{"id":22086,"date":"2024-02-25T15:55:29","date_gmt":"2024-02-25T10:25:29","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=22086"},"modified":"2024-02-26T09:54:01","modified_gmt":"2024-02-26T04:24:01","slug":"selina-concise-mathematics-class-8-icse-solutions-area-of-trapezium-and-a-polygon","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/selina-concise-mathematics-class-8-icse-solutions-area-of-trapezium-and-a-polygon\/","title":{"rendered":"Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon"},"content":{"rendered":"
Selina Publishers Concise Mathematics Class 8 ICSE Solutions\u00a0Chapter 20 Area of Trapezium and a Polygon<\/strong><\/p>\n ICSE Solutions<\/a>Selina ICSE Solutions<\/a>ML Aggarwal Solutions<\/a><\/p>\n ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 20 Area of Trapezium and a Polygon. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.<\/p>\n Selina Class 8 Maths Solutions<\/a>Physics<\/a>Chemistry<\/a>Biology<\/a>Geography<\/a>History & Civics<\/a><\/p>\n Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> Question 7.<\/strong><\/span> Question 8.<\/strong><\/span> Question 9.<\/strong><\/span> Question 10.<\/strong><\/span> Question 11.<\/strong><\/span> Question 12.<\/strong><\/span> Question 13.<\/strong><\/span> Question 14.<\/strong><\/span> Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> Question 7.<\/strong><\/span> Question 8.<\/strong><\/span> Question 9.<\/strong><\/span> Question 10.<\/strong><\/span> Question 11.<\/strong><\/span> Question 12.<\/strong><\/span> Question 13.<\/strong><\/span> Question 14.<\/strong><\/span> Question 15.<\/strong><\/span> Question 16.<\/strong><\/span> Question 17.<\/strong><\/span> Question 18.<\/strong><\/span> Question 19.<\/strong><\/span> Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span>Area of Trapezium and a Polygon\u00a0Exercise 20A –\u00a0Selina Concise Mathematics Class 8 ICSE Solutions<\/h3>\n
\nFind the area of a triangle, whose sides are :
\n(i) 10 cm, 24 cm and 26 cm
\n(ii) 18 mm, 24 mm and 30 mm
\n(iii) 21 m, 28 m and 35 m
\nSolution:<\/strong><\/span>
\n(i) Sides of \u2206 are
\na = 10 cm
\nb = 24 cm
\nc = 26 cm
\n
\n(ii) Sides of \u2206 are
\na = 18 mm
\nb = 24 mm
\nc = 30 mm
\n
\n(iii) Sides of \u2206 are
\na = 21 m
\nb = 28 m
\nc = 35 m
\n
\n<\/p>\n
\nTwo sides of a triangle are 6 cm and 8 cm. If height of the triangle corresponding to 6 cm side is 4 cm ; find :
\n(i) area of the triangle
\n(ii) height of the triangle corresponding to 8 cm side.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nThe sides of a triangle are 16 cm, 12 cm and 20 cm. Find :
\n(i) area of the triangle ;
\n(ii) height of the triangle, corresponding to the largest side ;
\n(iii) height of the triangle, corresponding to the smallest side.
\nSolution:<\/strong><\/span>
\nSides of \u2206 are
\na = 20 cm
\nb = 12 cm
\nc = 16 cm
\n
\n<\/p>\n
\nTwo sides of a triangle are 6.4 m and 4.8 m. If height of the triangle corresponding to 4.8 m side is 6 m; find :
\n(i) area of the triangle ;
\n(ii) height of the triangle corresponding to 6.4 m side.
\nSolution:<\/strong><\/span>
\n
\n= 9\/2 = 4.5 m
\nHence (i) 14.4 m2<\/sup> (ii) 4.5 m<\/p>\n
\nThe base and the height of a triangle are in the ratio 4 : 5. If the area of the triangle is 40 m2<\/sup>; find its base and height.
