{"id":44059,"date":"2022-05-29T10:00:14","date_gmt":"2022-05-29T04:30:14","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=44059"},"modified":"2023-01-25T14:18:14","modified_gmt":"2023-01-25T08:48:14","slug":"ml-aggarwal-class-8-solutions-for-icse-maths-chapter-18-ex-18-1","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-18-ex-18-1\/","title":{"rendered":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1"},"content":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1<\/h2>\n

Question 1.
\nThe length and breadth of a rectangular field are in the ratio 9 : 5. If the area of the field is 14580 square metres, find the cost of surrounding the field with a fence at the rate of \u20b93\u00b725 per metre.
\nSolution:
\nLet the length = 9x and the breadth = 5x
\nArea = l \u00d7 b \u21d2 14580 = 9x \u00d7 5x
\n\u21d2 45x2<\/sup> = 14580
\n\u2234 x2<\/sup> = \\(\\frac{14580}{45}\\) = 324 \u21d2 x = \\(\\sqrt{324}\\)
\n\u21d2 x = \\(\\sqrt{18 \\times 18}\\)
\nor x = 18
\nLength = 9 \u00d7 18 = 162 m
\nBreadth = 5 \u00d7 18 = 90 m
\nPerimeter = 2(l + b)
\n= 2 (162 + 90) = 2(252)
\n= 504 m.
\n\u2234 Cost for 504 m fencing the surrounding
\nat the rate of \u20b93\u00b725 per metre = \u20b9(504 \u00d7 3\u00b725) = \u20b91638<\/p>\n

Question 2.
\nA rectangle is 16 m by 9 m. Find a side of the square whose area equals the area of the rectangle. By how much does the perimeter of the rectangle exceed the perimeter of the square?
\nSolution:
\nArea of rectangle = (16 \u00d7 9) m2<\/sup> = 144 m2<\/sup>
\nArea of square = Area of rectangle (given)
\n\u2234 (side)2<\/sup> = 144
\nSide = \\(\\sqrt{144}=\\sqrt{12 \\times 12}\\) = 12 m
\nPerimeter of square = 4 \u00d7 12 = 48 m
\nPerimeter of rectangle = 2(l + b) = 2 (16 + 9) = 50 m
\nDifference in their perimeters = 50 – 48 = 2 m<\/p>\n

Question 3.
\nTwo adjacent sides of a parallelogram are 24 cm and 18 cm. If the distance between longer sides is 12 cm, find the distance between shorter sides.
\nSolution:
\nTaking 24 cm as a base of parallelogram, its height is 12 cm.
\n\"ML
\n\u2234 Area of parallelogram = b \u00d7 h = 24 \u00d7 12 = 288 cm2<\/sup>
\nLet d cm be the distance between the shortest sides.
\n\u2234 Area of parallelogram = (18 \u00d7 d) cm2<\/sup>
\n\u21d2 18 \u00d7 d = 288
\n\u21d2 d = \\(\\frac{288}{18}\\) = 16 cm<\/p>\n

Question 4.
\nRajesh has a square plot with the measurement as shown in the given figure. He wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of \u20b950 per m2<\/sup>.
\n\"ML
\nSolution:
\nSide of square plot = 24 m
\nLength of house (l) = 18 m
\nand breadth (b) = 12m
\n\"ML
\nNow area of square plot = (24)2<\/sup> m2<\/sup> = 24 \u00d7 24 = 576 m2<\/sup>
\nand area of hosue = 18 \u00d7 12 = 216 m2<\/sup>
\nRemaining area of the garden = 576 – 216 = 360 m2<\/sup>
\nCost of developing the garden = \u20b950 per m2<\/sup>
\nTotal cost = \u20b950 \u00d7 360 = \u20b918000<\/p>\n

Question 5.
\nA flooring tile has a shape of a parallelogram whose base is 18 cm and the corresponding height is 6 cm. How many such tiles are required to cover a floor of area 540 m2<\/sup>? (If required you can split the tiles in whatever way you want to fill up the comers).
\nSolution:
\nBase of the parallelogram-shaped flooring tile = 18 cm
\nand height = 6 cm
\n\"ML
\n\u2234 Area of one tile = Base \u00d7 Height = 18 \u00d7 6 = 108 cm2<\/sup>
\nArea of floor = 540 m2<\/sup>
\n\u2234 Number of tiles = \\(\\frac{\\text { Total area }}{\\text { Area of one tile }}\\)
\n= \\(\\frac{540 \\times 100 \\times 100}{108}\\) = 50000<\/p>\n

