\nThe surface area of a cube is 384 cm2<\/sup>. Find
\n(i) the length of an edge
\n(ii) volume of the cube.
\nSolution:
\nSurface area of a cube = 384 cm2<\/sup>
\n
![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-18-Mensuration-Ex-18.4-Q1.1.png\")
\n(ii) Volume = (Edge)3<\/sup> = (8)3\u00a0<\/sup>= 8 \u00d7 8 \u00d7 8 cm3<\/sup> = 512 cm3<\/sup><\/p>\n
Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Question 16. Question 17. Question 18.
\nFind the total surface area of a solid cylinder of radius 5 cm and height 10 cm. Leave your answer in terms of n.
\nSolution:
\nRadius of a solid cylinder (r) = 5 cm
\nHeight (h) = 10 cm
\nTotal surface area = 2\u03c0rh + 2\u03c0r2<\/sup>
\n= 2r\u03c0(h + r)
\n= 2\u03c0 \u00d7 5(10 + 5)
\n= \u03c0 \u00d7 10 \u00d7 15 = 150\u03c0 cm2<\/sup><\/p>\n
\nAn aquarium is in the form of a cuboid whose external measures are 70 cm \u00d7 28 cm \u00d7 35 cm. The base, side faces and back face are to be covered with coloured paper. Find the area of the paper needed.
\nSolution:
\nA cuboid shaped aquarium,
\nLength (l) = 70 cm
\nBreadth (b) = 28 cm
\nand height (h) = 35 cm
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-18-Mensuration-Ex-18.4-Q3.1.png\")
\nArea of base = 70 \u00d7 28 cm3<\/sup> = 1960 cm3<\/sup>
\nArea of side face = (28 \u00d7 35) \u00d7 2 cm2\u00a0<\/sup>= 1960 cm2<\/sup>
\nArea of back face = 70 \u00d7 35 cm2<\/sup> = 2450 cm
\n\u2234. Total area = 1960 + 1960 + 2450 = 6370 cm2<\/sup>
\n\u2234 Area of paper required = 6370 cm2<\/sup><\/p>\n
\nThe internal dimensions of rectangular hall are 15 m \u00d7 12 m \u00d7 4 m. There are 4 windows each of dimension 2 m \u00d7 1.5 m and 2 doors each of dimension 1.5 m \u00d7 2.5 m. Find the cost of white washing all four walls of the hall, if the cost of white washing is \u20b95 per m2<\/sup>. What will be the cost of white washing if the ceiling of the hall is also white washed?
\nSolution:
\nInternal dimension of rectangular hall = 15m \u00d7 12 m \u00d7 4 m
\nArea of 4-walls = 2(l + b) \u00d7 h
\n= 2(15 + 12) \u00d7 4
\n= 2 \u00d7 27 \u00d7 4 m2<\/sup>
\n= 216 m2<\/sup>
\nArea of 4 windows of size = 2 \u00d7 1.5 = 2 \u00d7 1.5 \u00d7 4 = 12 m2<\/sup>
\nArea of 2 door of size = 1.5 \u00d7 2.5 = 2 \u00d7 1.5 \u00d7 2.5 = 7.5 m2<\/sup>
\n\u2234 Area of remaining hall = 216 – (12 + 7.5) = 216 – 19.5 m2<\/sup> = 196.5 m2<\/sup>
\nArea of ceiling = l \u00d7 b = 15 \u00d7 12= 180 m2<\/sup>
\nCost of white washing the walls at the rate of \u20b95 per m2
\n<\/sup>= 196.5 \u00d7 5 = \u20b9982.50
\nArea of ceiling = l \u00d7 b = 15 \u00d7 12= 180 m2<\/sup>
\nCost of white washing = 180 \u00d7 5 = \u20b9900
\n\u2234 Total cost = \u20b9982.50 + 900.00 = \u20b91882.50<\/p>\n
\nA swimming pool is 50 m in length, 30 m in breadth and 2\u00b75 m in depth. Find the cost of cementing its floor and walls at the rate of \u20b927 per square metre.
