CAT 2021 Slot 3 · Quant · Question 12

The fencing gives the perimeter, so each side is \frac{40}{4}=10 m.

Let the diagonals be {d}_{1} and {d}_{2}. The area of a rhombus is \frac{{d}_{1}{d}_{2}}{2}=96, so {d}_{1}{d}_{2}=192.

The diagonals bisect each other at right angles, so the half-diagonals and a side form a right triangle: {\left( \frac{{d}_{1}}{2}\right)}^{2}+{\left( \frac{{d}_{2}}{2}\right)}^{2}={10}^{2}, giving {d}_{1}^{2}+{d}_{2}^{2}=400.

We want the total length {d}_{1}+{d}_{2}, and {\left({d}_{1}+{d}_{2}\right)}^{2}=400+2\left( 192 \right)=784, so {d}_{1}+{d}_{2}=28 m.

The cost is 28\times 125=3500, that is ₹3500.

There is no need to find the diagonals separately, though they turn out to be 16 m and 12 m.