NCERT Exemplar Solutions
Class 6 - Mathematics - Unit 6: Mensuration - Problems and Solutions

Question 39

1. The length of the outer boundary of the park (this is the perimeter).
2. The cost of fencing this boundary at Rs 20 per metre.
3. The cost of manuring the rectangular flower bed in the centre at Rs 50 per square metre.

Step 2: Read measurements from the figure
From Fig. 6.16:

• The park is a rectangle with length (400 ext{m}) and breadth (270 ext{m}).
• The flower bed is a rectangle with length (100 ext{m}) and breadth (80 ext{m}).

Step 3: Find the perimeter (outer boundary) of the park
For a rectangle, perimeter (P = 2 imes ( ext{length} + ext{breadth})).
First add the sides: (400 + 270 = 670).
Now multiply by 2: (2 imes 670 = 1340).
So, the outer boundary (= 1340 ext{m}).

Step 4: Find the cost of fencing
Rate of fencing (= ext{Rs }20) per metre.
Total length to fence (= 1340 ext{m}).
Cost (= 1340 imes 20 = 26{,}800).
So, cost of fencing (= ext{Rs }26{,}800).

Step 5: Find the area of the flower bed
Area of a rectangle (= ext{length} imes ext{breadth}).
(100 imes 80 = 8000).
So, area of the flower bed (= 8000 ext{m}^2).

Step 6: Find the cost of manuring
Rate of manuring (= ext{Rs }50) per square metre.
Cost (= 8000 imes 50 = 400{,}000).
So, cost of manuring (= ext{Rs }4{,}00{,}000).

Final Answers:
Outer boundary (= 1340 ext{m}).
Cost of fencing (= ext{Rs }26{,}800).
Cost of manuring (= ext{Rs }4{,}00{,}000).