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B BYJU'S, NCERT Solution For Class 8 Maths Chapter 11 Mensuration, The Learning App, Exercise 11.1, Page No: 171, 1. A square and a rectangular field with measurements as given in the figure have the, same perimeter. Which field has a larger area?, K-60 m, 80 m, (b), • (a), Solution:, Side of a square = 60 m (Given), And the length of rectangular field, 1 = 80 m (Given), According to question,, Perimeter of rectangular field = Perimeter of square field, 2(1+b) = 4xSide (using formulas), 2(80+b) = 4x60, 160+2b = 240, b = 40, Breadth of the rectangle is 40 m., Now, Area of Square field, = (side), = (60) = 3600 m?, And Area of Rectangular field, = lengthxbreadth = 80x40, = 3200 m?, Hence, area of square field is larger., 2. Mrs.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 Rs. 55 per m2., https://byjus.com, 1
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B BYJU'S, NCERT Solution For Class 8 Maths Chapter 11 Mensuration, The Learning App, 20 m, 25 m, House, Garden, 25 m, Solution:, Side of a square plot = 25 m, Formula: Area of square plot = square of a side (side), = (25) = 625, Therefore the area of a square plot is 625 m, Length of the house 20 m and, 5m, 20 m, 25, IV, 25, Breadth of the house 15 m, ..Area of the house lengthxbreadth, = 20x15 = 300 m2, Area of garden = Area of square plot - Area of house, %3D, = 625-300 = 325 m?, : Cost of developing the garden per sq. m is Rs. 55, https://byjus.com, 2
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B BYJU'S, NCERT Solution For Class 8 Maths Chapter 11 Mensuration, The Learning App, . Cost of developing the garden 325 sq. m Rs. 55x325, = Rs. 17,875, Hence total cost of developing a garden around is Rs. 17,875., 3. The 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 meters], 20 m, Solution::, Given: Total length 20 m, Diameter of semi circle = 7 m, :. Radius of semi circle = 7/2 = 3.5 m, Length of rectangular field, = 20-(3.5+3.5) = 20-7 = 13 m, Breadth of the rectangular field = 7 m, .. Area of rectangular field = 1xb, = 13x7= 91m?, https://byjus.com, 3
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B BYJU'S, NCERT Solution For Class 8 Maths Chapter 11 Mensuration, The Learning App, Area of two semi circles = 2x(1/2)xTxr2, = 2x(1/2)x22/7x3.5x3.5, = 38.5 m2, Area of garden = 91+38.5 = 129.5 m², Now Perimeter of two semi circles 2nr 2x(22/7)x3.5 22 m, And Perimeter of garden = 22+13+13, = 48 m. Answer, 4. A 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 m?? [If required you can split the tiles in whatever way you want to fill up the corners], Solution:, Given: Base of flooring tile 24 cm 0.24 m, Corresponding height of a flooring tile= 10 cm 0.10 m, Now Area of flooring tile= BasexAltitude, = 0.24x0.10, = 0.024, Area of flooring tile is 0.024m?, . Number of tiles required to cover the floor= Area of floor/Area of one tile = 1080/0.024, = 45000 tiles, https://byjus.com, 4
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B BYJU'S, NCERT Solution For Class 8 Maths Chapter 11 Mensuration, The Learning App, Hence 45000 tiles are required to cover the floor., 5. An 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 = 2nr,where r is the radius of the circle., (a), (b), 2.8 cm, (c), 1.5 cm, 2.8 em-, -2.8 em-, Solution:, (a) Radius = Diameter/2 = 2.8/2 cm 1.4 cm, Circumference of semi-circle = mr, = (22/7)x1.4 = 4.4, Circumference of semi-circle is 4.4 cm, Total distance covered by the ant= Circumference of semi -circle+Diameter, = 4.4+2.8 = 7.2 cm, (b) Diameter of semi-circle = 2.8 cm, Radius = Diameter/2 2.8/2 = 1.4 cm, Circumference of semi-circle = ar, = (22/7)x1.4 = 4.4 cm, https://byjus.com, -2em-, -2 cm-
