+355 Triangle $ABC$ has sides of length $11$ inches, $15$ inches and $20$ inches. What is the length of the altitude to the side of length $15$ inches? Express your answer in simplest radical form.
+394 By Heron's formula, the area of the triangle is
\(\sqrt{23*(23-11)*(23-15)*(23-20)}=\sqrt{23*12*8*3}=\sqrt{6624}=12\sqrt{46}\).
Because side * altitude / 2 = area, then.
\(\frac{20x}{2}=12\sqrt{46}\)
\(10x=12\sqrt{46}\)
\(x=\frac{6\sqrt{46}}{5}\).
