Calculate the average speed of a car that travels 490 km in 5¼ hours.
K Stephanie rides her bike for 20 minutes at an average speed of 0.75 km/min. How far did she travel in the 20 minutes?
What top speed was reached by the astronauts’ moon buggy in 1972?
150.8
2.464
0.26
2464
47.24
47.24
93.33
15
2.464
150 800
260
950.56
47.24
2464
105.68
93.33
385
260
139.4
408
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
408
47.24
150.8
Heinemann Maths Zone 8 © Reed International Books Australia Pty Ltd
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