Junior Certificate Ordinary Level Perimeter, Area
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Junior Certificate Ordinary Level Perimeter, Area and Volume (a) A small pizza has diameter 20 cm. A large pizza has diameter 30 cm. (i) What is the area of the base of a small pizza, to the nearest cm2. 20 cm 10 cm Area of circle π r 2
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01
Junior Certificate
Ordinary Level
Perimeter, Area and Volume<br>
Ordinary Level
Perimeter, Area and Volume<br>
02
(a) A small pizza has diameter 20 cm.
A large pizza has diameter 30 cm. (i) What is the area of the base of a small
pizza, to the nearest cm2. 20 cm 10 cm Area of circle = π r 2 = π × 102 = 314·159.. cm 2 (ii) What is the area of the base of a large
pizza, to the nearest cm2. 15 cm Area of circle = π × 152 = 706·858.. 7 cm 2<br>
A large pizza has diameter 30 cm. (i) What is the area of the base of a small
pizza, to the nearest cm2. 20 cm 10 cm Area of circle = π r 2 = π × 102 = 314·159.. cm 2 (ii) What is the area of the base of a large
pizza, to the nearest cm2. 15 cm Area of circle = π × 152 = 706·858.. 7 cm 2<br>
03
314 × 2 628 cm2 (a) A small pizza has diameter 20 cm.
A large pizza has diameter 30 cm. (iii) What is the difference in area between
one large pizza and two small pizzas? Area of large pizza = 707 cm2 Area of 2 small pizzas = – ––––––– 79 cm2<br>
A large pizza has diameter 30 cm. (iii) What is the difference in area between
one large pizza and two small pizzas? Area of large pizza = 707 cm2 Area of 2 small pizzas = – ––––––– 79 cm2<br>
04
r = 14 = 44 m 28 m l = 28 m (b) An athletics track has a total length of 400 m.
The track is made up of two parallel sides [ab] and [cd ], and two
semicircular ends as shown in the diagram. The diameters of the
ends, [ac] and [bd ], measure 28 m each. a b c d one of the semicircular ends. 2π r<br>
The track is made up of two parallel sides [ab] and [cd ], and two
semicircular ends as shown in the diagram. The diameters of the
ends, [ac] and [bd ], measure 28 m each. a b c d one of the semicircular ends. 2π r<br>
05
(c) A rectangular carton full of fruit juice measures 12 cm by
6 cm by 33 cm.
(i) Find the volume of juice in the carton. = 2376 cm3 12 cm = 12 × 6 × 33 Volume = length × width × height 33 cm 6 cm<br>
6 cm by 33 cm.
(i) Find the volume of juice in the carton. = 2376 cm3 12 cm = 12 × 6 × 33 Volume = length × width × height 33 cm 6 cm<br>
06
π (3)2 9π (c) A rectangular carton full of orange juice measures 12 cm by
6 cm by 33 cm.
(iii) The radius of each glass is 3 cm. Calculate the height of each
glass, correct to the nearest centimetre. 33 cm 12 cm Volume = 132 cm3 3 cm 6 cm h Volume of cylinder = π r2h 132 h = ___ h = 4·668… = 5 cm<br>
6 cm by 33 cm.
(iii) The radius of each glass is 3 cm. Calculate the height of each
glass, correct to the nearest centimetre. 33 cm 12 cm Volume = 132 cm3 3 cm 6 cm h Volume of cylinder = π r2h 132 h = ___ h = 4·668… = 5 cm<br>
07
(a) A swimming pool is 50 m in length.
Mary swims 25 lengths of the pool.
What distance, in kilometres, does Mary swim? = 1·25 km 50 × 25 = 1250 m 1000 m = 1 km<br>
Mary swims 25 lengths of the pool.
What distance, in kilometres, does Mary swim? = 1·25 km 50 × 25 = 1250 m 1000 m = 1 km<br>
08
(b) A garden is made up of a rectangular lawn that is surrounded
by a path. The garden is 16 m long and 10 m wide.
The path is 2 m wide. (i) Find, in m2, the area of the garden. 16 m = 160 m2 10 m 2 m 2 m 2 m 2 m Area = 16 × 10 Length × Breadth<br>
by a path. The garden is 16 m long and 10 m wide.
The path is 2 m wide. (i) Find, in m2, the area of the garden. 16 m = 160 m2 10 m 2 m 2 m 2 m 2 m Area = 16 × 10 Length × Breadth<br>
09
Length × Breadth (b) A garden is made up of a rectangular lawn that is surrounded
by a path. The garden is 16 m long and 10 m wide.
