Answer:
(C) 50π
Step-by-step explanation:
For this problem, we need to find a way to relate the inner circle, to the square, to the outer circle.
Given that we know the area of a circle is πr^2, and our inner circle has a area of 25π, we can find the radius.
25π = πr^2
25 = r^2
5 = r
Note, that the diameter of the inner circle is parallel to the side of the square, meaning that the diameter of the inner circle is the length of the side of the square.
diameter = 2 * radius
d = 2 * 5 = 10.
Now that we know the value of the side of the square, we can find the length of the diagonal of the square, which is the diameter of the outer circle.
Using the properties of the 45-45-90 right triangle, we can say that the diagonal of the square is the length of the side times sqrt(2).
Outer_Diameter = 10 * sqrt(2)
Now to find the outer area, we need the formula for the area of a circle. Note that the diameter is twice the radius, so we will simply divide by 2.
A = πr^2
A = π * [ ( 10 * sqrt(2) ) / 2 ]^2
A = π * [ 5 * sqrt(2) ] ^2
A = π * 25 * 2
A = 50π
So the area of the outer circle is 50π.
Cheers.