\nSolution:
\nLength of swimming pool = 50 m
\nBreadth of swimming pool = 30 m
\nDepth (Height) of swimming pool = 2\u00b75 m
\nArea of floor = 50 \u00d7 30 = 1500 m2<\/sup>
\nArea of four walls = 2 (50 + 30) \u00d7 2\u00b75 = 160 \u00d7 2\u00b75 = 400 m2<\/sup>
\nArea to be cemented = 1500 m2<\/sup> + 400 m2<\/sup> = 1900 m2<\/sup>
\nCost of cementing 1m2 = \u20b927
\nCost of cementing 1900m2\u00a0<\/sup>= \u20b927 \u00d7 1900 = \u20b951300<\/p>\n
\nThe floor of a rectangular hall has a perimeter 236 m. Its height is 4\u00b75 m. Find the cost of painting its four walls (doors and windows be ignored) at the rate of Rs. 8.40 per square metre.
\nSolution:
\nPerimeter of Hall = 236 m.
\nHeight = 4\u00b75 m
\nPerimeter = 2 (l + b) = 236 m
\nArea of four walls = 2 (l + b) \u00d7 h = 236 \u00d7 4\u00b75 = 1062 m2<\/sup>
\nCost of painting 1 m2<\/sup> = \u20b98\u00b740
\nCost of painting 1062 m2<\/sup> = \u20b98\u00b740 \u00d7 1062 = \u20b98920\u00b780<\/p>\n
\nA cuboidal fish tank has a length of 30 cm, a breadth of 20 cm and a height of 20 cm. The tank is placed on a horizontal table and it is three-quarters full of water. Find the area of the tank which is in contact with water.
\nSolution:
\nLength of tank = 30 cm
\nBreadth of tank = 20 cm
\nHeight of tank = 20 cm
\nAs the tank is three-quarters full of water
\n\u2234 Height of water in the tank = \\(\\frac{20 \\times 3}{4}\\) = 15 cm
\nArea of the tank in contact with the water = Area of floor of Tank
\n+ Area of 4 walls upto 15 cm
\n= 30 \u00d7 20 + 2 (30 + 20) \u00d7 15
\n= 600 + 2 \u00d7 50 \u00d7 15
\n= 600 + 1500 = 2100 cm2<\/sup><\/p>\n
\nThe volume of a cuboid is 448 cm3<\/sup>. Its height is 7 cm and the base is a square. Find
\n(i) a side of the square base
\n(ii) surface area of the cuboid.
\nSolution:
\nVolume of a cuboid = 448 cm3<\/sup>
\nHeight = 7 cm
\n\u2234 Area of base = \\(\\frac{448}{7}\\) = 64 cm2<\/sup>
\n\u2235 Base is a square.
\n(i) \u2234 Side of square base = \\(\\sqrt{64}\\) = 8 cm
\n(ii) Surface area of the cuboid = 2 [lb + bh + hl]
\n= 2[8 \u00d7 8 + 8 \u00d7 7 + 7 \u00d7 8] cm2<\/sup>
\n= 2[64 + 56 + 56]
\n= 2 \u00d7 176 = 352 cm2<\/sup><\/p>\n
\nThe length, breadth and height of a rectangular solid are in the ratio 5 : 4 : 2. If its total surface area is 1216 cm2<\/sup>, find the volume of the solid.
