To find the rate of change of the area of the triangle, we can use the formula for the area of a triangle:

$A = \frac{1}{2} \times \text{base} \times \text{height}$

Let $b$ represent the base of the triangle and $h$ represent the height. Then, the area $A$ can be expressed as:

$A = \frac{1}{2} \times b \times h$

Given:

• The height $h = 9$ cm (constant)
• The rate of change of the base $\frac{db}{dt} = -16$ cm per hour (negative because the base is decreasing)

We need to find the rate of change of the area, $\frac{dA}{dt}$.

Differentiate the area formula with respect to time $t$:

$\frac{dA}{dt} = \frac{1}{2} \times \left( \frac{db}{dt} \times h + b \times \frac{dh}{dt} \right)$

Since the height $h$ is constant, $\frac{dh}{dt} = 0$. Therefore, the equation simplifies to:

$\frac{dA}{dt} = \frac{1}{2} \times \frac{db}{dt} \times h$

Substitute the given values:

$\frac{dA}{dt} = \frac{1}{2} \times (-16) \times 9 = \frac{1}{2} \times (-144) = -72 \text{ cm}^2 \text{ per hour}$

So, the rate of change of the area of the triangle is $-72$ square centimeters per hour.

Would you like further details or have any questions?

Here are 5 related questions you might find interesting:

1. How would the rate of change of the area be affected if the height was also changing?
2. What if the base was increasing instead of decreasing? How would that change the rate?
3. How do you interpret the negative sign in the rate of change of the area?
4. What happens to the area if both the base and height are decreasing?
5. How does the rate of change of the area compare if the base was changing at a slower rate?

Tip: When dealing with rates of change, it's crucial to keep track of whether quantities are increasing or decreasing, as this affects the sign of the rate.