To solve the problem of finding the area of the shaded region in the diagram, let's break it down step-by-step:

Step 1: Understand the Geometry

The shaded region is the intersection of two identical circles with radii $r = 10 \, \text{cm}$, whose centers are $16 \, \text{cm}$ apart.

The area of the shaded region (intersection) can be calculated as the area of two circle segments. Each segment is part of a circle that lies on either side of the line joining the two circle centers.

Step 2: Segment Area Formula

The area of a segment of a circle is given by: $\text{Segment Area} = r^2 \cos^{-1}\left(\frac{d}{2r}\right) - \frac{d}{2} \sqrt{4r^2 - d^2}$ where:

• $r$ is the radius of the circle ($10 \, \text{cm}$),
• $d$ is the distance between the centers of the circles ($16 \, \text{cm}$).

Step 3: Calculate One Segment Area

Let us substitute $r = 10 \, \text{cm}$ and $d = 16 \, \text{cm}$ into the formula:

1. Calculate $\cos^{-1}\left(\frac{d}{2r}\right)$: $\cos^{-1}\left(\frac{d}{2r}\right) = \cos^{-1}\left(\frac{16}{20}\right) = \cos^{-1}(0.8)$ Using a calculator or reference, $\cos^{-1}(0.8) \approx 0.6435 \, \text{radians}$.

2. Calculate the term $\frac{d}{2} \sqrt{4r^2 - d^2}$: $\frac{d}{2} = \frac{16}{2} = 8$ $\sqrt{4r^2 - d^2} = \sqrt{4(10)^2 - 16^2} = \sqrt{400 - 256} = \sqrt{144} = 12$ So: $\frac{d}{2} \sqrt{4r^2 - d^2} = 8 \times 12 = 96$

3. Segment area for one circle: $\text{Segment Area} = 10^2 \times 0.6435 - 96 = 100 \times 0.6435 - 96 = 64.35 - 96 = 28.35 \, \text{cm}^2$

Step 4: Total Intersection Area

Since there are two identical circle segments contributing to the intersection: $\text{Total Shaded Area} = 2 \times 28.35 = 56.7 \, \text{cm}^2$

Final Answer:

The area of the shaded region is: $\boxed{56.7 \, \text{cm}^2}$

Would you like a detailed explanation of any step or visual representation? Here are related questions for further exploration:

1. How does the segment area formula account for both arc and triangular portions?
2. Can the shaded area be derived using integration instead of geometry?
3. What happens if the circle centers are closer or farther than 16 cm?
4. How can the formula be generalized for circles of different radii?
5. How can approximations affect the result in practical applications?

Tip: Always double-check the angle in radians when working with trigonometric functions, as incorrect units can lead to errors.