There is a rectangular field of length 300 m and breadth 240 m. Three roads, each of width 3.2 m pass
through the field such that two roads are parallel to the breadth and third road is parallel to the length.
Find the total area of three roads and the cost of gravelling the roads at the rate of 2.25 per m².
Answers
Answer:
area= 1717.76m^2
cost=Rs 3864.96
Step-by-step explanation:
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Answer:
The total area of three roads is 2466.52 m²
Rs. 5549.67 will be required to gravel the roads at the rate of Rs. 2.25 per m².
Step-by-step explanation:
Given , the length of the field = 300 m
The breadth of the field = 240 m
Width of the road = 3.2 m
Let road 1 and 2 be the ones that are parallel to the breadth of the field.
We need to find the Total Area of the roads
= Area of Road 1 + Area of Road 2 + Area of Road 3 - Common Area
= (240 * 3.2) + (240 * 3.2) + (300 * 3.2) - (3.2 * 3.2) - (3.2 * 3.2)
= 768 + 768 + 960 - 10.24 - 10.24
= 2496 - 29.48
= 2466.52 m²
Therefore, the total area of three roads is 2466.52 m²
We know that Cost = Rate * Area
=> Cost of gravelling = 2.25 * 2466.52
= Rs. 5549.67
Therefore, Rs. 5549.67 will be required to gravel the roads at the rate of Rs. 2.25 per m².