## PATH WAYS

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 m2

Example 2 :

Find the area of the shaded portion in the below figure. Solution :

=  Area of rectangle ABCD - Area of rectangle EFGH

Area of rectangle ABCD :

Length (AB)  =  120 m

Area of rectangle ABCD  =  Length x width

=  120 x 80

=  9600 m²

Area of rectangle EFGH :

Length (EF)  =  110 m

Area of rectangle EFGH  =  Length x width

=  110 x 70

=  7700 m²

=  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 m2

Area of the hall  =  40 x 30

=  1200 m2

Area of verandahs  =  2000 - 1200

=  800 m2

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 outer circle - area of inner circle

=  ΠR² -  Πr²

=  Π (R2 - r2)

=  Π (202 - 152)

=  Π x (400 - 225)

=  Π x 175

=  (22/7) x 175

=  22 x 25

=  550 cm²

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