Calculating the area of a triangle assuming you don't know Heron's formula I was wondering - let's say you know the lengths of the sides of a triangle and you want to calculate its area. This could be performed very easily if you know Heron's formula but... What if you don't?
What could one do to calculate the area of a triangle if they know the side lengths assuming they completely forgot how the Heron's formula went? Is deriving the formula yourself the only way or is there a simpler way of solving such a problem?
 A: As hinted in the comments, Euclid does indeed mention something that is very close to the law of cosines. In Book II, Proposition 12 covers obtuse-angled triangles and, as quoted below, Proposition 13 covers acute-angled triangles:

In acute-angled triangles the square on the side subtending the acute
  angle is less than the squares on the sides containing the acute angle
  by twice the rectangle contained by one of the sides about the acute
  angle, namely that on which the perpendicular falls, and the straight
  ine cut off within by the perpendicular towards the acute angle.

Let the following ABC acute-angled triangle having the angle B acute, and let AD be drawn from point A perpendicular to BC:

The conclusion of the Proposition is:
$$
c^2 = b^2 + a^2 + 2aCD
$$
The height of the triangle can then be calculated as:
$$
CD = \frac{c^2 - b^2 - a^2}{2a} \\
h = \sqrt{b^2 - CD^2} \\
$$
The area of the triangle can finally be calculated without Heron's formula nor trigonometric functions. For example, given $c=25$, $b=17$, $a=26$ using Heron's formula:
$$
s = \frac{25 + 17 + 26}{2} = 34 \\
A = \sqrt{34 (34-25) (34-17) (34-26)} = 204 \\
$$
And using this formula:
$$
CD = \frac{25^2 - 17^2 - 26^2}{2 \times 26} = -\frac{85}{13} \\
h = \sqrt{17^2 - (-\frac{85}{13})^2} = \frac{204}{13} \\
A = \frac{26 \times \frac{204}{13}}{2} = 204 \\
$$