The path is 2 m wide. (ii) Find, in m2, the area of the lawn. 16 m = 72 m2 10 m Area = 12 × 6 16 – 2 – 2 12 m 10 – 2 – 2 6 m 2 m 2 m 2 m 2 m<br>
by a path. The garden is 16 m long and 10 m wide.
The path is 2 m wide. (ii) Find, in m2, the area of the lawn. 16 m = 72 m2 10 m Area = 12 × 6 16 – 2 – 2 12 m 10 – 2 – 2 6 m 2 m 2 m 2 m 2 m<br>
10
(b) A garden is made up of a rectangular lawn that is surrounded
by a path. The garden is 16 m long and 10 m wide.
The path is 2 m wide. (iii) Find, in m2, the area of the path. 16 m 10 m 2 m 2 m 2 m 2 m Area of garden = 160 m2 Area of lawn = 72 m2 –––––– – Area of path = 88 m2<br>
by a path. The garden is 16 m long and 10 m wide.
The path is 2 m wide. (iii) Find, in m2, the area of the path. 16 m 10 m 2 m 2 m 2 m 2 m Area of garden = 160 m2 Area of lawn = 72 m2 –––––– – Area of path = 88 m2<br>
11
(a) A circle has a radius of 3·5 cm. 3·5 cm Circumference = 2π r = 22 cm of the circle.<br>
12
(b) A cube has side of length 2 cm.
(i) Find the volume of this cube in cm3. = 8 cm3 = 2 × 2 × 2 Volume = length × length × length 2 cm (ii) A rectangular block is built using 18 of these cubes.
Find the volume of the rectangular block in cm3. Volume of block = 18 × 8 = 144 cm3<br>
(i) Find the volume of this cube in cm3. = 8 cm3 = 2 × 2 × 2 Volume = length × length × length 2 cm (ii) A rectangular block is built using 18 of these cubes.
Find the volume of the rectangular block in cm3. Volume of block = 18 × 8 = 144 cm3<br>
13
4 cm 6 cm 6 cm Surface Area = 2(l × b + l × h + b × h)
= 2(6 × 6 + 6 × 4 + 6 × 4)
= 2(36 + 24 + 24)
= 2(84)
= 168 cm2 (b) A cube has side of length 2 cm.
(iii) This rectangular block is 6 cm long, 6 cm wide and 4 cm high.
Find its surface area in cm2.<br>
= 2(6 × 6 + 6 × 4 + 6 × 4)
= 2(36 + 24 + 24)
= 2(84)
= 168 cm2 (b) A cube has side of length 2 cm.
(iii) This rectangular block is 6 cm long, 6 cm wide and 4 cm high.
Find its surface area in cm2.<br>
14
110 m 110 – 30 75 m (c) A field has shape and measurements as shown in the diagram. Perimeter = (i) Find, in metres, the length of the perimeter of the field. 110 m 75 m 30 m 25 m 25 m 30 m 80 m 75 + 80 + 25 + 30 + 100 + 110 m = 460 m<br>
15
(c) A field has shape and measurements as shown in the diagram. Area = (ii) Find, in m2, the area of the field. 110 m 75 m 30 m 25 m 110 × 75 + 30 × 25 = 8250 + 750 = 9000 m2<br>
16
(c) A field has shape and measurements as shown in the diagram. 1 hectare = 10000 m2 (iii) Mary bought the field at a cost of €20 000 per hectare.
How much did Mary pay for the field? 110 m 75 m 30 m 25 m = 0·9 hectares Cost = 0·9 × 20000 Area = 9000 m2 Area = = €18 000<br>
How much did Mary pay for the field? 110 m 75 m 30 m 25 m = 0·9 hectares Cost = 0·9 × 20000 Area = 9000 m2 Area = = €18 000<br>
17
(a) A rectangular box has measurements as shown.
Find the volume of the box in cm3. = 5000 cm3 = 50 × 20 × 5 Volume = length × width × height 5 cm 50 cm 20 cm<br>
Find the volume of the box in cm3. = 5000 cm3 = 50 × 20 × 5 Volume = length × width × height 5 cm 50 cm 20 cm<br>
18
56 cm (b) The front wheel of a bicycle has a diameter of 56 cm.
