Q. 7.     Through a rectangular field of length 90\; mand breadth60\; m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3\; m.find

            (ii) the cost of constructing the roads at the rate of  Rs. 110\; per\; m^{2}.

Answers (1)
R Riya



It is given that the width of each road is 3m and the length of rectangular park is 90\; m and breadth is 60\; m

Now, We know that area of rectangle is = length \times breadth

Area of the total park is 

\Rightarrow 90 \times 60 = 5400 \ m^2                   -(i)

Area of road parallel to the width of the park ( ABCD ) is

\Rightarrow 60 \times 3 = 180 \ m^2                           -(ii)

Area of road parallel to the length of park ( PQRS )  is

\Rightarrow 3 \times 90 = 270 \ m^2                          -(iii)

The common area of both the roads ( KMLN ) is 

\Rightarrow 3 \times 3 = 9 \ m^2                               -(iv)

Area of roads = [(ii)+(iii)-(iv)]

\Rightarrow 180+270-9 = 441 \ m^2                

Now, the cost of constructing the roads at the rate of  Rs. 110\; per\; m^{2} is

\Rightarrow 441 \times 110 = 48510 \ Rs

Therefore, the cost of constructing the roads at the rate of  Rs. 110\; per\; m^{2} is Rs \ 48510

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