{"id":28126,"date":"2021-07-03T13:30:41","date_gmt":"2021-07-03T08:00:41","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=28126"},"modified":"2021-07-03T13:30:41","modified_gmt":"2021-07-03T08:00:41","slug":"ncert-solutions-for-class-8-maths-chapter-11-ex-11-1","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-11-ex-11-1\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 11 Mensuration Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.1<\/h2>\n

Question 1.
\nA square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
\n\"NCERT
\nAnswer:
\n(a) Side of the square = 60 m.
\n\u2234 Perimeter of the square = 4 \u00d7 side
\n= 4 \u00d7 60 = 240 m
\nArea of the square = side \u00d7 side
\n= 60 \u00d7 60 m2<\/sup> = 3600 m2<\/sup><\/p>\n

(b) Length of the rectangle = 80 m
\nPerimeter of the rectangle = Perimeter of a square
\n2 (l + b) = 240
\n2 (80 + b) = 240
\n160 + 2b = 240
\n2b = 240 – 160 = 80 ,
\nb = \\(\\frac { 80 }{ 2 }\\) = 40 m
\nArea of the rectangle = l \u00d7 b = 80 \u00d7 40m2<\/sup>
\n= 3200 m2<\/sup>
\n3600 m2<\/sup> > 3200 m2<\/sup>
\n\u2234 Area of the square field is greater.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nMrs Kaushik has a square plot with the measurement as shown in the figure. She 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 \u20b9 55 per m2<\/sup>.
\n\"NCERT
\nAnswer:
\nSide of a square = 25 m
\nArea of the plot = side \u00d7 side = 25 \u00d7 25 m2<\/sup> = 625 m2<\/sup><\/p>\n

In the given diagram, the construction portion is a rectangle with length = 20 m and breadth = 15 m<\/p>\n

Area of the construction portion = Area of a rectangle = 20 \u00d7 15 m2<\/sup> = 300 m2<\/sup><\/p>\n

Area of the garden = (625 – 300) m2<\/sup> = 325 m2<\/sup> Cost of developing the garden = \u20b9 55 \u00d7 325 = \u20b9 17,875<\/p>\n

Question 3.
\nThe shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden (Length of rectangle is 20 – (3.5 + 3.5) metres].
\n\"NCERT
\nAnswer:
\nRadius of the semicircles = \\(\\frac { 7 }{ 2 }\\)
\n\"NCERT
\nLength of the rectangle = 20 – (3.5 + 3.5) m = 20 – 7 = 13 m
\nBreadth of the rectangle = 7m
\nArea of the rectangle = l \u00d7 b = 13 \u00d7 7 = 91m2<\/sup>
\nPerimeter of the rectangle = 2 (l + 0) (Breadth will not be consider)
\n= 2 (13 + 0) m = 2 \u00d7 13 = 26 m
\nArea of the garden = 38.5 + 91m2<\/sup> = 129.5 m2<\/sup>
\nPerimeter of the garden = 22 + 26 m = 48 m<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nA flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2<\/sup>? (If required you can split the tiles in whatever way you want to fill up the corners).
\nAnswer:
\nBase of the parallelogram = 24 cm = \\(\\frac{24}{100}\\) m
\nCorresponding height = 10 cm = \\(\\frac{100}{10}\\)
\nArea of a tile = base \u00d7 height
\n= \\(\\frac{24}{100} \\times \\frac{10}{100}\\) m2<\/sup> = \\(\\frac{24}{1000}\\) m2<\/sup>
\nArea of the floor = 1080 m2<\/sup>
\nNumber of tiles = \\(\\frac{\\text { Area of the floor }}{\\text { Area of one tile }}\\)
\n= \\(\\frac{1080 \\times 1000}{24}\\) = 45 \u00d7 1000 = 45000 tiles<\/p>\n

Question 5.
\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? Remember, circumference of a circle can be obtained by using the expression c = 2\u03c0r, where r is the radius of the circle.
\n\"NCERT
\nAnswer:
\n(a) Radius of the semicircle = \\(\\frac{2.8}{2}\\) = 1.4 cm
\nPerimeter of the semicircle = \\(\\frac{2 \\pi \\mathrm{r}}{2}\\) + 2r
\n= \\(\\frac{22}{7}\\) \u00d7 1.4 + 2 \u00d7 1.4 = 4.4 + 2.8cm
\n= 7.2 cm<\/p>\n

(b) Perimeter of the semicircular part
\n= \\(\\frac{1}{2}\\) \u00d7 2\u03c0r = \u03c0r = \\(\\frac{22}{7}\\) \u00d7 1.4 = 4.4 cm
\nPerimeter of the rectangular part
\n= 1.5 + 2.8 + 15 cm = 5.8 cm
\nPerimeter of the given diagram
\n= 4.4 cm + 5.8 cm = 10.2 cm<\/p>\n

(c) Perimeter of the semicircular part
\n= \\(\\frac{2 \\pi r}{2}\\) \u03c0r = \u03c0r = \\(\\frac{22}{7}\\) \u00d7 1.4 cm = 4.4 cm
\nPerimeter of the given diagram
\n= (4.4 + 2 + 2) cm = 8.4 cm
\n\u2234 7.2 cm < 8.4cm < 10.2 cm
\nAnt has to take a longer round for (b).<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.1 Question 1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field …<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1<\/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":{"spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1 - NCERT Solutions<\/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:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-11-ex-11-1\/\" \/>\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 11.1 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts. 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