This problem is about related rates in calculus. Let's solve it step by step.

Problem Analysis:

The formula for the area of a right triangle is: $A = \frac{1}{2}ab$ where $a$ and $b$ are the two legs of the triangle.

The rates of change are:

• Short leg ($a$) increasing at $\frac{da}{dt} = 9 \, \text{in/sec}$,
• Long leg ($b$) shrinking at $\frac{db}{dt} = -9 \, \text{in/sec}$.

We want the rate of change of the area ($\frac{dA}{dt}$).

Step 1: Differentiate the Area Formula

Differentiating $A = \frac{1}{2}ab$ with respect to $t$ gives: $\frac{dA}{dt} = \frac{1}{2} \left( a \frac{db}{dt} + b \frac{da}{dt} \right)$

Step 2: Substitute Known Values

At the given moment:

• $a = 12 \, \text{in}$,
• $b = 16 \, \text{in}$,
• $\frac{da}{dt} = 9 \, \text{in/sec}$,
• $\frac{db}{dt} = -9 \, \text{in/sec}$.

Substituting these into the formula: $\frac{dA}{dt} = \frac{1}{2} \left( 12(-9) + 16(9) \right)$

Step 3: Simplify

First, calculate the terms inside the parentheses: $12(-9) = -108, \quad 16(9) = 144$ $\frac{dA}{dt} = \frac{1}{2} \left( -108 + 144 \right)$ $\frac{dA}{dt} = \frac{1}{2} (36)$ $\frac{dA}{dt} = 18 \, \text{in}^2/\text{sec}.$

Final Answer:

The rate of change of the area is: $\boxed{18 \, \text{in}^2/\text{sec}}.$

Would you like me to explain any step in more detail or solve another related problem?

Here are 5 related questions to consider:

1. What happens to $\frac{dA}{dt}$ if both legs are increasing?
2. How would you calculate $\frac{dA}{dt}$ if the hypotenuse was also changing?
3. How does this problem relate to optimization concepts?
4. What is the effect of one leg’s rate of change dominating the other?
5. Can you visualize how the area changes as time progresses?

Tip: Always note the sign of rates of change carefully; negative signs indicate shrinking dimensions!