CHAPTER 11 AREAS OF PLANE FIGURES
CHAPTER 11 AREAS OF PLANE FIGURES EXERCISE 45, Page 106 1. Find the angles p and q in diagram (a) below.
p = 180 - 75 = 105 (interior opposite angles of a parallelogram are equal) q = 180 - 105 - 40 = 35 2. Find the angles r and s in diagram (b) above.
r = 180 - 38 = 142 (the 38 angle is the alternate angle between parallel lines) s = 180 - 47 - 38 = 95 3. Find the angle t in diagram (c) above.
t = 360 - 62 - 95 - 57 = 146
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EXERCISE 46, Page 110 1. Name the types of quadrilateral shown in diagrams (i) to (iv) below, and determine for each (a) the area, and (b) the perimeter.
(i) Rhombus (a) Area = 4 3.5 = 14 cm 2 (b) Perimeter = 4 + 4 + 4 + 4 = 16 cm (ii) Parallelogram (a) Area = 30 6 = 180 mm 2 (b) Perimeter = 30 + 30 + 2
6
2
82 = 80 mm
(iii) Rectangle (a) Area = 120 30 = 3600 mm 2 (b) Perimeter = (2 120) + (2 30) = 300 mm (iv) Trapezium (a) Area =
1 26 12 10 = 190 cm 2 2
(b) Perimeter = 26 + 12 + 102 102
4
2
102
= 26 + 12 + 14.14 + 10.77 = 62.91 cm
171 © John Bird Published by Taylor and Francis
2. A rectangular plate is 85 mm long and 42 mm wide. Find its area in square centimetres. 85 mm = 8.5 cm and 42 mm = 4.2 cm
Area of plate = 8.5 × 4.2 = 35.7 cm 2 3. A rectangular field has an area of 1.2 hectares and a length of 150 m. If 1 hectare = 10000 m2 find (a) its width, and (b) the length of a diagonal. Area of field = 1.2 ha = 1.2 10000 m 2 = 12000 m 2 (a) Area = length width from which, width =
(b) By Pythagoras, length of diagonal =
150
2
area 12000 = 80 m length 150
802 = 170 m
4. Find the area of a triangle whose base is 8.5 cm and perpendicular height 6.4 cm.
Area of triangle =
=
1 × base × perpendicular height 2
1 × 8.5 × 6.4 = 27.2 cm 2 2
5. A square has an area of 162 cm 2 . Determine the length of a diagonal. A square ABCD is shown below of side x cm. The diagonal is given by length AC
Area of square = x 2 = 162
172 © John Bird Published by Taylor and Francis
By Pythagoras, (AC) 2 x 2 x 2 2x 2 = 2 × 162 from which,
diagonal, AC =
2x 2 [2 162] = 18 cm
6. A rectangular picture has an area of 0.96 m 2 . If one of the sides has a length of 800 mm, calculate, in millimetres, the length of the other side.
Area = 0.96 m 2 = 0.96 106 mm 2 and area = length × breadth, i.e. 0.96 106 mm 2 = 800 mm × breadth from which,
breadth =
0.96 106 = 1200 mm 800
7. Determine the area of each of the angle iron sections shown in below.
(a) Area = 2 (7 2) + (1 1) = 28 + 1 = 29 cm 2 (b) Area = (30 8) + 10(25 – 8 – 6) + (6 50) = 240 + 110 + 300 = 650 mm 2
8. The diagram below shows a 4 m wide path around the outside of a 41 m by 37 m garden. Calculate the area of the path.
173 © John Bird Published by Taylor and Francis
Area of garden = 41 × 37 m 2 Area of garden, neglecting the path = (41 – 8) × (37 – 8) = 33 × 29 m 2 Hence, area of path = (41 × 37) – (33 × 29) = 1517 – 957 = 560 m 2
9. The area of a trapezium is 13.5 cm2 and the perpendicular distance between its parallel sides is 3 cm. If the length of one of the parallel sides is 5.6 cm, find the length of the other parallel side.
Area of a trapezium =
i.e. i.e.
13.5 =
1 × (sum of parallel sides) × (perpendicular distance between the 2 parallel sides) 1 × (5.6 + x) × (3) 2
where x is the unknown parallel side
27 = 3(5.6 + x)
i.e.
