9.1 a) Given a triangle with heights equal to 4, 5, and 6. What type of triangle is it: acute, right, or obtuse? b) Does a triangle exist with heights equal to 2, 3, and 6?
Answer: a) acute-angled; b) does not exist. Hint: From the formula for the area of a triangle $S=a h / 2$, it follows that the sides of the given triangle are $\frac{2 S}{4} ; \frac{2 S}{5} ; \frac{2 S}{6}$ (such a triangle exists, since $\frac{1}{4}<\frac{1}{5}+\frac{1}{6}$). To determine what kind of triangle it is, we need to compare the square of the largest side with the sum of the squares of the other sides. In this case, from the inequality $\left(\frac{1}{4}\right)^{2}<\left(\frac{1}{5}\right)^{2}+\left(\frac{1}{6}\right)^{2}$, it follows that the triangle is acute-angled. b) Since $\frac{1}{2}=\frac{1}{3}+\frac{1}{6}$, the triangle inequality does not hold, and therefore, such a triangle does not exist.