Question 6.
\nAn ant is moving around a few food pieces of different shapes scattered on the floor. For which food piece would the ant have to take a longer round?
\n\"ML
\n\"ML
\nSolution:
\n(a) Diameter of semicircle = 2.8 cm
\n\u2234 Perimeter = \u03c0r + 2r
\n= \\(\\frac{22}{7}\\) \u00d7 2.8 + 2 \u00d7 2.8
\n= 8.8 + 5.6 cm = 14.4 cm
\n\"ML<\/p>\n

(b) Total perimeters
\n= 1.5 + 1.5 + 2.8 + Semi circular
\n= 5.8 + 8.8 = 14.6 cm
\n\"ML
\n(c) Total perimeter = 2 + 2 + Semi circumference
\n= 4 + 8.8 = 12.8 cm
\nIt is clear that distance of (b) i.e. 14.6 is longer.
\n\"ML<\/p>\n

Question 7.
\nIn the adjoining figure, the area enclosed between the concentric circles is 770 cm2. If the radius of the outer circle is 21 cm, calculate the radius of the inner circle.
\n\"ML
\nSolution:
\nRadius of outer circle (R) = 21 cm.
\nradius of inner circle (r) = r cm.
\nArea of shaded portion = 770 cm2<\/sup>
\n\u21d2 \u03c0 (R2<\/sup> – r2<\/sup>) = 770
\n\u21d2 \\(\\frac{22}{7}\\) (212<\/sup> – r2<\/sup>) = 770
\n\u21d2 441 – r2<\/sup> = 770 \u00d7 \\(\\frac{7}{22}\\) = 35 \u00d7 7 = 245
\n\u21d2 r2<\/sup> = 441 – 245
\n\u21d2 r2<\/sup> = 196
\n\u21d2 r2<\/sup> = 196
\n\u21d2 r = \\(\\sqrt{196}=\\sqrt{14 \\times 14}\\)
\n\u21d2 r = 14 cm<\/p>\n

Question 8.
\nA copper wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent into the form of a circle, find the area of the circle.
\nSolution:
\nArea of the square = 121 cm2<\/sup>
\n\u2234 Side = \\(\\sqrt{121}=\\sqrt{11 \\times 11}\\) = 11 cm
\nPerimeter = 4 a = 4 \u00d7 11= 44 cm
\nNow, circumference of the circle = 44 cm
\n\u2234 Radius = \\(\\frac{44 \\times 7}{2 \\times 22}\\) = 7cm
\nand area of the circle = \u03c0r2<\/sup> = \\(\\frac{22}{7}\\)(7)2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 7 = 154 cm2<\/sup><\/p>\n

Question 9.
\nFrom the given figure, find
\n(i) the area of \u2206 ABC
\n(ii) length of BC
\n(iii) the length of altitude from A to BC
\n\"ML
\nSolution:
\n(i) Base = 3 cm, height = 4 cm.
\nArea = \\(\\frac{1}{2}\\) \u00d7 base \u00d7 height
\n= \\(\\frac{1}{2}\\) \u00d7 3 \u00d7 4 = 6 cm2<\/sup>
\n(ii) By pythagoras theorem,
\nBC2<\/sup> = AB2<\/sup> + AC2<\/sup>
\n\u2234 BC2<\/sup> = (3)2<\/sup> + (4)2<\/sup>
\n= 9 + 16 = 25
\n\u21d2 BC = \\(\\sqrt{25}\\) cm = 5 cm
\n(iii) Now, Base = BC = 5 cm., h = AD = ?
\nArea = \\(\\frac{1}{2}\\) \u00d7 b \u00d7 h
\n\u21d2 6 = \\(\\frac{1}{2}\\) \u00d7 5 \u00d7 h
\n[\u2235 Area = 6 cm2<\/sup> as in part (i)]
\n\u21d2 h = \\(\\frac{12}{5}\\) = 2\u00b74 cm.<\/p>\n