(i) Calculate, in cm, the length of the radius of the wheel. Radius = × diameter 56 = 28 cm . (ii) Calculate, in cm, the length of the circumference of the wheel. 28 cm Circumference = 2π r = 176 cm<br>
(i) Calculate, in cm, the length of the radius of the wheel. Radius = × diameter 56 = 28 cm . (ii) Calculate, in cm, the length of the circumference of the wheel. 28 cm Circumference = 2π r = 176 cm<br>
19
cm –––––– (b) The front wheel of a bicycle has a diameter of 56 cm. . One complete turn = 176 cm (iii) How far does the bicycle travel when the wheel
makes 250 complete turns?
Give your answer in metres. 250 complete turns = 176 × 250 = 1 metre = 100 cm = 440 m 44 000 100 m<br>
makes 250 complete turns?
Give your answer in metres. 250 complete turns = 176 × 250 = 1 metre = 100 cm = 440 m 44 000 100 m<br>
20
14 cm (c) A solid metal cylinder has radius 4 cm and height 14 cm.
(i) Find the volume of the cylinder in terms of π. Volume of cylinder = π r 2 h = π = 224π cm3 × 42 × 14 4 cm = π × 16 × 14<br>
(i) Find the volume of the cylinder in terms of π. Volume of cylinder = π r 2 h = π = 224π cm3 × 42 × 14 4 cm = π × 16 × 14<br>
21
(c) A solid metal cylinder has radius 4 cm and height 14 cm.
(ii) Find the curved surface area of the cylinder in terms of π. Curved surface area of cylinder = 2π r h = 2π = 112π cm2 × 4 × 14 = π × 8 × 14 14 cm 4 cm CSA<br>
(ii) Find the curved surface area of the cylinder in terms of π. Curved surface area of cylinder = 2π r h = 2π = 112π cm2 × 4 × 14 = π × 8 × 14 14 cm 4 cm CSA<br>
22
(c) A solid metal cylinder has radius 4 cm and height 14 cm.
(iii) Find the total surface area of the cylinder in terms of π. Total surface area of cylinder = 2π r h Area of circles = 2π r2 = 112π cm2 TSA = 112π + 32π cm2 CSA = π × 2 × 16 14 cm 4 cm + 2π r2 = 2 × π × 42 Area of circles = 32π cm2 TSA = 144π cm2<br>
(iii) Find the total surface area of the cylinder in terms of π. Total surface area of cylinder = 2π r h Area of circles = 2π r2 = 112π cm2 TSA = 112π + 32π cm2 CSA = π × 2 × 16 14 cm 4 cm + 2π r2 = 2 × π × 42 Area of circles = 32π cm2 TSA = 144π cm2<br>
23
1·5 m (b) The gable-end of a house has measurements
as shown in the diagram.
(i) Find, in m2, the area of the bottom rectangle
section of the gable-end. Area of rectangle = Length × breadth = 7 × 8
= 56 m2 7 m 8 m<br>
as shown in the diagram.
(i) Find, in m2, the area of the bottom rectangle
section of the gable-end. Area of rectangle = Length × breadth = 7 × 8
= 56 m2 7 m 8 m<br>
24
1·5 m (b) The gable-end of a house has measurements
as shown in the diagram.
(ii) Find, in m2, the area of the top triangular
section of the gable-end. 8 m 5 = 6 m2 7 m<br>
as shown in the diagram.
(ii) Find, in m2, the area of the top triangular
section of the gable-end. 8 m 5 = 6 m2 7 m<br>
25
💧 = €46 1·5 m 8 m 6 m2 7 m 56 m2 62 m2 5 litres = 31 m2 5 litres = 31 m2 €23 €23 💧 Cost of paint = 2 × 23 (b) The gable-end of a house has measurements
as shown in the diagram.
(iii) The cost of 5 litres of paint is €23.
5 litres of this paint will cover an area of 31 m2.
Find the cost of painting the gable-end with this paint.<br>
as shown in the diagram.
(iii) The cost of 5 litres of paint is €23.
5 litres of this paint will cover an area of 31 m2.
Find the cost of painting the gable-end with this paint.<br>
26
(a) The length of each side of a square tile is 9 cm.
What area, in cm2, will 12 of these tiles cover? Area of square = length2 = 972 cm2 Area of 12 squares = 81 × 12 9 cm = 92 = 81 cm2<br>
What area, in cm2, will 12 of these tiles cover? Area of square = length2 = 972 cm2 Area of 12 squares = 81 × 12 9 cm = 92 = 81 cm2<br>
27
(b)
(i) A circular disc has a radius of 5 cm.
