The length of the shorter altitude and the shorter leg of a parallelogram are 9 cm and sqrt82 cm, respectively. The length of the longer diagonal is 15 cm. What is the area of this parallelogram?
Let $h$ be the shorter altitude of the parallelogram and $d$ the shorter diagonal. Using the formula for the area of a parallelogram, we see that the area of the parallelogram is $h\cdot d$. Thus, we need to find $h$ in order to solve the problem.
Let $a$ and $b$ be the sides of the parallelogram with $a > b$. Then, the shorter leg of the parallelogram is $b= \sqrt{82}$ cm.
Since the diagonals of a parallelogram bisect each other, we know that the shorter diagonal is equal to half the sum of the sides of the parallelogram. So we have the equation $\frac{15}{2}=a+b$. We can solve this equation for $a$ in terms of $b$ by substituting $b = \sqrt{82}$.
\begin{align*} \frac{15}{2}&=a + \sqrt{82} \\ \frac{15}{2}-\sqrt{82}&=a. \end{align*}
Since the altitudes of a parallelogram are perpendicular to the bases, we see that $h^{2} + (15/2-\sqrt{82})^{2} = 9^{2}$. Solving for $h$, we find that $h = 3\sqrt{2}$. Thus, the area of the parallelogram is $h\cdot d = 3\sqrt{2}\cdot\sqrt{82} = \boxed{6\sqrt{41}}$ square centimeters.