$A$ field in the form of a parallelogram has sides $60 \, m$ and $40 \, m$ and one of its diagonals is $80 \, m$ long. Find the area of the parallelogram.

(N/A) Let the field be $ABCD.$
Area of the parallelogram $ABCD = 2 \times (\text{Area of } \Delta ABC) \quad ...(1)$
Now,the sides of $\Delta ABC$ are $a = 40 \, m, b = 60 \, m$ and $c = 80 \, m$.
Semi-perimeter of $\Delta ABC$,$s = \frac{a+b+c}{2} = \frac{40+60+80}{2} = \frac{180}{2} = 90 \, m$.
Using Heron's Formula,Area of $\Delta ABC = \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{90(90-40)(90-60)(90-80)}$
$= \sqrt{90 \times 50 \times 30 \times 10}$
$= \sqrt{1350000} = 300\sqrt{15} \, m^2 \approx 1161.895 \, m^2$.
From equation $(1)$,
Area of parallelogram $ABCD = 2 \times 300\sqrt{15} = 600\sqrt{15} \, m^2 \approx 2323.79 \, m^2$.