3. [5] Let $D E F$ be a triangle and $H$ the foot of the altitude from $D$ to $E F$. If $D E=60, D F=35$, and $D H=21$, what is the difference between the minimum and the maximum possible values for the area of $D E F$ ?
Answer: 588 Observe that the two possible configurations come from $D E F$ obtuse and $D E F$ acute. In either case, we have that $H F=\sqrt{35^{2}-21^{2}}=28$ and $E H=\sqrt{60^{2}-21^{2}}=9 \sqrt{39}$. This means that $H F-E H$, so in the two cases the values of $E F$ are $F H+E H$ and $E H-F H$. The difference in area is hence $\frac{1}{2} \cdot 2 \cdot F H \cdot D H=28 \cdot 21=588$.