{"id":28134,"date":"2021-07-03T15:02:45","date_gmt":"2021-07-03T09:32:45","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=28134"},"modified":"2022-03-02T10:30:25","modified_gmt":"2022-03-02T05:00:25","slug":"ncert-solutions-for-class-8-maths-chapter-11-ex-11-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-8-maths-chapter-11-ex-11-2\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 11 Mensuration Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.2<\/h2>\n

Question 1.
\nThe shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.
\n\"NCERT
\nAnswer:
\nHere, = a = 1.2m b = 1m and h = 0.8m
\nArea of a trapezium = \\(\\frac{1}{2}\\) \u00d7 h \u00d7 (a + b)
\n= \\(\\frac{1}{2}\\) \u00d7 0.8 \u00d7 (1.2 + 1)m2<\/sup> = 0.4 \u00d7 2.2 = 0.88 m2<\/sup><\/p>\n

Question 2.
\nThe area of trapezium is 34 cm2<\/sup> and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.
\nAnswer:
\nLet the length of the other parallel side be \u2018X\u2019, height of a trapezium = 4 m. Length of one parallel side = 10 cm
\n\"NCERT
\nArea of a trapezium = 34 cm2<\/sup>
\n\\(\\frac{1}{2}\\) \u00d7 h \u00d7 (a + b) = 34
\n\\(\\frac{1}{2}\\) \u00d7 4 \u00d7 (10 + x) = 34
\n2 (10 + x) = 34
\n20 + 2x = 34
\n2x = 34 – 20
\n2x = 14
\nx = \\(\\frac{14}{2}\\) = 7 cm
\nHence, length of the other side = 7 cm<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nLength of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.
\nAnswer:
\nGiven BC = 48m,CD = 17mandAD = 40m
\nPerimeter of a trapezium = 120 m
\nAB + BC + CD + DA = 120
\nAB + 48 + 17 + 40 = 120
\nAB + 105 = 120
\nAB = 120 – 105 = 15 m
\nAB is the height of the trapezium.
\nArea of a trapezium = \\(\\frac{1}{2}\\) \u00d7 h \u00d7 (a + b)
\n= \\(\\frac{1}{2}\\) \u00d7 15 \u00d7 (48 + 40) m2<\/sup>
\n= \\(\\frac{1}{2}\\) \u00d7 15 \u00d7 88 m2<\/sup> = 15 \u00d7 44 m2<\/sup> = 660 m2<\/sup>
\nArea of the given trapezium = 660 m2<\/sup><\/p>\n

Question 4.
\nThe diagonal of quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
\n\"NCERT
\nAnswer:
\nIn the given quadrilateral, d = 24 m, h1<\/sub> = 13 m and h2<\/sub> = 8 m
\nArea of the quadrilateral = \\(\\frac{1}{2}\\) \u00d7 d \u00d7 (h1<\/sub> + h2<\/sub>)
\n= \\(\\frac{1}{2}\\) \u00d7 24 \u00d7 (13 + 8) m2<\/sup>
\n= 12 \u00d7 21 m2<\/sup> = 252 m2<\/sup><\/p>\n

Question 5.
\nThe diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
\nAnswer:
\nHere d1<\/sub> = 7.5 cm, d2<\/sub> = 12 cm
\nArea of the rhombus = \\(\\frac{1}{2}\\) \u00d7 d1<\/sub> \u00d7 d2<\/sub>
\n= \\(\\frac{1}{2}\\) \u00d7 7.5 \u00d7 12 cm2<\/sup> = 7.5 \u00d7 6 cm2<\/sup> = 45 cm2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 6.
\nFind the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
\nAnswer
\nWe know that rhombus is also a parallelogram.
\nArea of the rhombus = Area of a parallelogram = bh = 5 \u00d7 4.8 cm2<\/sup> \u00d7 = 24 cm2<\/sup>
\nLet the other diagonal be \u2018x\u2019
\nArea of a rhombus = 24 cm2<\/sup>
\n\\(\\frac{1}{2}\\) \u00d7 d1<\/sub> \u00d7 d2<\/sub> = 24
\n\\(\\frac{1}{2}\\) \u00d7 8 \u00d7 x = 24
\n4 \u00d7 x = 24
\nx = \\(\\frac{24}{4}\\) = 6 cm
\n\u2234 The required diagonal = 6 cm<\/p>\n

Question 7.
\nThe floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonal are 45 cm and 30 cm length.
\nFind the total cost of polishing the floor if the cost per m2<\/sup> is \u20b9 4.
\nAnswer:
\nTiles are in the shape of a rhombus.
\nHere d1<\/sub> = 45 cm and d2<\/sub> = 30 cm
\nArea of the rhombus (one tile) = \\(\\frac{1}{2}\\) \u00d7 d1<\/sub> \u00d7 d2<\/sub>
\n= \\(\\frac{1}{2}\\) \u00d7 45 \u00d7 30 cm2<\/sup> = 45 \u00d7 15 = 675 cm2<\/sup>
\nTotal number of tiles = 3000
\nArea of the floor = 3000 \u00d7 675 cm2<\/sup>
\n= \\(\\frac{3000 \\times 675}{100 \\times 100}\\)m2<\/sup> = \\(\\frac{2025}{10}\\)m2<\/sup>
\nCost of polishing the floor = \\(\\frac{2025}{10}\\) \u00d7 4
\n= 405 \u00d7 2 = \u20b9 810<\/p>\n