\nSolution:<\/strong><\/span>
\nLet base of \u2206 = 4x m
\nand height of \u2206 = 5x m
\narea of \u2206 =40 m2<\/sup>
\n<\/p>\n
\nThe base and the height of a triangle are in the ratio 5 : 3. If the area of the triangle is 67.5 m2<\/sup>; find its base and height.
\nSolution:<\/strong><\/span>
\nLet base = 5x m
\nheight = 3x m
\n
\nbase = 5x = 5 x 3 = 15 m
\nheight = 3x = 3 x 3 = 9 m<\/p>\n
\nThe area of an equilateral triangle is 144\u221a3 cm2<\/sup>; find its perimeter.
\nSolution:<\/strong><\/span>
\nLet each side of an equilateral triangle = x cm
\n<\/p>\n
\nThe area of an equilateral triangle is numerically equal to its perimeter. Find its perimeter correct to 2 decimal places.
\nSolution:<\/strong><\/span>
\nLet each side of the equilateral traingle = x
\n<\/p>\n
\nA field is in the shape of a quadrilateral ABCD in which side AB = 18 m, side AD = 24 m, side BC = 40m, DC = 50 m and angle A = 90\u00b0. Find the area of the field.
\nSolution:<\/strong><\/span>
\nSince \u2220A = 90\u00b0
\nBy Pythagorus Theorem,
\nIn \u2206ABD,
\n<\/p>\n
\nThe lengths of the sides of a triangle are in the ratio 4 : 5 : 3 and its perimeter is 96 cm. Find its area.
\nSolution:<\/strong><\/span>
\nLet the sides of the triangle ABC be 4x, 5x and 3x
\nLet AB = 4x, AC = 5x and BC = 3x
\nPerimeter = 4x + 5x + 3x = 96
\n=> 12x = 96
\n<\/p>\n
\nOne of the equal sides of an isosceles triangle is 13 cm and its perimeter is 50 cm. Find the area of the triangle.
\nSolution:<\/strong><\/span>
\nIn isosceles \u2206ABC
\nAB = AC = 13 cm But perimeter = 50 cm
\n<\/p>\n
\nThe altitude and the base of a triangular field are in the ratio 6 : 5. If its cost is \u20b9 49,57,200 at the rate of \u20b9 36,720 per hectare and 1 hectare = 10,000 sq. m, find (in metre) dimensions of the field,
\nSolution:<\/strong><\/span>
\nTotal cost = \u20b9 49,57,200
\nRate = \u20b9 36,720 per hectare
\nTotal area of the triangular field
\n<\/p>\n
\nFind the area of the right-angled triangle with hypotenuse 40 cm and one of the other two sides 24 cm.
\nSolution:<\/strong><\/span>
\nIn right angled triangle ABC Hypotenuse AC = 40 cm
\nOne side AB = 24 cm
\n<\/p>\n
\nUse the information given in the adjoining figure to find :
\n(i) the length of AC.
\n(ii) the area of a \u2206ABC
\n(iii) the length of BD, correct to one decimal place.
\n
\nSolution:<\/strong><\/span>
\n<\/p>\nArea of Trapezium and a Polygon\u00a0Exercise 20B –\u00a0Selina Concise Mathematics Class 8 ICSE Solutions<\/h3>\n
\nFind the length and perimeter of a rectangle, whose area = 120 cm2<\/sup> and breadth = 8 cm
\nSolution:<\/strong><\/span>
\narea of rectangle = 120 cm2<\/sup>
\nbreadth, b = 8 cm
\nArea = l x b
\nl x 8 = 120
\nl = 120\/8 = 15 cm
\nPerimeter = 2 (l+b) = 2(15+8) = 2 x 23 = 46 cm
\nLength = 15 cm; Perimeter = 46 cm<\/p>\n
\nThe perimeter of a rectangle is 46 m and its length is 15 m. Find its :
\n(i) breadth
\n(ii) area
\n(iii) diagonal.