Question 10.
\nA rectangular garden 80 m by 40 m is divided into four equal parts by two cross-paths 2.5 m wide. Find
\n(i) the area of the cross-paths.
\n(ii) the area of the unshaded portion.
\n\"ML
\nSolution:
\nLength of rectangular garden = 80 m
\nand breadth = 40 m
\nWidth of crossing path 2.5 m
\n\"ML
\nArea of length wise path
\n= 80 \u00d7 2.5 = 200 m2<\/sup>
\nArea of breadth wise path
\n= 40 \u00d7 2.5 = 100 m2<\/sup>
\n(i) Total area of both paths
\n= 200 + 100 – 2.5 \u00d7 2.5 m2<\/sup>
\n= 300 – 6.25 = 293.75 m2<\/sup>
\n(ii) Area of unshaded portion
\n= Area of garden – Area of paths
\n= 80 \u00d7 40 – 293.75 m2<\/sup>
\n= 3200 – 293.75 m2<\/sup>
\n= 2906.25 m2<\/sup><\/p>\n

Question 11.
\nIn the given figure, ABCD is a rectangle. Find the area of the shaded region.
\n\"ML
\nSolution:
\nIn the given figure.
\n\"ML
\nLength of rectangle = 18 cm
\nand breadth = 12 cm
\n\u2234 Area = l \u00d7 b = 18 \u00d7 12 cm2<\/sup> = 216 cm2<\/sup>
\nArea of triangle I = \\(\\frac{1}{2}\\) \u00d7 12 \u00d7 10 = 60 cm2<\/sup>
\nArea of triangle III = \\(\\frac{1}{2}\\) \u00d7 18 \u00d7 7 = 63 cm2<\/sup>
\n\u2234 Area of shaded portion
\n= Area of rectangle – Area of 3 triangles
\n= 216 – (60 + 63 + 20)
\n= 216 – 143 cm2
\n= 73 cm2<\/sup><\/p>\n

Question 12.
\nIn the adjoining figure, ABCD is a square grassy lawn of area 729 m2<\/sup>. A path of uniform width runs all around it. If the area of the path is 295 m2<\/sup>, find
\n(i) the length of the boundary of the square field enclosing the lawn and the path.
\n(ii) the width of the path.
\n\"ML
\nSolution:
\nArea of square ABCD = 729 m2<\/sup>
\nSide = \\(\\sqrt{729}=\\sqrt{27 \\times 27}\\) = 27 m
\n\"ML
\nLet the width of path = x m
\nThen side of outer field = 27 + x + x = (27 + 2x) m
\nArea of square PQRS = (27 + 2x)2<\/sup> m2<\/sup>
\nArea of PQRS – Area of ABCD = Area of path
\n\u2234 (27 + 2x)2<\/sup> m2<\/sup> – 729 m2<\/sup> = 295 m2<\/sup>
\n\u21d2 729 + 4x2<\/sup> + 108x – 729 = 295
\n\u21d2 4x2<\/sup> + 108x – 295 = 0
\n\u21d2 \\(x=\\frac{-108 \\pm \\sqrt{(108)^{2}-4 \\times(4) \\times(-295)}}{8}\\)
\n\"ML
\n\u2234 Width of the path is 2.5 m
\nNow, side of square field PQRS
\n= 27 + 2x = (27 + 2 \u00d7 2\u00b75) m = 32 m
\nLength of boundary = 4 \u00d7 side = 32 \u00d7 4 = 128<\/p>\n

ML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 Question 1. The length and breadth of a rectangular field are in the ratio 9 : 5. If the area of the field is 14580 square metres, find the cost of surrounding the field with a fence at the rate of \u20b93\u00b725 … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\nML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 - CBSE Library<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-18-ex-18-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1\" \/>\n<meta property=\"og:description\" content=\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Ex 18.1 Question 1. 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The length and breadth of a rectangular field are in the ratio 9 : 5. If the area of the field is 14580 square metres, find the cost of surrounding the field with a fence at the rate of \u20b93\u00b725 ... 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