9 = 5.6 + x
from which,
x = 9 – 5.6 = 3.4 cm
10. Calculate the area of the steel plate shown below.
2 1 Area of steel plate = (25 60) + (140 – 60)(25) + 25 + (50 25) + 55 50 2
= 1500 + 2000 + 625 + 1250 + 1375 = 6750 mm 2
174 © John Bird Published by Taylor and Francis
11. Determine the area of an equilateral triangle of side 10.0 cm.
An equilateral triangle ABC is shown below.
Perpendicular height, AD =
10.02 5.02
by Pythagoras
= 8.6603 cm Hence, area of triangle =
=
1 × base × perpendicular height 2
1 × 10.0 × 8.6603 2
= 43.30 cm 2 12. If paving slabs are produced in 250 mm by 250 mm squares, determine the number of slabs
required to cover an area of 2 m 2 .
Number of slabs =
2 106 mm 2 = 32 250 250
175 © John Bird Published by Taylor and Francis
EXERCISE 47, Page 111 1. A rectangular garden measures 40 m by 15 m. A 1 m flower border is made round the two shorter
sides and one long side. A circular swimming pool of diameter 8 m is constructed in the middle of the garden. Find, correct to the nearest square metre, the area remaining. A sketch of a plan of the garden is shown below.
Shaded area = (40 15) – [(15 1) + (38 1) + (15 1) + 42 ] = 600 – [15 + 38 + 15 + 16] = 600 – 118.27 = 481.73 m 2 = 482 m 2 , correct to the nearest square metre. 2. Determine the area of circles having (a) a radius of 4 cm (b) a diameter of 30 mm (c) a
circumference of 200 mm. (a) Area = r 2 4 = 50.27 cm 2 2
d 2 30 d = 706.9 mm 2 (b) Area = r 2 2 4 4 2
2
(c) Circumference = 2πr = 200 mm, from which, radius, r =
200 100 mm 2
2
100 2 Hence, area = r = 3183 mm 2
176 © John Bird Published by Taylor and Francis
3. An annulus has an outside diameter of 60 mm and an inside diameter of 20 mm. Determine its area. d 2 2 d12 60 20 602 202 = 2513 mm 2 Area of annulus = 4 4 4 4 4 2
2
4. If the area of a circle is 320 mm2, find (a) its diameter, and (b) its circumference.
(a) Area of circle, A =
d 2 4
i.e. 320 =
from which, diameter, d =
d 2 4
4 320 = 20.185 mm = 20.19 mm correct to 2 decimal places.
(b) Circumference of circle = 2r = d = 20.185 = 63.41 mm
5. Calculate the areas of the following sectors of circles: (a) radius 9 cm, angle subtended at centre 75 (b) diameter 35 mm, angle subtended at centre 4837'
2 75 2 2 (a) Area of sector = r 9 = 53.01 cm 360 360 37 2 48 2 60 35 2 (b) Area of sector = = 129.9 mm r 360 2 360 6. Determine the shaded area of the template shown below.
177 © John Bird Published by Taylor and Francis
Area of template = shaded area = (120 90) -
1 2 80 = 10800 – 5026.55 = 5773 mm 2 4
7. An archway consists of a rectangular opening topped by a semi-circular arch as shown below.
Determine the area of the opening if the width is 1 m and the greatest height is 2 m.
The semicircle has a diameter of 1 m, i.e. a radius of 0.5 m. Hence, the archway shown is made up of a rectangle of sides 1 m by 1.5 m and a semicircle of radius 0.5 m. Thus, area of opening = (1.5 1) +
1 2 0.5 = 1.5 + 0.393 = 1.89 m 2 2
178 © John Bird Published by Taylor and Francis
EXERCISE 48, Page 114 1. Calculate the area of a regular octagon if each side is 20 mm and the width across the flats is
48.3 mm. The octagon is shown sketched below and is comprised of 8 triangles of base length 20 mm and perpendicular height 48.3/2
48.3 1 = 1932 mm 2 Area of octagon = 8 20 2 2
2. Determine the area of a regular hexagon which has sides 25 mm.
The hexagon is shown sketched below and is comprised of 6 triangles of base length 25 mm and perpendicular height h as shown.