\nSolution:<\/strong><\/span>
\n(i) Perimeter of rectangle = 46 m
\nlength, l = 15 m
\n2 (l+b) = 46
\n2(15 + b) = 46
\n15+b = 46\/2 = 23
\nb = 23 – 15
\nb = 8 m
\n<\/p>\n
\nThe diagonal of a rectangle is 34 cm. If its breadth is 16 cm; find its :
\n(i) length
\n(ii) area
\nSolution:<\/strong><\/span>
\n
\nAC2<\/sup> = AB2<\/sup>+BC2<\/sup> (By Pythagoras theorem)
\n(34)2<\/sup> = l2<\/sup> + (16)2<\/sup>
\n1156 = l2<\/sup> + 256
\nl2<\/sup> = 1156 – 256
\nl2<\/sup> = 900
\nl = \u221a900 = 30 cm
\narea = l x b = 30 x 16 = 480 cm2<\/sup>
\n(i) 30 cm (ii) 480 cm2<\/sup><\/p>\n
\nThe area of a small rectangular plot is 84 m2<\/sup>. If the difference between its length and the breadth is 5 m; find its perimeter.
\nSolution:<\/strong><\/span>
\nArea of a rectangular plot = 84 m2<\/sup>
\nLet breadth = x m
\nThen length = (x + 5) m
\nArea = l x b
\nx(x + 5) = 84
\nx2<\/sup> + 5x – 84 = 0
\n=> x2<\/sup>+ 12x – 7x – 84 = 0
\n=> x(x + 12) – 7(x + 12) = 0
\n=> (x + 12) (x – 7) = 0
\nEither x + 12 = 0, then x = -12 which is not possible being negative
\nor x – 7 = 0, then x = 7
\nLength = x + 5 = 7 + 5 = 12m
\nand breadth = x = 7 m
\nPerimeter = 2(l + b) = 2(12+ 7) = 2 x 19 m = 38 m<\/p>\n
\nThe perimeter of a square is 36 cm; find its area
\nSolution:<\/strong><\/span>
\nPerimeter of Square = 36 cm
\n<\/p>\n
\nFind the perimeter of a square; whose area is : 1.69 m2<\/sup>
\nSolution:<\/strong><\/span>
\nArea of square= 1.69 m2<\/sup>
\nSide = \u221aarea = \u221a1.69 = 1.3 m
\nPerimeter = 4 x side = 4 x 1.3 = 5.2 m<\/p>\n
\nThe diagonal of a square is 12 cm long; find its area and length of one side.
\nSolution:<\/strong><\/span>
\nLet side of square = a cm
\ndiagonal = 12 cm
\nBy Pythagoras Theorem, a2<\/sup> + a2<\/sup> = (12)2<\/sup>
\n2a2<\/sup> = 144
\na2<\/sup> = 72
\nArea of square = a2<\/sup> = 72 cm2<\/sup>
\na2<\/sup> = 72
\na = \u221a72 = 8.49 cm<\/p>\n
\nThe diagonal of a square is 15 m; find the length of its one side and perimeter.
\nSolution:<\/strong><\/span>
\nDiagonal of square = 15 m
\nLet side of square = a
\na2<\/sup> + a2<\/sup> = (15)2<\/sup> = 225
\na2<\/sup> = 225\/2 = 112.50
\na = \u221a112.50 = 10.6 m
\nPerimeter = 4 x a = 10.6 x 4 = 42.4 m<\/p>\n
\nThe area of a square is 169 cm2<\/sup>. Find its:
\n(i) one side
\n(ii) perimeter
\nSolution:<\/strong><\/span>
\nLet each side of the square be x cm.
\nIts area = x2<\/sup> = 169 (given)
\nx = \u221a169
\nx = 13 cm
\n(i) Thus, side of the square = 13 cm
\n(ii) Again perimeter = 4 (side) = 4 x 13 = 52 cm<\/p>\n
\nThe length of a rectangle is 16 cm and its perimeter is equal to the perimeter of a square with side 12.5 cm. Find the area of the rectangle.
