Uhh, Heron's Formula?

$11\times13-\dfrac{11\times5+13\times4+8\times7}2=61.5$

Just to see how Heron's formula works here:

Heron's formula can be written as $A = \frac{1}{4}\sqrt{4a^2b^2 - (a^2+b^2-c^2)^2}$ which is less symmetric than the usual form, but it allows us to remove the radicals in the given side lengths (and as a plus, note that this form easily reduces to $\frac{1}{2}ab$ with a right triangle)

That is, if we set $a=\sqrt{146},b=\sqrt{113},c=\sqrt{185},$ then $\begin{aligned} A &= \frac{1}{4}\sqrt{4(146)(113) - (146+113-185)^2} \\ &= \frac{1}{4}\sqrt{4(146)(113) - 74^2} \\ &= \frac{1}{2}\sqrt{(146)(113) - 37^2} \\ &= \frac{1}{2}\sqrt{15129} \\ &= \boxed{61.5} \end{aligned}$

Let $a = \sqrt{113}$ , $b = \sqrt{146}$ , and $c = \sqrt{185}$ .

Then by the law of cosines, $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{\sqrt{113}^2 + \sqrt{146}^2 - \sqrt{185}^2}{2\sqrt{113}\sqrt{146}} = \frac{137}{\sqrt{113}\sqrt{146}}$ .

And $\sin C = \sqrt{1 - \cos^2 C} = \sqrt{1 - (\frac{137}{\sqrt{113}\sqrt{146}})^2} = \frac{123}{\sqrt{113}\sqrt{146}}$

So the area of the triangle is $A = \frac{1}{2}ab \sin C = \frac{1}{2} \sqrt{113} \sqrt{146} \frac{123}{\sqrt{113} \sqrt{146}} = \frac{123}{2} = \boxed{61.5}$ .

It is simple.Just use your programing knowledge.I had done this question by using Turbo C++.