Tan 30 =
12.5 12.5 from which, h = = 21.65 mm h tan 30
1 Hence, area of hexagon = 6 25 21.65 = 1624 mm 2 2 179 © John Bird Published by Taylor and Francis
3. A plot of land is in the shape shown below. Determine (a) its area in hectares (1 ha = 104 m2), and
(b) the length of fencing required, to the nearest metre, to completely enclose the plot of land.
(a) Area of land = (30 10) +
1 1 1 2 30 + 70 40 + 70 100 80 45 15 2 2 2
= 300 + 450 + 1400 +[7000 – 937.5] = 9176 m 2 =
9176 ha = 0.918 ha 104
(b) Perimeter = 20 + 10 + 30 + 10 + 20 + 20 +
1 2 30 + 20 + 2 +
15
2
70
2
402 + 40
152 + 45 +
20
2
152 + 20
= 110 + 30 + 20 + 80.62 + 40 + 21.21 + 45 + 25 + 20 = 456 m, to the nearest metre.
180 © John Bird Published by Taylor and Francis
EXERCISE 49, Page 115 1. The area of a park on a map is 500 mm2. If the scale of the map is 1 to 40000 determine the true
area of the park in hectares (1 hectare = 104 m2) 500 106 40000 Area of park = 500 10 40000 m ha = 80 ha 104 2
6
2
2
2. A model of a boiler is made having an overall height of 75 mm corresponding to an overall height
of the actual boiler of 6 m. If the area of metal required for the model is 12500 mm 2 , determine, in square metres, the area of metal required for the actual boiler.
The scale is
6000 : 1 i.e. 80 : 1 75
Area of metal required for actual boiler = 12500 106 m 2 80 = 80 m 2 2
3. The scale of an Ordnance Survey map is 1:2500. A circular sports field has a diameter of 8 cm
on the map. Calculate its area in hectares, giving your answer correct to 3 significant figures. (1 hectare = 104 m 2 )
d 2 8 Area of sports field on map = 104 m 2 4 4 2
8 2 2 104 2500 8 2 104 2500 m 2 4 ha True area of sports field = 4 104 2
= 3.14 ha
181 © John Bird Published by Taylor and Francis
p = 180 - 75 = 105 (interior opposite angles of a parallelogram are equal) q = 180 - 105 - 40 = 35 2. Find the angles r and s in diagram (b) above.
r = 180 - 38 = 142 (the 38 angle is the alternate angle between parallel lines) s = 180 - 47 - 38 = 95 3. Find the angle t in diagram (c) above.
t = 360 - 62 - 95 - 57 = 146
170 © John Bird Published by Taylor and Francis
EXERCISE 46, Page 110 1. Name the types of quadrilateral shown in diagrams (i) to (iv) below, and determine for each (a) the area, and (b) the perimeter.
(i) Rhombus (a) Area = 4 3.5 = 14 cm 2 (b) Perimeter = 4 + 4 + 4 + 4 = 16 cm (ii) Parallelogram (a) Area = 30 6 = 180 mm 2 (b) Perimeter = 30 + 30 + 2
6
2
82 = 80 mm
(iii) Rectangle (a) Area = 120 30 = 3600 mm 2 (b) Perimeter = (2 120) + (2 30) = 300 mm (iv) Trapezium (a) Area =
1 26 12 10 = 190 cm 2 2
(b) Perimeter = 26 + 12 + 102 102
4
2
102
= 26 + 12 + 14.14 + 10.77 = 62.91 cm
171 © John Bird Published by Taylor and Francis
2. A rectangular plate is 85 mm long and 42 mm wide. Find its area in square centimetres. 85 mm = 8.5 cm and 42 mm = 4.2 cm
Area of plate = 8.5 × 4.2 = 35.7 cm 2 3. A rectangular field has an area of 1.2 hectares and a length of 150 m. If 1 hectare = 10000 m2 find (a) its width, and (b) the length of a diagonal. Area of field = 1.2 ha = 1.2 10000 m 2 = 12000 m 2 (a) Area = length width from which, width =