\nSolution:<\/strong><\/span>
\nLength of the rectangle = 16 cm
\nLet its breadth be x cm
\nPerimeter = 2 (16 + x) = 32 + 2x
\nAlso perimeter = 4(12.5) = 50 cm.
\nAccording to statement,
\n32 + 2x = 50
\n=> 2x = 50 – 32 = 18
\n=> x = 9
\nBreadth of the rectangle = 9 cm.
\nArea of the rectangle (l x b)= 16 x 9 = 144 cm2<\/sup><\/p>\n
\nThe perimeter of a square is numerically equal to its area. Find its area.
\nSolution:<\/strong><\/span>
\nLet each side of the square be x cm.
\nIts perimeter = 4x,
\nArea =x2<\/sup>
\nBy the given condition 4x = x2<\/sup>
\n=> x2<\/sup> – 4x = 0
\n=> x (x – 4) = 0
\n=> x = 4 [x \u2260 0]
\nArea = x2<\/sup> = (4)2<\/sup> = 4 x 4 = 16 sq.units.<\/p>\n
\nEach side of a rectangle is doubled. Find the ratio between :
\n(i) perimeters of the original rectangle and the resulting rectangle.
\n(ii) areas of the original rectangle and the resulting rectangle.
\nSolution:<\/strong><\/span>
\nLet length of the rectangle = x
\nand breadth of the rectangle = y
\n(i) Perimeter P = 2(x + y)
\nAgain, new length = 2x
\nNew breadth = 2y
\n<\/p>\n
\nIn each of the following cases ABCD is a square and PQRS is a rectangle. Find, in each case, the area of the shaded portion.
\n(All measurements are in metre).
\n
\nSolution:<\/strong><\/span>
\n(i) Area of the shaded portion
\n= Area of the rectangle PQRS – Area of square ABCD
\n= 3.2 x 1.8 – (1.4)2<\/sup> (\u2235 PQ = 3.2 and PS = 1.8) Side of square AB = 1.4
\n= 5.76 – 1.96 = 3.80 = 3.8 m2<\/sup>
\n(ii) Area of the shaded portion = Area of square ABCD – Area of rectangle PQRS
\n= 6 x 6 – (3.6) (4.8) = 36 – 17.28 = 18.72 m2<\/sup><\/p>\n
\nA path of uniform width, 3 m, runs around the outside of a square field of side 21 m. Find the area of the path.
\nSolution:<\/strong><\/span>
\nAccording to the given information the figure will be as shown alongside.
\nClearly, length of the square field excluding path = 21 m.
\n
\nArea of the square side excluding the path = 21 x 21 = 441 m2<\/sup>
\nAgain, length of the square field including the path = 21 + 3 + 3 = 27 m
\nArea of the square field including the path = 27 x 27 = 729 m2<\/sup>
\nArea of the path = 729 – 441 = 288 m2<\/sup><\/p>\n
\nA path of uniform width, 2.5 m, runs around the inside of a rectangular field 30 m by 27 m. Find the area of the path.
\nSolution:<\/strong><\/span>
\nAccording to the given statement the figure will be as shown alongside.
\n
\nClearly, the length of the rectangular field including the path = 30 m.
\nBreadth = 27 m.
\nIts Area = 30 x 27 = 810 m2<\/sup>
\nWidth of the path = 2.5 m
\nLength of the rectangular field including the path = 30 – 2.5 – 2.5 = 25 m.
\nBreadth = 27 – 2.5 – 2.5 = 22m
\nArea of the rectangular field including the path = 25 x 22 = 550 m2<\/sup>
\nHence, area of the path = 810 – 550 = 260 m2<\/sup>.<\/p>\n
\nThe length of a hall is 18 m and its width is 13.5 m. Find the least number of square tiles, each of side 25 cm, required to cover the floor of the hall,
\n(i) without leaving any margin.
\n(ii) leaving a margin of width 1.5 m all around. In each case, find the cost of the tiles required at the rate of Rs. 6 per tile
\nSolution:<\/strong><\/span>
\n(i) Length of hall (l) = 18 m and breadth (b) = 13.5 m
\n
\n<\/p>\n
\nA rectangular field is 30 m in length and 22m in width. Two mutually perpendicular roads, each 2.5 m wide, are drawn inside the field so that one road is parallel to the length of the field and the other road is parallel to its width. Calculate the area of the crossroads.
