CHAPTER 27 AREAS OF COMMON SHAPES
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1 EXERCISE 113 Page 65 CHAPTER 7 AREAS OF COMMON SHAPES 1. Find the angles p and q in the diagram below: p = = 105 (interior opposite angles of a parallelogram are equal) q = = 35. Find the angles r and s in the diagram below: r = = 1 (the 38 angle is the alternate angle between parallel lines) s = = Find the angle t in the diagram below: t = = , John Bird
2 EXERCISE 11 Page Name the types of quadrilateral shown in (i) to (iv) below, and determine for each (a) the area, and (b) the perimeter. (i) Rhombus (a) Area = 3.5 = 1 cm (b) Perimeter = = 16 cm (ii) Parallelogram (a) Area = 30 6 = 180 mm (b) Perimeter = ( ) = 80 mm (iii) Rectangle (a) Area = = 3600 mm (b) Perimeter = ( 10) + ( 30) = 300 mm (iv) Trapezium = 190 cm (a) Area = ( ) (b) Perimeter = ( ) + ( + 10 ) = = 6.91 cm 39 01, John Bird
3 . A rectangular plate is 85 mm long and mm wide. Find its area in square centimetres. Area of plate = length width = 85 mm = 8.5. cm = 35.7 cm 3. A rectangular field has an area of 1. hectares and a length of 150 m. If 1 hectare = m, find (a) its width, and (b) the length of a diagonal. Area of field = 1. ha = m = m (a) Area = length width from which, width = (b) By Pythagoras, length of diagonal = ( ) area 1000 = = 80 m length = 170 m. Find the area of a triangle whose base is 8.5 cm and perpendicular height 6. cm. Area of triangle = 1 base perpendicular height = = 7. cm 5. A square has an area of 16 cm. Determine the length of a diagonal. Let each side of square = x, then area = x = 16 from which, x = 16 Using Pythagoras theorem, x + x = d where d is the length of the diagonal i.e. ( 16 ) + ( 16 ) = d i.e = d from which, diagonal, d = 3 = 18 cm 6. A rectangular picture has an area of 0.96 m. If one of the sides has a length of 800 mm, calculate, in millimetres, the length of the other side. 0 01, John Bird
4 Area of picture = length width from which, width = area length mm = = 100 mm 800 mm 7. Determine the area of each of the angle iron sections shown below. (a) Area = (7 ) + (1 1) = = 9 cm (b) Area = (30 8) + 10(5 8 6) + (6 50) = = 650 mm 8. The diagram shows a m wide path around the outside of a 1 m by 37 m garden. Calculate the area of the path. Area of path = (1 37) [(1 8) (37 8)] = = 560 m 9. The area of a trapezium is 13.5 cm 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 = 1 (sum of parallel sides) (perpendicular distance between the parallel sides) Hence, area of trapezium = 13.5 = 1 (5.6 + b) 3 where b is the unknown parallel side i.e. 7 = b 1 01, John Bird
5 i.e = 3b i.e. 10. = 3b and the unknown parallel side, b = 10. = 3. cm Calculate the area of the steel plate shown. Area of steel plate = (5 60) + (10 60)(5) + ( ) 5 + (50 5) + = = 6750 mm Determine the area of an equilateral triangle of side 10.0 cm. An equilateral triangle is shown below. Area of triangle = 1 base height By Pythagoras, h = 10.0 and h = = 8.66 cm 01, John Bird
6 Area of triangle = = 3.30 cm 1. If paving slabs are produced in 50 mm by 50 mm squares, determine the number of slabs required to cover an area of m. Number of slabs = 10 mm = The diagram shows a plan view of an office block to be built. The walls will have a height of 8 m, and it is necessary to make an estimate of the number of bricks required to build the walls. Assuming that any doors and windows are ignored in the calculation and that 8 bricks are required to build 1 m of wall, calculate the number of external bricks required. Length of outside wall = = 600 m Area of wall = = 800 m Number of external bricks required = = , John Bird
