The altitudes of a triangle are 12, 15 and 20. What is the area of the triangle?

Answer: 150.
Denote the sides of the triangle by \(a, b\) and \(c\) and its altitudes by \(h_{a}, h_{b}\) and \(h_{c}\). Then we know that \(h_{a}=12, h_{b}=15\) and \(h_{c}=20\). By the well known relation \(a: b = h_{b}: h_{a}\) it follows \(b = \frac{h_{a}}{h_{b}} a = \frac{12}{15} a = \frac{4}{5} a\). Analogously, \(c = \frac{h_{a}}{h_{c}} a = \frac{12}{20} a = \frac{3}{5} a\). Thus half of the triangle's circumference is \(s = \frac{1}{2}(a + b + c) = \frac{1}{2}\left(a + \frac{4}{5} a + \frac{3}{5} a\right) = \frac{6}{5} a\). For the area \(\Delta\) of the triangle we have \(\Delta = \frac{1}{2} a h_{a} = \frac{1}{2} a \cdot 12 = 6 a\), and also by the well known Heron formula \(\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{6}{5} a \cdot \frac{1}{5} a \cdot \frac{2}{5} a \cdot \frac{3}{5} a} = \sqrt{\frac{6^{2}}{5^{4}} a^{4}} = \frac{6}{25} a^{2}\). Hence, \(6 a = \frac{6}{25} a^{2}\), and we get \(a = 25 (b = 20, c = 15)\) and consequently \(\Delta = 150\).