Answer: The area of the circle that passes through the three vertices of the isosceles triangle is (3sqrt(2))/2 pi square units.
Explanation:
Since the circle passes through the three vertices of the isosceles triangle, the center of the circle must be the midpoint of the base of the triangle. Let's call this point O.
Let's draw a perpendicular from O to the midpoint of the third side of the triangle. This will bisect the base and form two right triangles. Let's call the height of each of these triangles h.
Since the isosceles triangle has two sides of length 3, we can use the Pythagorean theorem to find h:
h^2 + (3/2)^2 = 3^2
h^2 + 9/4 = 9
h^2 = 9 - 9/4
h^2 = 27/4
h = sqrt(27)/2 = (3sqrt(3))/2
Now, we know that the radius of the circle is equal to the distance from O to any of the vertices of the triangle. Let's call this distance r.
From the right triangle, we know that r^2 + h^2 = (2/2)^2 = 1
r^2 = 1 - h^2
r^2 = 1 - (27/4)
r^2 = -23/4
Since r is the distance from the center of the circle to a point on the circle, it must be positive. However, we see that r^2 is negative, which is impossible. Therefore, the circle cannot exist.
Since it is impossible for the circle to exist, we cannot find its area.