(b) By Pythagoras, length of diagonal =
150
2
area 12000 = 80 m length 150
802 = 170 m
4. Find the area of a triangle whose base is 8.5 cm and perpendicular height 6.4 cm.
Area of triangle =
=
1 × base × perpendicular height 2
1 × 8.5 × 6.4 = 27.2 cm 2 2
5. A square has an area of 162 cm 2 . Determine the length of a diagonal. A square ABCD is shown below of side x cm. The diagonal is given by length AC
Area of square = x 2 = 162
172 © John Bird Published by Taylor and Francis
By Pythagoras, (AC) 2 x 2 x 2 2x 2 = 2 × 162 from which,
diagonal, AC =
2x 2 [2 162] = 18 cm
6. A rectangular picture has an area of 0.96 m 2 . If one of the sides has a length of 800 mm, calculate, in millimetres, the length of the other side.
Area = 0.96 m 2 = 0.96 106 mm 2 and area = length × breadth, i.e. 0.96 106 mm 2 = 800 mm × breadth from which,
breadth =
0.96 106 = 1200 mm 800
7. Determine the area of each of the angle iron sections shown in below.
(a) Area = 2 (7 2) + (1 1) = 28 + 1 = 29 cm 2 (b) Area = (30 8) + 10(25 – 8 – 6) + (6 50) = 240 + 110 + 300 = 650 mm 2
8. The diagram below shows a 4 m wide path around the outside of a 41 m by 37 m garden. Calculate the area of the path.
173 © John Bird Published by Taylor and Francis
Area of garden = 41 × 37 m 2 Area of garden, neglecting the path = (41 – 8) × (37 – 8) = 33 × 29 m 2 Hence, area of path = (41 × 37) – (33 × 29) = 1517 – 957 = 560 m 2
9. The area of a trapezium is 13.5 cm2 and the perpendicular distance between its parallel sides is 3 cm. If the length of one of the parallel sides is 5.6 cm, find the length of the other parallel side.
Area of a trapezium =
i.e. i.e.
13.5 =
1 × (sum of parallel sides) × (perpendicular distance between the 2 parallel sides) 1 × (5.6 + x) × (3) 2
where x is the unknown parallel side
27 = 3(5.6 + x)
i.e.
9 = 5.6 + x
from which,
x = 9 – 5.6 = 3.4 cm
10. Calculate the area of the steel plate shown below.
2 1 Area of steel plate = (25 60) + (140 – 60)(25) + 25 + (50 25) + 55 50 2
= 1500 + 2000 + 625 + 1250 + 1375 = 6750 mm 2
174 © John Bird Published by Taylor and Francis
11. Determine the area of an equilateral triangle of side 10.0 cm.
An equilateral triangle ABC is shown below.
Perpendicular height, AD =
10.02 5.02
by Pythagoras
= 8.6603 cm Hence, area of triangle =
=
1 × base × perpendicular height 2
1 × 10.0 × 8.6603 2
= 43.30 cm 2 12. If paving slabs are produced in 250 mm by 250 mm squares, determine the number of slabs
required to cover an area of 2 m 2 .
Number of slabs =
2 106 mm 2 = 32 250 250
175 © John Bird Published by Taylor and Francis
EXERCISE 47, Page 111 1. A rectangular garden measures 40 m by 15 m. A 1 m flower border is made round the two shorter
sides and one long side. A circular swimming pool of diameter 8 m is constructed in the middle of the garden. Find, correct to the nearest square metre, the area remaining. A sketch of a plan of the garden is shown below.
Shaded area = (40 15) – [(15 1) + (38 1) + (15 1) + 42 ] = 600 – [15 + 38 + 15 + 16] = 600 – 118.27 = 481.73 m 2 = 482 m 2 , correct to the nearest square metre. 2. Determine the area of circles having (a) a radius of 4 cm (b) a diameter of 30 mm (c) a
circumference of 200 mm. (a) Area = r 2 4 = 50.27 cm 2 2
d 2 30 d = 706.9 mm 2 (b) Area = r 2 2 4 4 2
2
(c) Circumference = 2πr = 200 mm, from which, radius, r =
200 100 mm 2
2
100 2 Hence, area = r = 3183 mm 2
176 © John Bird Published by Taylor and Francis
3. An annulus has an outside diameter of 60 mm and an inside diameter of 20 mm. Determine its area. d 2 2 d12 60 20 602 202 = 2513 mm 2 Area of annulus = 4 4 4 4 4 2
2
4. If the area of a circle is 320 mm2, find (a) its diameter, and (b) its circumference.
(a) Area of circle, A =
d 2 4
i.e. 320 =
from which, diameter, d =
d 2 4
4 320 = 20.185 mm = 20.19 mm correct to 2 decimal places.