Taking π as 3·14, find, in cm2, the area of the disc. Area of circle = π r 2 5 cm = 3·14 = 78·5 cm2 × 52<br>
(i) A circular disc has a radius of 5 cm.
Taking π as 3·14, find, in cm2, the area of the disc. Area of circle = π r 2 5 cm = 3·14 = 78·5 cm2 × 52<br>
28
(b)
(ii) A rectangular piece of cardboard has measurements as shown.
Two circular pieces, each of radius length 5 cm, are cut out of
this rectangular piece of cardboard as shown.
Find, in cm2, the area of the remaining piece of cardboard. Area of rectangle 12 cm = 24 × 12 = 288 cm2 24 cm 5 cm = length × breadth From part (i) Area of one disc = 78·5 cm2 Area of two discs = 78·5 × 2 cm2 = 157 cm2 – 157 cm2 Remaining area 288 cm2 –––––– 131 cm2 I can see you, It’s 11:52 AM , Sunday, January 24, 2021 and as usual, you are not paying attention!<br>
(ii) A rectangular piece of cardboard has measurements as shown.
Two circular pieces, each of radius length 5 cm, are cut out of
this rectangular piece of cardboard as shown.
Find, in cm2, the area of the remaining piece of cardboard. Area of rectangle 12 cm = 24 × 12 = 288 cm2 24 cm 5 cm = length × breadth From part (i) Area of one disc = 78·5 cm2 Area of two discs = 78·5 × 2 cm2 = 157 cm2 – 157 cm2 Remaining area 288 cm2 –––––– 131 cm2 I can see you, It’s 11:52 AM , Sunday, January 24, 2021 and as usual, you are not paying attention!<br>
29
15 cm (c) A solid metal cylinder has radius 10 cm and height 15 cm.
(i) Taking π as 3·14, find, in cm3, the volume of the metal
cylinder. Volume of cylinder = π r 2 h = 3·14 = 4710 cm3 × 102 10 cm × 15<br>
(i) Taking π as 3·14, find, in cm3, the volume of the metal
cylinder. Volume of cylinder = π r 2 h = 3·14 = 4710 cm3 × 102 10 cm × 15<br>
30
height –––– 2355 length 15 cm and width 14 cm.
Calculate, in cm, its height,
correct to one decimal place. (c) A solid metal cylinder has radius 10 cm and height 15 cm.
(ii) The cylinder was melted down and half of the metal was
recast as a rectangular solid. This rectangular solid has 15 cm Volume of cylinder = 4710 cm 3 2355 = 15 × 14 × height 15 cm 14 cm Volume = length × width × height from part (i) = 210 height × height = 11·214… = 11·2 cm 10 cm<br>
Calculate, in cm, its height,
correct to one decimal place. (c) A solid metal cylinder has radius 10 cm and height 15 cm.
(ii) The cylinder was melted down and half of the metal was
recast as a rectangular solid. This rectangular solid has 15 cm Volume of cylinder = 4710 cm 3 2355 = 15 × 14 × height 15 cm 14 cm Volume = length × width × height from part (i) = 210 height × height = 11·214… = 11·2 cm 10 cm<br>
31
Area of circle = π r 2 12 m 6 m 3·14 × 62 113·04 (c) A garden with a semicircular lawn and two flowerbeds has
measurements as shown in the diagram.
(ii) Taking π as 3·14, find the area of the lawn, in m2. radius = 6 cm 113·04 Area of lawn = 56·52 m2<br>
measurements as shown in the diagram.
(ii) Taking π as 3·14, find the area of the lawn, in m2. radius = 6 cm 113·04 Area of lawn = 56·52 m2<br>
32
(c) A garden with a semicircular lawn and two flowerbeds has
measurements as shown in the diagram.
(iii) Find the area of the flowerbeds, in m2. 12 m 6 m Area of lawn = 56·52 m2 Area of garden = 72 m2 _____ – 15·48 m2 Area of flowerbeds =<br>
measurements as shown in the diagram.
(iii) Find the area of the flowerbeds, in m2. 12 m 6 m Area of lawn = 56·52 m2 Area of garden = 72 m2 _____ – 15·48 m2 Area of flowerbeds =<br>
33
(c) A garden with a semicircular lawn and two flowerbeds has
measurements as shown in the diagram.
