1.A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
Solution:
Given,
side of a square (s) = 60 m
Length of rectangular field, l = 80 m
Now,
According to question:
Perimeter of rectangular field = Perimeter of square field
By using formula:
2 (l + b) = 4 × Side
2(80+b) = 4×60
160+2b = 240
b = 40
Breadth of the rectangle is 40 m.
Now, Area of Square field
= (side)2
= (60)2 = 3600 m2
And Area of Rectangular field
= length × breadth = 80×40
= 3200 m2
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.
Solution:
From given figure:
Side (s) of a square plot = 25 m
By formula:
Area of square plot = square of a side = (s)2
(25)2 = 625
Therefore, the area of a square plot is 625 m2
From figure,
length of the house = 20 m and
Breadth of the house = 15 m
Now,
Area of the house = length × breadth
= 20×15 = 300 m2
Now,
Area of garden = Area of square plot – Area of house
325 m2 = 625 – 300
Cost of developing the garden per sq. m is Rs. 55
Cost of developing the garden 325 sq. m = Rs. 55×325
= 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]
Solution:
According to given figure:
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
Now,
Area of rectangular field = length × breadth
= 13×7
= 91m2
Area of two semi circles = 2 × (1/2) × π × r2
= 2× (1/2) × 22/7 × 3.5 × 3.5
= 38.5 m2
Area of garden = 91 + 38.5
= 129.5 m2
So, Perimeter of two semi circles = 2πr
= 2 × (22/7) × 3.5
= 22 m
And Perimeter of garden = 22 + 13 + 13
= 48 m
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 m2? [If required you can split the tiles in whatever way you want to fill up the corners]
Solution:
Given in question:
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 = Base × Altitude
= 0.24×0.10
= 0.024
So, area of flooring tile is 0.024m2
Number of tiles required to cover the floor= Area of floor/Area of one tile
= 1080/0.024
= 45000 tiles
Hence, no. required tiles to cover the floor = 45000
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 = 2πr ,where r is the radius of the circle.
Solution:
(a)
We know that radius = diameter/2
= 2.8/2 cm = 1.4 cm
Now,
Circumference of semi-circle = πr
= (22/7) × 1.4 = 4.4
So, circumference of semi-circle is 4.4 cm
Hence, total distance covered by the ant= circumference of semi-circle + diameter
= 4.4+2.8 = 7.2 cm
(b) Given, diameter of semi-circle = 2.8 cm
So,
Radius = Diameter/2 = 2.8/2 = 1.4 cm
Circumference of semi-circle = π r
= (22/7) × 1.4 = 4.4 cm
Hence, total distance covered by the ant= 1.5+2.8+1.5+4.4 = 10.2 cm
(c) Diameter of semi-circle = 2.8 cm
Radius = Diameter/2 = 2.8/2
= 1.4 cm
Circumference of semi-circle = πr
= (22/7)×1.4
= 4.4 cm
Hence, total distance covered by the ant = 2+2+4.4 = 8.4 cm
After analyzing results of three figures, we concluded that the ant would take a longer round for figure (b) food piece.
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