5. [7] A triangle has altitudes of length 15,21 , and 35 . Find its area.
Answer: $\quad 245 \sqrt{3}$ If $A$ is the area of the triangle, the sides are $\frac{2 A}{15}, \frac{2 A}{21}$, and $\frac{2 A}{35}$. So the triangle is similar to a $\frac{1}{15}, \frac{1}{21}, \frac{1}{35}$ triangle, which is similar to a $3,5,7$ triangle. Let the sides be $3 k, 5 k$, and $7 k$. Then the angle between the sides of length $3 k$ and $7 k$ is $120^{\circ}$, so the area is $\frac{15 \sqrt{3}}{4} k^{2}$. But the area can also be calculated as $\frac{(3 k)(35)}{2}=\frac{105 k}{2}$. Setting these values equal, $k=\frac{14 \sqrt{3}}{3}$ and the area is $245 \sqrt{3}$.