Question 8.
\nwants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m2<\/sup> and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
\n\"NCERT
\nAnswer:
\nLet the length of the side be \u2018x\u2019 (along the river)
\nLength of the other side = \\(\\frac{1}{2}\\) \u00d7 x m (along the road)
\nPerpendicular distance between two parallel sides = 100 m
\nArea of the trapezium field = 10500 m2<\/sup>
\n\\(\\frac{1}{2}\\) \u00d7 h (a + b) = 10500
\n\\(\\frac{1}{2}\\) \u00d7 100 (x + \\(\\frac{1}{2}\\)x) = 10500
\n\\(\\frac{1}{2}\\) \u00d7 100 \u00d7 \\(\\frac{3 x}{2}\\) = 10500
\n\\(\\frac{300}{4}\\) x = 10500
\nx = \\(\\frac{10500}{4}\\) \u00d7 4 = 35 \u00d7 4m = 140 m
\n\u2234 Length of the side along the river = 140 m<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nTop surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.
\n\"NCERT
\nAnswer:
\nThe regular octagon can be divided into two trapeziums and one rectangle.
\nArea of the octagonal surface
\n= Area of rectangular portion + 2 (Area of trapezoidal portion)
\n= 11 \u00d7 5 + 2[\\(\\frac { 1 }{ 2 }\\) \u00d7 4(5 + 11)]
\n= 55 + 2 \u00d7 \\(\\frac { 1 }{ 2 }\\) \u00d7 4 \u00d7 16 m
\n= 55 + 64 m2<\/sup> = 119 m2<\/sup>
\n\u2234 Area of the octagonal surface =119 cm2<\/sup><\/p>\n

Question 10.
\nThere is a pentagonal shaped park as shown in the figure. For finding its area, Joyti and Kavita divided it in two different ways.
\n\"NCERT
\nFind the area of this park using both ways. Can you suggest some other way of finding its area?
\nAnswer:
\nFor Joyti\u2019s diagram
\nThe given shape is divided into two
\ncongruent trapeziums
\nHere a = 30m, b = 15m (parallel sides),
\nh = \\(\\frac { 15 }{ 2 }\\) m.
\nArea of a trapezium = \\(\\frac { 1 }{ 2 }\\) \u00d7 h(a + b)squnits.
\nArea of one trapezium
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 \\(\\frac { 15 }{ 2 }\\) \u00d7 (30 + 15) m2<\/sup>
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 \\(\\frac { 15}{ 2 }\\) \u00d7 45 m2<\/sup> = \\(\\frac { 675 }{ 4 }\\)2<\/sup>
\nArea of the pentagonal shape = 2 \\(\\frac { 675}{ 4 }\\)
\n= \\(\\frac { 675}{ 2 }\\)m2<\/sup> = 337.5 m2<\/sup><\/p>\n

For Kavita\u2019s diagram
\nThe given shape is divided into a square and a triangle.
\nHere, side of a square is 15 m; height of the triangle is30-15 = 15m
\nArea of the pentagonal shape = Area of the square + Area of the triangle
\n= 15 \u00d7 15 + \\(\\frac { 1 }{ 2 }\\) \u00d7 15 \u00d7 15 m2<\/sup>
\n= 225 + \\(\\frac { 225 }{ 2 }\\)m2<\/sup>
\n= 225 + 112.5 m2<\/sup>
\n= 337.5 m2<\/sup>
\n\"NCERT<\/p>\n

Another method
\nDivide the pentagonal shape into three triangles of side 15 m and height 15 m. Area of the pentagonal shape
\n= 3 \u00d7 Area of the triangle
\n= 3 \u00d7 \\(\\frac { 1 }{ 2 }\\) \u00d7 15 \u00d7 15 = 337.5 m2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 11.
\nDiagram of the adjacent picture frame has outer dimensions = 24 cm \u00d7 28 cm and inner dimensions 16 cm \u00d7 20 cm. Find the area of each section of the frame, if the width of each section is same.
\n\"NCERT
\nAnswer:
\nIn trapezium (1) parallel sides are 24 cm and 16 cm.
\nHeight = \\(\\frac{28-20}{2}\\) = 4 cm
\nAreaofitstrapezium(1) = \\(\\frac{1}{2}\\) \u00d7 4 \u00d7 (24 + 16)
\n= 2 \u00d7 40 = 80 cm2<\/sup>
\nIn trapezium (2) parallel sides are 28 cm and 20 cm.
\nHeight = \\(\\frac{24-16}{2}=\\frac{8}{2}\\) = 4 cm
\nArea of the trapezium (2)
\n= \\(\\frac{1}{2}\\) \u00d7 4 \u00d7 (28 + 20)cm2<\/sup> = 2 \u00d7 48cm2<\/sup> = 96cm2<\/sup>
\nArea of each section of the frame is 80 cm2<\/sup>, 96 cm2<\/sup>, 80 cm2<\/sup> and 96 cm2<\/sup>
\nNote: [Area of 1 section = Area of 3rd section and Area of 2nd section = Area of 4th section]<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.2 Question 1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides …<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2 - MCQ Questions<\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-8-maths-chapter-11-ex-11-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts. 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