What is the area of a triangle with side lengths $13$, $14$, and $15$?
Herons formula is \(\sqrt{s(s-a)(s-b)(s-c)}\), \(s\) is the perimiter divided by 2, and a, b, and c are the side lengths. We sum 13, 14, and 15 to get 42, then we plug in the lengths. \(\sqrt{42(42-13)(42-14)(42-15)}\)\(=\boxed{126\sqrt{58}}\)
QM....if you ACTUALLY looked at jcx516' s origianl solution....YOU should have been able to spot the minor error made in the calculation .... then you would not have to have been so rude/curt and unappreciative : "that is wrong"
You can split a 13-14-15 triangle into a 9-12-15 triangle and a 5-12-13 triangle, so the area is \(\frac{9 \cdot 12}{2} + \frac{5 \cdot 12}{2} = 84\).