(iv) Taking π as 3·14, find the total perimeter of the semicircular
lawn, in m. 6 m = 3·14 × 6 = 18·84 12 m Total perimeter = 12 + 30·84 m<br>
measurements as shown in the diagram.
(iv) Taking π as 3·14, find the total perimeter of the semicircular
lawn, in m. 6 m = 3·14 × 6 = 18·84 12 m Total perimeter = 12 + 30·84 m<br>
34
(b) A solid rectangular block of wood has length 16 cm,
width 4 cm and height 6 cm.
(i) Find, in cm3, the volume of the block of wood. Volume = length × width × height 4 cm = 384 cm3 = 16 × 4 × 6 6 cm 16 cm<br>
width 4 cm and height 6 cm.
(i) Find, in cm3, the volume of the block of wood. Volume = length × width × height 4 cm = 384 cm3 = 16 × 4 × 6 6 cm 16 cm<br>
35
(b) A solid rectangular block of wood has length 16 cm,
width 4 cm and height 6 cm.
(ii) Cubes with sides of length 2 cm, as shown, are made from
the block of wood. Find the number of cubes that can be
made from the block of wood. 2 2 cm 2 4 6 16 Volume of cube = length × length × length = 8 cm3 = 2 × 2 × 2 From part (i) volume of block = 384 cm3 Number of blocks = = 48 Number of blocks = 6 × 8 = 48<br>
width 4 cm and height 6 cm.
(ii) Cubes with sides of length 2 cm, as shown, are made from
the block of wood. Find the number of cubes that can be
made from the block of wood. 2 2 cm 2 4 6 16 Volume of cube = length × length × length = 8 cm3 = 2 × 2 × 2 From part (i) volume of block = 384 cm3 Number of blocks = = 48 Number of blocks = 6 × 8 = 48<br>
36
(b) A solid rectangular block of wood has length 16 cm,
width 4 cm and height 6 cm.
(iii) Calculate, in cm2, the surface area of the block of wood. Surface Area = 2 × l × b + 2 × l × h + 2 × b × h
= 2 × 16 × 6
= 192 + 48 + 128
= 368 cm2 + 2 × 6 × 4 + 2 × 16 × 4 4 cm 6 cm 16 cm<br>
width 4 cm and height 6 cm.
(iii) Calculate, in cm2, the surface area of the block of wood. Surface Area = 2 × l × b + 2 × l × h + 2 × b × h
= 2 × 16 × 6
= 192 + 48 + 128
= 368 cm2 + 2 × 6 × 4 + 2 × 16 × 4 4 cm 6 cm 16 cm<br>
37
= 47·1 metres r = 15 m S The length of the two semicircular ends is the circumference of the circle 153 m C = 2πr (c) An athletics track has two equal parallel sides [PQ] and [SR]
and two equal semi-circular ends with diameters [PS] and [QR].
| PQ| = |SR | = 153 metres, and | PS| = |QR | = 30 metres. R Q P 30 m = 2 × 3·14 × 15 One semicircular end = 3·14 × 15 (i) Taking π as 3·14, calculate the length of one of the
semi-circular ends, correct to the nearest metre.<br>
and two equal semi-circular ends with diameters [PS] and [QR].
| PQ| = |SR | = 153 metres, and | PS| = |QR | = 30 metres. R Q P 30 m = 2 × 3·14 × 15 One semicircular end = 3·14 × 15 (i) Taking π as 3·14, calculate the length of one of the
semi-circular ends, correct to the nearest metre.<br>
38
153 m (ii) Calculate the total length of one lap of the track,
correct to the nearest metre. 30 m 153 m 47 m 153 m 47 m The length of one lap = 153 + 47 + 153 + 47 = 400 metres (c) An athletics track has two equal parallel sides [PQ] and [SR]
and two equal semi-circular ends with diameters [PS] and [QR].
| PQ| = |SR | = 153 metres, and | PS| = |QR | = 30 metres. S R Q P<br>
correct to the nearest metre. 30 m 153 m 47 m 153 m 47 m The length of one lap = 153 + 47 + 153 + 47 = 400 metres (c) An athletics track has two equal parallel sides [PQ] and [SR]
and two equal semi-circular ends with diameters [PS] and [QR].
| PQ| = |SR | = 153 metres, and | PS| = |QR | = 30 metres. S R Q P<br>
39
S R Q P (iii) Noirín ran a 5000 metre race on the above track in 15 minutes.
