Two adjacent sides of a parallelogram are $74\text{cm}$ and $40\text{cm}$. If one of its diagonals is $102\text{cm}$, then the area of the parallelogram is

A diagonal splits a parallelogram into two triangles of equal area. Let the sides of one triangle be $a=74\text{cm}$, $b=40\text{cm}$, and $c=102\text{cm}$.Find the semi-perimeter ($s$):
$$s=\frac{a+b+c}{2}=\frac{74+40+102}{2}=\frac{216}{2}=108\text{cm}$$
Calculate the area of this triangle using Heron's formula:
$$\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$$
$$\text{Area}=\sqrt{108(108-74)(108-40)(108-102)}$$
$$\text{Area}=\sqrt{108\times 34\times 68\times 6}$$
$$\text{Area}=\sqrt{(36\times 3)\times 34\times (34\times 2)\times 6}$$
$$\text{Area}=\sqrt{36\times 34^2\times (3\times 2\times 6)}=\sqrt{36\times 34^2\times 36}$$
$$\text{Area}=36\times 34=1224\text{sq.cm}$$
Total area of the parallelogram is twice the area of this triangle:
$$\text{Total Area}=2\times 1224=2448\text{sq.cm}$$