so what if a triangle has altitudes of 10, 15, and 25?

Then use the general approach given by Altheman.

oh okay. good idea. so the sides would be 2A/10, 2A/15, 2A/25 then use heron's formula.

Yup. The general formula is

haha that formula's pretty long. heron's is probably easier to do

That is the simplified result of Heron.

oh.. oops And the H stands for.. oh yeah

See here also: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&t=112464

Much simpler: Let sides be $a$, $b$, $c$. Equating areas we get $a = 5b/4$, $c=3b/4$. Now apply cosine rule to $\angle A$. By substituting $a$ and $c$ as above, we get $Cos\angle A = 0$. This means triangle is right angled at $A$! This implies $c = 15$ and $b = 20$. Thus area = $1/2*15*20$ = $150$.

I think this is middle school math

The problem itself is probably msm level but the later variations definitely aren’t.

This problem was also on Baltic Way 2006. Here is simple solution. We know that a $ah_a=bh_b=ch_c$ and now get $b=\frac{ah_a}{h_b}=\frac{12a}{15}=\frac{4a}{5}$ and $c=\frac{ah_a}{h_c}=\frac{12a}{20}=\frac{3a}{5}$. We will now use Heron formula to find the area. $P=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}=\frac{6a}{5}$ Finally, $P=\sqrt{\frac{36a^4}{625}}=\frac{6a^2}{25}$, and area is also $P=\frac{ah_a}{2}=6a$ Now, $\frac{6a^2}{25}=6a$ or $a=25 \implies b=20, c=15$, so $P=150$