CAT Exam 2021 Slot 3 Solved Paper

We can say 40m is the perimeter of the park
so side of rhombus $= 10 $
Now $1/2 × \d_1 × \d_2 = 96 $
so we get $\d_1 × \d_2 = 192 $
And as we know diagonals of a rhombus are perpendicular bisectors of each other :
So $\d_1^2/4 + \d_2^2/4 =100 $
so we get $\d_1^2 + \d_2^2 =400$
Solving (1) and (2)
We get $\d_1 = 12 $ and $\d_2 = 16 $
Now the cost, in INR, of laying electric wires along its two diagonals, at the rate of $₹125 \per \m$, is $= (12+16) (125) =3500$