Area of an acute triangle, given two sides and an altitude 
$12$, $13$, $15$ are the lengths (perhaps not in order) of two sides of an acute-angled triangle and of the height over the third side of triangle. Find the area of the triangle (no calculator allowed).


Misread the question. 
Draw the triangle and the height. Denote by $x,y$ the two segments determined by the height on the triangle. 
First, since the perpendicular to a line is smaller than any of the secants, the altitude must be 12. 
You have
$$12^2+x^2=13^2 \Rightarrow x^2=13^2-12^2= (13-12)(13+12)=25 \\
12^2+y^2=15^2 \Rightarrow y^2=15^2-12^2= (15-12)(15+12)=3*=81 $$
Therefore the last side is $\sqrt{25}+\sqrt{81}=14$
Then the area is 
$$\frac{12 * 14}{2}= 84$$