7 EXERCISE 115 Page A rectangular garden measures 0 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 = (0 15) [(15 1) + (38 1) + (15 1) + π ( ) ] = 600 [ π] = = m = 8 m, correct to the nearest square metre. Determine the area of circles having (a) a radius of cm, (b) a diameter of 30 mm and (c) a circumference of 00 mm. (a) Area of circle = πr = π () = 16π = 50.7 cm π d (b) Area of circle = π (30) = = 900 π = mm c (c) Circumference, c = πr hence radius, r = π = 00 π = 100 π mm Area of circle = πr 100 = π π = 100 π = 3183 mm or cm 3. An annulus has an outside diameter of 60 mm and an inside diameter of 0 mm. Determine its area. 01, John Bird
8 Area of shaded part = area of large circle area of small circle = π D π d = π (D d ) = π (60 0 ) = 513 mm. If the area of a circle is 30 mm, find (a) its diameter and (b) its circumference. π d (a) Area of circle = π d hence 30 = 30 i.e. d 30 = and diameter, d = π π = = 0.19 mm (b) Circumference, c = πr = πd = π = 63.1 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 8 37' (a) Area of sector = θ (πr ) = (π9 ) = 75 π = cm (b) If diameter = 35 mm, then radius, r = 35/ =17.5 mm, and 8 37 ' area of sector = ( π 17.5 ) ( 17.5 ) 360 π = ( π 17.5 ) = 19.9 mm 360 = 5 01, John Bird
9 6. Determine the shaded area of the template shown. Area of template = shaded area = (10 90) 1 ( 80) π = = 5773 mm 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 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 π 0.5 = = 1.89 m 8. The major axis of an ellipse is 00 mm and the minor axis 100 mm. Determine the area and perimeter of the ellipse. If the major axis = 00 mm, then the semi-major axis, a = 100 mm If the minor axis = 100 mm, then the semi-minor axis, b = 50 mm Hence, area of ellipse = πab = π(100)(50) = cm and perimeter of ellipse π(a + b) = π( ) = 150π = 71 cm 6 01, John Bird
10 9. If fencing costs 15 per metre, find the cost (to the nearest pound) of enclosing an elliptical plot of land which has major and minor diameter lengths of 10 m and 80 m Perimeter of ellipse = π(a + b) = π + = 100π m Hence, cost of fencing = π = 71 to the nearest pound 10. A cycling track is in the form of an ellipse, the axes being 50 m and 150 m, respectively, for the inner boundary, and 70 m and 170 m for the outer boundary. Calculate the area of the track Area of an ellipse = πab, hence area of cycle track = π π = π(135)(85) π(15)(75) = 6597 m 7 01, John Bird
11 EXERCISE 116 Page Calculate the area of a regular octagon if each side is 0 mm and the width across the flats is 8.3 mm. The octagon is shown sketched below and comprises eight triangles of base length 0 mm and perpendicular height 8.3/ Area of octagon = = 193 mm. Determine the area of a regular hexagon which has sides 5 mm. The hexagon is shown sketched below and comprises six triangles of base length 5 mm and perpendicular height h as shown. Tan 30 = 1.5 h from which, h = 1.5 tan 30 = 1.65 mm Hence, area of hexagon = = 16 mm 8 01, John Bird
12 3. A plot of land is in the shape shown below. Determine (a) its area in hectares (1 ha = 10 m ), 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 ( 30) π + ( )( ) 1 + ( ) ( )( 15) = π [ ] = 9176 m = ha = ha 1 30 (b) Perimeter = ( π ) ( 70 0) = π = 56 m, to the nearest metre ( ) ( 0 15) , John Bird
13 EXERCISE 117 Page 7 1. The area of a park on a map is 500 mm. If the scale of the map is 1 to 0 000, determine the true area of the park in hectares (1 hectare = 10 m ) Area of park = ( ) ( ) m = ha = 80 ha 10. 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 mm, determine, in square metres, the area of metal required for the actual boiler. The scale is :1 i.e. 80:1 Area of metal required for actual boiler = m ( 80 ) = 80 m 3. The scale of an Ordnance Survey map is 1:500. 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 = 10m ) Area of sports field on map = ( 8) π d π = 10 m True area of sports field = ( 8) π π ( 8) 10 ( 500) 10 ( 500) m = ha 10 = 3.1 ha 50 01, John Bird
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