Calculate, in seconds, the average time it took Noirín to
complete one lap of the track during that race. 153 m 47 m 153 m 47 m One lap = 400 m = 12·5 laps 5000 m Time for one lap = = 72 seconds (c) An athletics track has two equal parallel sides [PQ] and [SR]
and two equal semi-circular ends with diameters [PS] and [QR].
| PQ| = |SR | = 153 metres, and | PS| = |QR | = 30 metres.<br>
Calculate, in seconds, the average time it took Noirín to
complete one lap of the track during that race. 153 m 47 m 153 m 47 m One lap = 400 m = 12·5 laps 5000 m Time for one lap = = 72 seconds (c) An athletics track has two equal parallel sides [PQ] and [SR]
and two equal semi-circular ends with diameters [PS] and [QR].
| PQ| = |SR | = 153 metres, and | PS| = |QR | = 30 metres.<br>
40
(a) A disc has a radius of 2·5 cm.
Taking π as 3·14, calculate, in cm2, the area of the disc. Area of circle = π r 2 2·5 cm Area = 3·14 = 19·625 cm2 × 2·52<br>
Taking π as 3·14, calculate, in cm2, the area of the disc. Area of circle = π r 2 2·5 cm Area = 3·14 = 19·625 cm2 × 2·52<br>
41
(ii) The garden is to be covered completely with square concrete
slabs each of side 50 cm.
Find the number of slabs required to cover the garden. (b) A rectangular garden has measurements as shown.
(i) Find, in m2, the area of the garden. Area of rectangle = length × breadth 9 m = 18 × 9 = 162 m2 18 m Area = 162 m2 = 0·25 m2 Area of slab = 0·5 × 0·5 Number of slabs = = 648<br>
slabs each of side 50 cm.
Find the number of slabs required to cover the garden. (b) A rectangular garden has measurements as shown.
(i) Find, in m2, the area of the garden. Area of rectangle = length × breadth 9 m = 18 × 9 = 162 m2 18 m Area = 162 m2 = 0·25 m2 Area of slab = 0·5 × 0·5 Number of slabs = = 648<br>
42
(c) A rectangular carton full of orange juice measures 10 cm by
7 cm by 25 cm.
(i) Find, in cm3, the volume of orange juice in the carton. = 1750 cm3 10 cm = 10 × 7 × 25 Volume = length × width × height 25 cm 7 cm<br>
7 cm by 25 cm.
(i) Find, in cm3, the volume of orange juice in the carton. = 1750 cm3 10 cm = 10 × 7 × 25 Volume = length × width × height 25 cm 7 cm<br>
43
25 cm = 1750 cm3 10 cm = 10 × 7 × 25 Volume = length × width × height (c) A rectangular carton full of orange juice measures 10 cm by
7 cm by 25 cm.
(ii) The orange juice fills 14 cylindrical glasses exactly.
Find, in cm3, the volume of each glass. Volume of each glass = 125 cm3 7 cm<br>
7 cm by 25 cm.
(ii) The orange juice fills 14 cylindrical glasses exactly.
Find, in cm3, the volume of each glass. Volume of each glass = 125 cm3 7 cm<br>
44
(c) A rectangular carton full of orange juice measures 10 cm by
7 cm by 25 cm.
(iii) The radius of each glass is 2·4 cm. Taking π as 3·14, calculate
the height of each glass, correct to the nearest centimetre. 25 cm 10 cm Volume = 125 cm3 2·4 cm 7 cm h Volume of cylinder = π r2h 125 h = 3·14(2·4)2 18·0864 _______ h = 6·911… = 7 cm<br>
7 cm by 25 cm.
(iii) The radius of each glass is 2·4 cm. Taking π as 3·14, calculate
the height of each glass, correct to the nearest centimetre. 25 cm 10 cm Volume = 125 cm3 2·4 cm 7 cm h Volume of cylinder = π r2h 125 h = 3·14(2·4)2 18·0864 _______ h = 6·911… = 7 cm<br>
45
(c) A field has shape and measurements as shown in the diagram. (i) Find, in metres, the length of the perimeter of the field. 80 m 30 m 120 m 35 m 80 – 30 85 m 50 m Perimeter = 80 + 120 + 50 + 85 + 30 + 35 m = 400 m 120 – 35<br>
46
(c) A field has shape and measurements as shown in the diagram. Area = 35 × 30 + 50 × 120 = 1050 + 6000 = 7050 m2 (ii) Find, in m2, the area of the field. 80 m 30 m 120 m 35 m 50 m<br>
47
(c) A field has shape and measurements as shown in the diagram. 80 m 30 m 120 m 35 m (iii) Tim bought the field at a cost of €41 000 per hectare.