\nSolution:
\nRatio in length, breadth and height of a rectangular solid = 5 : 4 : 2
\nTotal surface area =1216 cm2<\/sup>
\nLet Length = 5x,
\nBreadth = 4x
\nand height = 2x
\nTotal surface area = 2[5x \u00d7 4x + 4x \u00d7 2x + 2x \u00d7 5x]
\n= 2[20x2<\/sup> + 8x2<\/sup> + 10x2<\/sup> ]
\n= 2 \u00d7 38x2<\/sup> = 76x2<\/sup>
\n\u2234 94x2<\/sup> = 1216
\n\u21d2 x2<\/sup> = \\(\\frac{1216}{76}\\) = 16 = (4)2<\/sup>
\n\u2234 x = 4
\n\u2234 Length = 5 \u00d7 4 = 20 cm
\nBreadth = 4 \u00d7 4 = 16 cm
\nHeight = 2 \u00d7 4 = 8 cm
\nand volume = lbh = 20 \u00d7 16 \u00d7 8 = 2560 cm3<\/sup><\/p>\n
\nA rectangular room is 6 m long, 5 m wide and 3\u00b75 m high. It has 2 doors of size 1\u00b71 m by 2 m and 3 windows of size 1\u00b75 m by 1\u00b74 m. Find the cost of whitewashing the walls and the ceiling of the room at the rate of \u20b95\u00b730 per square metre.
\nSolution:
\nLength of room = 6 m
\nBreadth of room = 5 m
\nHeight of room = 3\u00b75 m
\nArea of four walls = 2 (l + b) \u00d7 h
\n= 2 (6 + 5) \u00d7 3\u00b75 = 77 m2<\/sup>
\nArea of 2 doors and 3 windows
\n= (2 \u00d7 1\u00b71 \u00d7 2 + 3 \u00d7 1\u00b75 \u00d7 1\u00b74)
\n= (44 + 6\u00b73) m2<\/sup> = 10\u00b77 m2<\/sup>
\nArea of ceiling = l \u00d7 b = 6 \u00d7 5 = 30 m2<\/sup>
\nTotal area for white washing
\n= (77 – 10\u00b77 + 30) m2<\/sup> = 96\u00b73 m2<\/sup>
\nCost of white washing = \u20b9(96\u00b73 \u00d7 5\u00b730) = \u20b9510\u00b739<\/p>\n
\nA cuboidal block of metal has dimensions 36 cm by 32 cm by 0\u00b725 m. It is melted and recast into cubes with an edge of 4 cm.
\n(i) How many such cubes can be made?
\n(ii) What is the cost of silver coating the surfaces of the cubes at the rate of \u20b90\u00b775 per square centimetre?
\nSolution:
\n(i) Length of cuboid = 36 cm
\nBreadth of cuboid = 32 cm
\nHeight of cuboid = 0\u00b725 \u00d7 100 = 25 cm
\nVolume of cubiod = lbh = (36 \u00d7 32 \u00d7 25) cm3<\/sup> = 28800 cm3<\/sup>.
\nVolume of cube = (side)2\u00a0<\/sup>= (4)2<\/sup> = 64 cm2<\/sup>
\nNumber of cubes recasting from cubiod = \\(\\frac{28800}{64}\\) = 450
\n(ii) Surface area of 1 cube = 6 \u00d7 a2\u00a0<\/sup>= 6 \u00d7 16 = 96 cm2<\/sup>
\nSurface area of 450 cubes = 96 \u00d7 450 = 43200 cm2<\/sup>
\nCost of silver coating on cubes = \u20b90\u00b775 \u00d7 43200 = \u20b932400<\/p>\n
\nThree cubes of silver with edges 3 cm, 4 cm and 5 cm are melted and recast into a single cube, find the cost of coating the surface of the new cube with gold at the rate of \u20b93\u00b750 per square centimetre?
\nSolution:
\nLet a cm be the edge of new cube
\n\u2234 According to question,
\na3<\/sup> = 33<\/sup> + 43<\/sup> + 53<\/sup> = 27 + 64 + 125 = 216 cm3<\/sup>
\na = \\(\\sqrt[3]{216}\\)
\n\u21d2 a = 6 cm.
\nSurface area of new cube = 6 \u00d7 (side)2\u00a0<\/sup>= 6 \u00d7 (6)2<\/sup> = 216 cm2<\/sup>
\nCost of coating the surface of new cube = \u20b93\u00b750 \u00d7 216 = \u20b9156<\/p>\n
\nThe curved surface area of a hollow cylinder is 4375 cm2<\/sup>, it is cut along its height and formed a rectangular sheet of width 35 cm. Find the perimeter of the rectangular sheet.
