For problem 2:

Let's denote the following:

O: Center of both circles

A and B: Endpoints of the chord (chord length AB = 10)

P: Point of tangency between the chord and the smaller circle

r: Radius of the smaller circle

R: Radius of the larger circle

Since the chord is tangent to the smaller circle at point P, we know that OP is perpendicular to AB. Additionally, since O is the center of both circles, line segment OP bisects chord AB (AP = BP = 5).

Finding the Radius of the Smaller Circle (r):

Using the Pythagorean Theorem in right triangle APO:

AP^2 + OP^2 = r^2 (since AP = BP = 5)

5^2 + OP^2 = r^2

We don't have the value of OP yet, but we can find it using the information about the larger circle and the chord.

Finding the Radius of the Larger Circle (R):

In right triangle OBP:

OB^2 + OP^2 = R^2 (since OB = half the chord length = 5)

5^2 + OP^2 = R^2

Relating the Radii (R and r):

Since the chord is tangent to the smaller circle, the distance between the center (O) and the point of tangency (P) is the radius of the smaller circle (r).

Additionally, this distance (OP) is also the difference between the radii of the larger and smaller circles (R - r).

Therefore, we have: OP = r and OP = R - r

Setting these two expressions equal to each other:

r = R - r

Solving for r (the smaller radius):

2r = R r = R/2

Substitute r in terms of R back into Equation 1:

5^2 + (R/2)^2 = R^2

Expand and rearrange:

R^2 - 5R - 25 = 0

Factor the equation:

(R - 10)(R + 2.5) = 0

Since the radius cannot be negative, we have R = 10.

Therefore, the radius of the smaller circle (r) is r = R/2 = 5.

Area of the Ring-Shaped Region:

The area of the ring is the difference between the area of the larger circle and the area of the smaller circle:

Area of Ring = π * R^2 - π * r^2

Substitute the values of R and r:

Area of Ring = π * (10)^2 - π * (5)^2

Area of Ring = π * (100 - 25)

Area of Ring = 75π

Therefore, the area of the ring-shaped region is 75π square units.