\nSolution:<\/strong><\/span>
\nLength of rectangular field (l) = 30 m and breadth (b) = 22m
\nwidth of parallel roads perpendicular to each other inside the field =2.5m
\n
\nArea of cross roads = width of roads (Length + breadth) – area of middle square
\n= 2.5 (30 + 22) – (2.5)2<\/sup>
\n= 2.5 x 52 – 6.25 m2<\/sup>
\n= (130 – 6.25) m = 123.75 m2<\/sup><\/p>\n
\nThe length and the breadth of a rectangular field are in the ratio 5 : 4 and its area is 3380 m2<\/sup>. Find the cost of fencing it at the rate of \u20b975 per m.
\nSolution:<\/strong><\/span>
\nRatio in length and breadth = 5 : 4
\nArea of rectangular field = 3380 m2<\/sup>
\nLet length = 5x and breadth = 4x
\n5x x 4x = 3380
\n=> 20x2<\/sup>= 3380
\nx2<\/sup> = 3380\/20 = 169 = (13)2
\nx = 13
\nLength = 13 x 5 = 65 m
\nBreadth =13 x 4 = 52 m
\nPerimeter = (l + b) = 2 x (65 + 52) m = 2 x 117 = 234 m
\nRate of fencing = \u20b9 75 per m
\nTotal cost = 234 x 75 = \u20b9 17550<\/p>\n
\nThe length and the breadth of a conference hall are in the ratio 7 : 4 and its perimeter is 110 m. Find:
\n(i) area of the floor of the hall.
\n(ii) number of tiles, each a rectangle of size 25 cm x 20 cm, required for flooring of the hall.
\n(iii) the cost of the tiles at the rate of \u20b9 1,400 per hundred tiles.
\nSolution:<\/strong><\/span>
\nRatio in length and breadth = 7 : 4
\nPerimeter = 110 m
\n
\n<\/p>\nArea of Trapezium and a Polygon\u00a0Exercise 20C –\u00a0Selina Concise Mathematics Class 8 ICSE Solutions<\/h3>\n
\nThe following figure shows the cross-section ABCD of a swimming pool which is trapezium in shape.
\nIf the width DC, of the swimming pool is 6.4cm, depth (AD) at the shallow end is 80 cm and depth (BC) at deepest end is 2.4m, find Its area of the cross-section.
\n
\nSolution:<\/strong><\/span>
\nArea of the cross-section = Area of trapezium ABCD
\n
\n= 1024 cm2<\/sup> or = 10.24 sq.m.<\/p>\n
\nThe parallel sides of a trapezium are in the ratio 3 : 4. If the distance between the parallel sides is 9 dm and its area is 126 dm2<\/sup> ; find the lengths of its parallel sides.
\nSolution:<\/strong><\/span>
\n
\nLet parallel sides of trapezium be
\na = 3x
\nb = 4x
\nDistance between parallel sides, h = 9 dm
\narea of trapezium = 126 dm2<\/sup>
\n<\/p>\n
\nThe two parallel sides and the distance between them are in the ratio 3 : 4 : 2. If the area of the trapezium is 175 cm2<\/sup>, find its height.
\nSolution:<\/strong><\/span>
\nLet the two parallel sides and the distance between them be 3x, 4x, 2x cm respectively
\nArea = \\(\\frac { 1 }{ 2 }\\) (sum of parallel sides) x (distance between parallel sides)
\n= \\(\\frac { 1 }{ 2 }\\) (3x + 4x) x 2x = 175 (given)
\n=> 7x x x = 175
\n=> 7x2<\/sup> = 175
\n=> x2<\/sup> = 25
\n=> x = 5
\nHeight i.e. distance between parallel sides = 2x = 2 x 5 = 10 cm<\/p>\n