How much did Tim pay for the field? [1 hectare = 10 000 m2] = 0·705 hectares Area = 7050 m2 Area = 50 m Cost = 0·705 × 41000 = €28 905<br>
How much did Tim pay for the field? [1 hectare = 10 000 m2] = 0·705 hectares Area = 7050 m2 Area = 50 m Cost = 0·705 × 41000 = €28 905<br>
48
60 cm (b) A bicycle wheel has a diameter of 60cm.
(i) Calculate, in cm, the radius of the bicycle wheel. 60 = 30 cm . (ii) Taking π as 3·142 calculate, in cm, the circumference of the
wheel. 30 cm Circumference = 2π r = 2 × 3·142 × 30 = 188·52 cm<br>
(i) Calculate, in cm, the radius of the bicycle wheel. 60 = 30 cm . (ii) Taking π as 3·142 calculate, in cm, the circumference of the
wheel. 30 cm Circumference = 2π r = 2 × 3·142 × 30 = 188·52 cm<br>
49
––––––– 64096·8 cm (b) A bicycle wheel has a diameter of 60cm. One complete turn = 188·52 cm (iii) How far does the bicycle travel when the wheel
makes 340 complete turns?
Give your answer to the nearest metre. 340 complete turns = 188·52 × 340 = 1 metre = 100 cm = 640·968 m 100 m = 641 m<br>
makes 340 complete turns?
Give your answer to the nearest metre. 340 complete turns = 188·52 × 340 = 1 metre = 100 cm = 640·968 m 100 m = 641 m<br>
50
(i) A rectangular piece of silver measures 4 cm by 24 cm.
Find, in cm2, the area of the piece of silver. (c) (ii) Brian wants to cut circular discs of radius 2 cm from the piece
of silver. What is the greatest number of discs that he can cut
from the piece? Area = length × breadth 24 cm 4 cm = 24 × 4 = 96 cm2 2 cm Diameter = 2 × 2 = 4 cm Number of discs = 6<br>
Find, in cm2, the area of the piece of silver. (c) (ii) Brian wants to cut circular discs of radius 2 cm from the piece
of silver. What is the greatest number of discs that he can cut
from the piece? Area = length × breadth 24 cm 4 cm = 24 × 4 = 96 cm2 2 cm Diameter = 2 × 2 = 4 cm Number of discs = 6<br>
51
(i) A rectangular piece of silver measures 4 cm by 24 cm.
Find, in cm2, the area of the piece of silver. (c) (iii) Taking π as 3·142, find in cm2, the area of the silver remaining
after the discs have been cut out.
Give your answer correct to one decimal place. 24 cm 4 cm 2 cm Area of circle = π r 2 = 3·142 × 22 = 12·568 × 6 = 75·408 cm2 Area of rectangle = 96 cm2 75·408 96 – –––––– 20·592 Area remaining = 20·6 cm2 r =<br>
Find, in cm2, the area of the piece of silver. (c) (iii) Taking π as 3·142, find in cm2, the area of the silver remaining
after the discs have been cut out.
Give your answer correct to one decimal place. 24 cm 4 cm 2 cm Area of circle = π r 2 = 3·142 × 22 = 12·568 × 6 = 75·408 cm2 Area of rectangle = 96 cm2 75·408 96 – –––––– 20·592 Area remaining = 20·6 cm2 r =<br>
52
(iii) Express the volume of the sphere in
(ii), as a percentage of the volume
of the cube in (i). (i) A cube has side of length 2 cm.
Find the volume of this cube in cm3. = 1000 cm3 = 10 × 10 × 10 V = length × length × length 10 cm (b) V = 1000 cm3 = 52·4 %<br>
(ii), as a percentage of the volume
of the cube in (i). (i) A cube has side of length 2 cm.
Find the volume of this cube in cm3. = 1000 cm3 = 10 × 10 × 10 V = length × length × length 10 cm (b) V = 1000 cm3 = 52·4 %<br>
53
(i) Taking π as 3·142, calculate the length of the semicircular end.
Give your answer to the nearest metre. r = 14 = 43·988 28 m l = (c) A park is in the shape of a rectangle with a semicircular end.