\nSolution:
\nCurved surface area of a hollow cylinder = 4375 cm2<\/sup>
\nBy cutting it from the height,
\nit becomes a rectangular sheet whose width = 35 cm
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-18-Mensuration-Ex-18.4-Q13.1.png\")
\n\u2234Height of cylinder = 35 cm
\n\u2234 Length of sheet = \\(\\frac{\\text { Area }}{\\text { Height }}\\)
\n= \\(\\frac{4375}{35}\\) = 125 cm
\nNow perimeter of the sheet = 2(l + b)
\n= 2 \u00d7 (125 + 35)
\n= 2 \u00d7 160 = 320 cm<\/p>\n
\nA road roller has a diameter 0.7 m and its width is 1.2 m. Find the least number of revolutions that the roller must take in order to level a playground of size 120 m \u00d7 44 m.
\nSolution:
\nDiameter of a road roller = 0.7 m = 70 cm
\n\u2234 Radius (r) = \\(\\frac{70}{2}\\) cm = 35 cm = \\(\\frac{35}{100}\\)m
\nand width (h) = 1.2 m
\nCurved surface area = 2\u03c0rh
\n= \\(2 \\times \\frac{22}{7} \\times \\frac{35}{100} \\times 1.2 \\mathrm{m}^{2}\\)
\n= \\(\\frac{264}{100} \\mathrm{m}^{2}\\)
\nArea of polyground = 120 m \u00d7 44 m
\n= 120 \u00d7 44 m2<\/sup>
\n= 5280 m2<\/sup>
\nNumber of revolution made by the road roller
\n= \\(\\frac{5280}{264}\\) \u00d7 100
\n= 2000 revolutions<\/p>\n
\nA company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label?
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-18-Mensuration-Ex-18.4-Q15.1.png\")
\nSolution:
\nDiameter of cylindrical container = 14 cm
\n\u2234 Radius (r) = \\(\\frac{14}{2}\\) = 7 cm
\nand height (h) = 20 cm
\nWidth of lable = 2 0 – (2 + 2) cm = 20 – 4 = 16 cm
\n\u2234 Area of lable = 2\u03c0rh = 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 16 = 704 cm2<\/sup><\/p>\n
\nThe sum of the radius and height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 cm2<\/sup>. Find the height and the volume of the cylinder.
\nSolution:
\nSum of height and radius of a cylinder = 37 cm
\nTotal surface area = 1628 cm2<\/sup>
\nLet radius be r, then height = (37 – r) cm
\nTotal surface area = 2\u03c0(h + r)
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-18-Mensuration-Ex-18.4-Q16.1.png\")
\n\u21d2 r = 7 cm
\nHeight = 37 – 7 = 30 cm
\nNow, volume = \u03c0r2<\/sup>h = \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 7 \u00d7 30 cm3\u00a0<\/sup>= 4620 cm3<\/sup><\/p>\n
\nThe ratio between the curved surface and total surface of a cylinder is 1 : 2. Find the volume of the cylinder, given that its total surface area is 616 cm3<\/sup>.
\nSolution:
\nRatio between curved surface and total surface area of a cylinder = 1 : 2
\nTotal surface area = 616 cm2<\/sup>
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-18-Mensuration-Ex-18.4-Q17.1.png\")
<\/p>\n
\nThe given figure shown a metal pipe 77 cm long. The inner diameter of cross section is 4 cm and the outer one is 4.4 cm.
\nFind its
\n(i) inner curved surface area
\n(ii) outer curved surface area
\n(iii) total surface area.
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-18-Mensuration-Ex-18.4-Q18.1.png\")
\nSolution:
\nLength of metal pipe (h) = 77 cm
\nInner diameter = 4 cm
\nand outer diameter = 4.4 cm
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-18-Mensuration-Ex-18.4-Q18.2.png\")
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-18-Mensuration-Ex-18.4-Q18.3.png\")
\nTofal surface area = 968 + 1064.8 + 5.28 = 2038.08 cm3<\/sup><\/p>\nML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"