(b) Circumference of circle = 2r = d = 20.185 = 63.41 mm
5. Calculate the areas of the following sectors of circles: (a) radius 9 cm, angle subtended at centre 75 (b) diameter 35 mm, angle subtended at centre 4837'
2 75 2 2 (a) Area of sector = r 9 = 53.01 cm 360 360 37 2 48 2 60 35 2 (b) Area of sector = = 129.9 mm r 360 2 360 6. Determine the shaded area of the template shown below.
177 © John Bird Published by Taylor and Francis
Area of template = shaded area = (120 90) -
1 2 80 = 10800 – 5026.55 = 5773 mm 2 4
7. An archway consists of a rectangular opening topped by a semi-circular arch as shown below.
Determine the area of the opening if the width is 1 m and the greatest height is 2 m.
The semicircle has a diameter of 1 m, i.e. a radius of 0.5 m. Hence, the archway shown is made up of a rectangle of sides 1 m by 1.5 m and a semicircle of radius 0.5 m. Thus, area of opening = (1.5 1) +
1 2 0.5 = 1.5 + 0.393 = 1.89 m 2 2
178 © John Bird Published by Taylor and Francis
EXERCISE 48, Page 114 1. Calculate the area of a regular octagon if each side is 20 mm and the width across the flats is
48.3 mm. The octagon is shown sketched below and is comprised of 8 triangles of base length 20 mm and perpendicular height 48.3/2
48.3 1 = 1932 mm 2 Area of octagon = 8 20 2 2
2. Determine the area of a regular hexagon which has sides 25 mm.
The hexagon is shown sketched below and is comprised of 6 triangles of base length 25 mm and perpendicular height h as shown.
Tan 30 =
12.5 12.5 from which, h = = 21.65 mm h tan 30
1 Hence, area of hexagon = 6 25 21.65 = 1624 mm 2 2 179 © John Bird Published by Taylor and Francis
3. A plot of land is in the shape shown below. Determine (a) its area in hectares (1 ha = 104 m2), and
(b) the length of fencing required, to the nearest metre, to completely enclose the plot of land.
(a) Area of land = (30 10) +
1 1 1 2 30 + 70 40 + 70 100 80 45 15 2 2 2
= 300 + 450 + 1400 +[7000 – 937.5] = 9176 m 2 =
9176 ha = 0.918 ha 104
(b) Perimeter = 20 + 10 + 30 + 10 + 20 + 20 +
1 2 30 + 20 + 2 +
15
2
70
2
402 + 40
152 + 45 +
20
2
152 + 20
= 110 + 30 + 20 + 80.62 + 40 + 21.21 + 45 + 25 + 20 = 456 m, to the nearest metre.
180 © John Bird Published by Taylor and Francis
EXERCISE 49, Page 115 1. The area of a park on a map is 500 mm2. If the scale of the map is 1 to 40000 determine the true
area of the park in hectares (1 hectare = 104 m2) 500 106 40000 Area of park = 500 10 40000 m ha = 80 ha 104 2
6
2
2
2. A model of a boiler is made having an overall height of 75 mm corresponding to an overall height
of the actual boiler of 6 m. If the area of metal required for the model is 12500 mm 2 , determine, in square metres, the area of metal required for the actual boiler.
The scale is
6000 : 1 i.e. 80 : 1 75
Area of metal required for actual boiler = 12500 106 m 2 80 = 80 m 2 2
3. The scale of an Ordnance Survey map is 1:2500. A circular sports field has a diameter of 8 cm
on the map. Calculate its area in hectares, giving your answer correct to 3 significant figures. (1 hectare = 104 m 2 )
d 2 8 Area of sports field on map = 104 m 2 4 4 2
8 2 2 104 2500 8 2 104 2500 m 2 4 ha True area of sports field = 4 104 2
= 3.14 ha
181 © John Bird Published by Taylor and Francis