The rectangle is 150 m long and 28 m wide.
The diameter of the semicircular end is also 28 m.
There is a path around the park which is used for walking and
jogging. Path 2π r = 3·142 × 14 150 m = 44 m<br>
Give your answer to the nearest metre. r = 14 = 43·988 28 m l = (c) A park is in the shape of a rectangle with a semicircular end.
The rectangle is 150 m long and 28 m wide.
The diameter of the semicircular end is also 28 m.
There is a path around the park which is used for walking and
jogging. Path 2π r = 3·142 × 14 150 m = 44 m<br>
54
(ii) Calculate the total length of the path around the park. 10 28 m Path 150 m 44 m 150 m 150 m 28 m The length of one lap = 372 m (iii) Barbara wishes to jog 2·5 km. How many laps of the path
must she complete to ensure that she jogs this distance? 2·5 km = 2500 m 150 + 44 + 150 + 28 = 6·72 Barbara must complete 7 laps (c) A park is in the shape of a rectangle with a semicircular end.
The rectangle is 150 m long and 28 m wide.
The diameter of the semicircular end is also 28 m.
There is a path around the park which is used for walking and
jogging.<br>
must she complete to ensure that she jogs this distance? 2·5 km = 2500 m 150 + 44 + 150 + 28 = 6·72 Barbara must complete 7 laps (c) A park is in the shape of a rectangle with a semicircular end.
The rectangle is 150 m long and 28 m wide.
The diameter of the semicircular end is also 28 m.
There is a path around the park which is used for walking and
jogging.<br>
55
65 m A 110 m 35 m B C 55 m (c) The shape and measurements of a field are shown in the
diagram below. (i) Find the length | AB |. | AB | = 110 – 65 = 45 m 110 m 65 m 45 m<br>
diagram below. (i) Find the length | AB |. | AB | = 110 – 65 = 45 m 110 m 65 m 45 m<br>
56
45 m 35 m 55 m 65 m A 110 m 35 m B C 55 m (c) The shape and measurements of a field are shown in the
diagram below. (ii) Find the length of the perimeter of the field. Perimeter = 65 + 20 + 45 + 35 + 110 + 55 m 20 m = 330 m<br>
diagram below. (ii) Find the length of the perimeter of the field. Perimeter = 65 + 20 + 45 + 35 + 110 + 55 m 20 m = 330 m<br>
57
20 m 65 m A 110 m 35 m B C 55 m (c) The shape and measurements of a field are shown in the
diagram below. (iii) The sections [AB] and [BC] are stone walls.
A farmer wishes to put fencing around the rest of the field.
The fencing costs €62·50 per 5 metres.
Find the cost of the fencing. Fencing length = 330 – 45 – 35 = 250 m Cost × 62·50 45 m = €3125<br>
diagram below. (iii) The sections [AB] and [BC] are stone walls.
A farmer wishes to put fencing around the rest of the field.
The fencing costs €62·50 per 5 metres.
Find the cost of the fencing. Fencing length = 330 – 45 – 35 = 250 m Cost × 62·50 45 m = €3125<br>
58
(a) A circular disc has a radius of 20 cm.
Taking π as 3·142 find, in cm2, the area of the disc. 20 cm Area = π r2 = 3·142 × (20)2
= 1256·8 cm2<br>
Taking π as 3·142 find, in cm2, the area of the disc. 20 cm Area = π r2 = 3·142 × (20)2
= 1256·8 cm2<br>
59
(b) A solid rectangular block of wood has
length 8 cm, width 4 cm and height 6cm.
(i) Find, in cm3, the volume of the
the block of wood. = 192 cm3 = 8 × 4 × 6 Volume = length × width × height 4 cm 8 cm 6 cm (ii) Find, in cm3, the volume of a cube of 2 cm. = 8 cm3 = 23 Volume = length3 (iii) How many solid cubes, each of side 2 cm,
can be made from the block of wood in (i)? Number of cubes = 24<br>
length 8 cm, width 4 cm and height 6cm.
(i) Find, in cm3, the volume of the
the block of wood. = 192 cm3 = 8 × 4 × 6 Volume = length × width × height 4 cm 8 cm 6 cm (ii) Find, in cm3, the volume of a cube of 2 cm. = 8 cm3 = 23 Volume = length3 (iii) How many solid cubes, each of side 2 cm,
can be made from the block of wood in (i)? Number of cubes = 24<br>