BTGmoderatorLU wrote:Source: Economist GMAT



If P is the center of the circle shown above, and BAC = 30º, and the area of triangle ABC is 6, what is the area of the circle?
$$A.\ \sqrt{3}\pi$$
$$B.\ 2\sqrt{3}\pi$$
$$C.\ 4\pi$$
$$D.\ 6\pi$$
$$E.\ 4\sqrt{3}\pi$$
The OA is E

Area of the circle = πr^2, where r = AC/2

If we get the value of AC, we get the answer.

Note that since AC is a diameter, it subtends 90º on the circumference of the circle. Thus, triangle ABC is a rightangled triangle of 30-60-90 type. A rightangled triangle of 30-60-90 type has its sides in the ratio of 1 : √3 : 2. Thus, BC : AB : AC :: 1 : √3 : 2.

Area of the triangle = 1/2*(AB*BC) = 6

=> AB*BC = 12

=> √3(BC)^2 = 12

BC = √(12/√3)

=> AC = 2*[√(12/√3)]

Radius = 1/2 of AC = 1/2 of 2*[√(12/√3)] = √(12/√3)

Area of the circle = πr^2 = π[√(12/√3)]^2 = 4√3π

The correct answer: E

Hope this helps!

-Jay
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BTGmoderatorLU wrote:Source: Economist GMAT



If P is the center of the circle shown above, and BAC = 30º, and the area of triangle ABC is 6, what is the area of the circle?
$$A.\ \sqrt{3}\pi$$
$$B.\ 2\sqrt{3}\pi$$
$$C.\ 4\pi$$
$$D.\ 6\pi$$
$$E.\ 4\sqrt{3}\pi$$
The OA is E

When a triangle is inscribed in a circle with a side coinciding with the diameter of the circle, that triangle must be a right triangle, and the side coinciding with the diameter is the hypotenuse. Therefore, triangle ABC is a right triangle with hypotenuse AC. Furthermore, since angle BAC = 30 degrees, triangle ABC is a 30-60-90 triangle. So if BC = x, AB = x√3, then we have:

1/2 * x * x√3 = 6

x^2√3 = 12

x^2 = 12/√3 = (12√3)/3 = 4√3

Notice that AC = 2x, so x is the length of the radius of the circumscribed circle and, therefore, the area of the circle is:

πx^2 = π * 4√3 = (4√3)π

Answer: E