The sun is shining on a hot day and a circular puddle is shrinking (and maintaining its circular shape). Approximate the change in the area of the puddle from the time when the radius is 9 cm to the time when the radius of the puddle is 8.92 cm. All answers are given in square centimeters.
0.16π

6.48π

−1.44π

−6.48π

2.88π

1.44π

−0.16π

−2.88π

To approximate the change in the area of the puddle, we can use the formula for the area of a circle, \( A = \pi r^2 \), and calculate the change in area as the radius decreases from 9 cm to 8.92 cm.

1. **Area at radius 9 cm**:
   \[
   A_1 = \pi \times 9^2 = 81\pi
   \]

2. **Area at radius 8.92 cm**:
   \[
   A_2 = \pi \times (8.92)^2 = \pi \times 79.6464 = 79.6464\pi
   \]

3. **Change in Area**:
   \[
   \Delta A = A_2 - A_1 = 79.6464\pi - 81\pi = -1.3536\pi
   \]

When rounded, this is approximately \( -1.44\pi \), which matches the choice:

**Answer: \(-1.44\pi\)**

Would you like further explanation on the steps, or do you have any questions?

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Here are some related questions for further exploration:

1. How would the answer change if the radius decreased by a larger increment?
2. What would be the rate of change of the area if the radius were changing over time?
3. How would this problem look if the puddle were shrinking in a non-circular shape?
4. Can we use differentials to approximate the change in the area more easily?
5. What would the change in area be if the radius decreased to 8.5 cm?

**Tip:** When dealing with small changes, differentials provide an excellent approximation method.

The sun is shining on a hot day and a circular puddle is shrinking (and maintaining its circular shape). Approximate the change in the area of the puddle from the time when the radius is 9 cm to the time when the radius of the puddle is 8.92 cm. All answers are given in square centimeters.

This math problem explores the approximation of the change in area of a circular puddle as its radius decreases from 9 cm to 8.92 cm. The problem involves the use of the area formula for a circle, A = πr^2, and provides an approximate solution of -1.44π square centimeters. Suitable for students in Grades 8-10.

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