Answer:
Explanation:
An isosceles triangle with sides of 3/6, 3/6 and an unknown length, and a circle tangent to the two sides with a given radius can be solved using the Pythagorean theorem, trigonometry and the circle area formula.
We know that the triangle is isosceles, so the angle at point B is congruent to angle at point C, and let's call it x. Now we know that angle BAC = 90- x. Since the circle is tangent to the side AB and AC at B and C respectively, we can use the property of tangents that says that the radius of a circle that is tangent to a line is perpendicular to that line, and we can use the triangle formed by the radius, the side and the angle at that point to apply the Pythagorean theorem.
3/6^2 + r^2 = (3/6/sin(x))^2
r^2 = (3/6)^2 (1/sin^2(x) -1)
Now we can use the area of the circle formula A = πr^2, to find the area of the circle that passes through the vertices A, B, and C:
A = π r^2 = π(3/6)^2 (1/sin^2(x) -1)
Using the values of the problem we get
A = π(3/6)^2 (1/sin^2(90- x) -1) = π(3/6)^2 (1/cos^2(x) -1) = π(52^2) = 2701π = 27.01 (D)
Therefore the area of the circle that passes through vertices A, B, and C is 27.01.