circumference of a circle can be obtained by using the expression c = 27tr, wherer
EXERCISE 11.1
80 m
60 m
(a)
1. A square and a rectangular field with
measurements as given in the figure have the same
perimeter. Which field has a larger area?
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 $ 55 per m².
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
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
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,
20 m
25 m
House
of this garden (Length of rectangle is
7m 20 - (3.5 +3.5) metres).
T
1
Garden
-25 m
the corners).
is the radius of the circle.
(b)
(a)
(c)
2.8 cm
T T
1.5 cm
2.8 cm
2.8 cm
2 cm
2 cm
Area of Trapezium
Answers
Answer:
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 has a larger area?
Solution:
Given,
Side of a square = 60 m
Length, l = 80 m
According to question,
Perimeter of rectangular field = Perimeter of square field
According to formula,
2(l + b) = 4 × Side
2(80 + b) = 4 × 60
160 + 2b = 240
b = 40
Therefore, 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.
Question 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:
Given,
Side of a square = 25 m
According to formula,
Area of square plot = square of a side = (side)2 = (25)2 = 625
Therefore, the area of a square plot is 625 m2
Length of the house = 20 m and
Breadth of the house = 15 m
Area of the house = length × breadth = 20 × 15 = 300 m2
Area of garden = Area of square plot – Area of house = 625 – 300 = 325 m2
Therefore, per unit cost of making garden is Rs. 55
Total Cost of developing the garden of 325 sq. m = Rs. 55 × 325 = Rs. 17,875
Question 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:
Given,
Total length = 20 m
Diameter of semi circle = 7 m
Therefore, 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 = l × b = 13 × 7= 91m2
Area of two semi circles = 2 × (1/2) × pi × r2
= 2 × (1/2) × 22/7 × 3.5 × 3.5
= 38.5 m2
Area of garden = 91 + 38.5 = 129.5 m2
Now Perimeter of two semi circles = 2.pi.r = 2 × (22/7) × 3.5 = 22 m
Therefore, Perimeter of garden = 22 + 13 + 13 = 48 m
Question 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,
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
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 45000 tiles are required to cover the floor.
Question 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, the circumference of a circle can be obtained by using the expression C = 2.pi.r, where r is the radius of the circle.
Solution:
(a)
Radius = Diameter/2 = 2.8/2 cm = 1.4 cm
Circumference of semi-circle = pi.r = (22/7)×1.4 = 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 = r = (22/7) × 1.4 = 4.4 cm
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 = pi.r = (22/7) × 1.4 = 4.4 cm
Total distance covered by the ant = 2 + 2 + 4.4 = 8.4 cm
After analyzing results of three figures, we concluded that for figure (b) food piece, the ant would take a longer round.
Answer:
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