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. Find the angles r and s in diagram (b) above. r = 180-38 = 14 (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 = 360-6 - 95-57 = 146 170
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 (b) Perimeter = 4 + 4 + 4 + 4 = 16 cm (ii) Parallelogram (a) Area = 30 6 = 180 (b) Perimeter = 30 + 30 + 6 8 = 80 (iii) Rectangle (a) Area = 10 30 = 3600 (b) Perimeter = ( 10) + ( 30) = 300 (iv) Trapezium 1 6 1 10 = 190 (a) Area = cm (b) Perimeter = 6 + 1 + 10 10 4 10 = 6 + 1 + 14.14 + 10.77 = 6.91 cm 171
. A rectangular plate is 85 long and 4 wide. Find its area in square centimetres. 85 = 8.5 cm and 4 = 4. cm Area of plate = 8.5 4. = 35.7cm 3. A rectangular field has an area of 1. hectares and a length of 150 m. If 1 hectare = 10000 m find (a) its width, and (b) the length of a diagonal. Area of field = 1. ha = 1. 10000 m = 1000 m (a) Area = length width from which, width = (b) By Pythagoras, length of diagonal = 150 80 area 1000 = 80 m length 150 = 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 = 1 8.5 6.4 = 7. cm 5. A square has an area of 16 cm. 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 = 16 17
By Pythagoras, (AC) x x x = 16 from which, diagonal, AC = x [ 16] = 18 cm 6. A rectangular picture has an area of 0.96 m. If one of the sides has a length of 800, calculate, in millimetres, the length of the other side. Area = 0.96 m = 6 0.96 10 and area = length breadth, i.e. 6 0.96 10 = 800 breadth from which, breadth = 0.9610 800 6 = 100 7. Determine the area of each of the angle iron sections shown in below. (a) Area = (7 ) + (1 1) = 8 + 1 = 9 cm (b) Area = (30 8) + 10(5 8 6) + (6 50) = 40 + 110 + 300 = 650 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
Area of garden = 41 37 m Area of garden, neglecting the path = (41 8) (37 8) = 33 9 m Hence, area of path = (41 37) (33 9) = 1517 957 = 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) i.e. 13.5 = 1 (5.6 + x) (3) where x is the unknown parallel side i.e. 7 = 3(5.6 + x) i.e. from which, 9 = 5.6 + x x = 9 5.6 = 3.4 cm 10. Calculate the area of the steel plate shown below. Area of steel plate = (5 60) + (140 60)(5) + 5 + (50 5) + = 1500 + 000 + 65 + 150 + 1375 = 6750 1 55 50 174
11. Determine the area of an equilateral triangle of side 10.0 cm. An equilateral triangle ABC is shown below. Perpendicular height, AD = 10 0 5 0.. by Pythagoras = 8.6603 cm Hence, area of triangle = 1 base perpendicular height = 1 10.0 8.6603 = 43.30 cm 1. If paving slabs are produced in 50 by 50 squares, determine the number of slabs required to cover an area of m. Number of slabs = 10 50 50 6 = 3 175
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 swiing 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) + 4 ] = 600 [15 + 38 + 15 + 16] = 600 118.7 = 481.73 m = 48m, correct to the nearest square metre.. Determine the area of circles having (a) a radius of 4 cm (b) a diameter of 30 (c) a circumference of 00. (a) Area = r 4 = 50.7cm (b) Area = d d 30 r 4 4 = 706.9 (c) Circumference = πr = 00, from which, radius, r = 00 100 Hence, area = 100 r = 3183 176
3. An annulus has an outside diameter of 60 and an inside diameter of 0. Determine its area. Area of annulus = d 60 0 1 60 0 = 513 d 4 4 4 4 4 4. If the area of a circle is 30, find (a) its diameter, and (b) its circumference. (a) Area of circle, A = d 4 i.e. 30 = d 4 from which, diameter, d = 430 = 0.185 = 0.19 correct to decimal places. (b) Circumference of circle = r = d = 0.185 = 63.41 5. Calculate the areas of the following sectors of circles: (a) radius 9 cm, angle subtended at centre 75 (b) diameter 35, angle subtended at centre 4837' 75 360 360 (a) Area of sector = r 9 = 53.01cm 37 48 35 (b) Area of sector = r 60 = 19.9 360 360 6. Determine the shaded area of the template shown below. 177
Area of template = shaded area = (10 90) - 1 80 4 = 10800 506.55 = 5773 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 0.5 = 1.5 + 0.393 = 1.89 m 178
EXERCISE 48, Page 114 1. Calculate the area of a regular octagon if each side is 0 and the width across the flats is 48.3. The octagon is shown sketched below and is comprised of 8 triangles of base length 0 and perpendicular height 48.3/ 1 48.3 Area of octagon = 8 0 = 193. Determine the area of a regular hexagon which has sides 5. The hexagon is shown sketched below and is comprised of 6 triangles of base length 5 and perpendicular height h as shown. Tan 30 = 1.5 h from which, h = 1.5 tan 30 = 1.65 Hence, area of hexagon = 1 6 51.65 = 164 179
3. A plot of land is in the shape shown below. Determine (a) its area in hectares (1 ha = 10 4 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 + 1 70 40 1 + 70100 80 4515 = 300 + 450 + 1400 +[7000 937.5] 9176 = 9176 m = ha = 0.918 ha 4 10 1 30 (b) Perimeter = 0 + 10 + 30 + 10 + 0 + 0 + + 0 + 70 40 = 110 + 30 + 0 + 80.6 + 40 + 1.1 + 45 + 5 + 0 = 456 m, to the nearest metre. + 40 + 15 15 + 45 + 0 15 + 0 180
EXERCISE 49, Page 115 1. The area of a park on a map is 500. If the scale of the map is 1 to 40000 determine the true area of the park in hectares (1 hectare = 10 4 m ) Area of park = 6 6 50010 40000 50010 40000 m ha = 80 ha 4 10. A model of a boiler is made having an overall height of 75 corresponding to an overall height of the actual boiler of 6 m. If the area of metal required for the model is 1500, determine, in square metres, the area of metal required for the actual boiler. The scale is 6000 75 : 1 i.e. 80 : 1 Area of metal required for actual boiler = 1500 6 10 m 80 = 80m 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 = 10 4 m ) Area of sports field on map = 8 d 10 m 4 4 4 True area of sports field = 8 4 8 10 500 4 10 500 m 4 ha 4 4 10 = 3.14 ha 181