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Let us find the area and perimeter of different shapes through the following examples.

Example 1: Given alongside is a rectangular park of length 40 m and width 30 m. A pavement of
one metre width is running inside along the boundary of the park and it has to be cemented.

Observe and answer the following questions. 1 m
(a) What is the boundary length of the park?

(b) How much land is occupied by the park?
30 m
(c) If 1 bag of cement is required to cover 4 sq. m area, how
many bags of cement would be required to construct the
pavement?
40 m
Solution: (a) Total length of the boundary = Perimeter of the park
= 2 × (length + breadth) units

= 2 × (40 m + 30 m) = 2 × 70 m = 140 m
(b) Land occupied by the park = Area of the rectangular park
= length × breadth

= 40 m × 30 m = 1200 sq. m
(c) Area of the pavement = Total area of the park – Area inside the park excluding
pavement = 40 m × 30 m – (38 m × 28 m) = 1200 – 1064 sq. m = 136 sq. m.

Therefore, the total area to be cemented = 136 sq. m
Since 1 bag of cement is required to cover 4 sq. m area Length of park excluding
of the pavement. pavement = (40 – 1 – 1) m
136 = 38 m
So, the total number of cement bags required = = 34 Breadth of park excluding
4
Thus, 34 bags of cement would be required to construct pavement = (30 – 1 – 1) m
= 28 m
the pavement.
Example 2: A circular ground with a path of 3 m width all around it is shown 3 m
in the figure. The radius of the whole ground is 35 m. Find the area occupied
by the path. 35 m
Solution: The radius of the circular ground including path, R = 35 m

Since, the width of the path = 3 m
So, the radius of the inner ground, r = (35 – 3) m = 32 m

 22 
2
Area of the whole ground = π × R where π= 7   Quick Check


22 22 A square and a rectangle with
2
2
×
= × 35 m = × 35 35 m 2
7 7 measurements as given in the
2
= 22 × 5 × 35 m = 3850 m 2 figure have the same perimeter.
Now, area of the ground (excluding the path) = π × r Which figure has a larger area
2
and by how much?
22 22
2
2
×
= × 32 m = × 32 32 m 2
7 7
2
= 3218.29 m 40 cm
2
2
So, the area of the path = 3850 m – 3218.29 m = 631.71 m 2 60 cm 80 cm
231 Mensuration