Path ways are laid in big plots, fields and grounds, along breadth wise, length wise and both breadth wise and length wise.
Example 1 :
A plot is 100 m long and 60 m broad. A path 5 m wide runs breadth wise across the ground in the middle. Find the area of the path.
Solution :
In the bellow figure the shaded portion is the path way. To find the area of pathway we can take the length as 60 m and breadth as 5 m.
Area of pathway = L x B
= 60 x 5
= 300 m^{2}
Example 2 :
Find the area of the shaded portion in the below figure.
Solution :
Area of shaded portion
= Area of rectangle ABCD - Area of rectangle EFGH
Area of rectangle ABCD :
Length (AB) = 120 m
Breadth (BC) = 80 m
Area of rectangle ABCD = Length x width
= 120 x 80
= 9600 m²
Area of rectangle EFGH :
Length (EF) = 110 m
Breadth (FG) = 70 m
Area of rectangle EFGH = Length x width
= 110 x 70
= 7700 m²
Area of shaded portion
= Area of rectangle ABCD - Area of rectangle EFGH
= 9600 - 7700
= 1900 m²
Example 3 :
There is a hall 40 m in length and 30 m in breadth. There is verandah 5 m wide surrounding the hall on all the four sides out side. Find the area of the verandah.
Solution :
First let us draw the diagram based on the given details.
The hall is a rectangle. The hall and verandahs which is out side is bigger rectangle.
Then length of bigger rectangle = 40 + 5 + 5
= 50 m
Breadth of bigger rectangle = 30 + 5 + 5
= 40 m
Area of hall and verandahs = 50 x 40
= 2000 m^{2}
Area of the hall = 40 x 30
= 1200 m^{2}
Area of verandahs = 2000 - 1200
= 800 m^{2}
Example 4 :
Find the area of a ring whose outer radius is 20 cm and inner radius is 15 cm respectively.
Solution :
Let R and r be the radius of the outer, inner circle.
Area of outer circle = ΠR²
Area of inner circle = Πr²
Radius of outer circle (R) = 20 cm
Radius of the inner circle (r) = 15 cm
Area of ring (shaded portion)
= Area of outer circle - area of inner circle
= ΠR² - Πr²
= Π (R^{2} - r^{2})
= Π (20^{2} - 15^{2})
= Π x (400 - 225)
= Π x 175
= (22/7) x 175
= 22 x 25
= 550 cm²
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