Question 12.

We can say 40m is the perimeter of the park  

so side of rhombus = 10

Now $\frac{1}{2} \times d_1 \times d_2 = 96$  

so we get $d_1 \times d_2 = 192$ (1)  

And as we know diagonals of a rhombus are perpendicular bisectors of each other:  

so $\frac{d_1^2}{4} + \frac{d_2^2}{4} = 100$  

so we get $d_1^2 + d_2^2 = 400$ (2)

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