<s>[INST] 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).

 [/INST] 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 
$$\frac{12 * 14}{2}= 84$$
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