rxs0005 wrote:

If a triangle inscribed in a circle has area 40, what is the area of the circle?

(1) The base of the triangle is equal to the diameter of the circle.

(2) The measure of one of the angles in the triangle is 30.

(1)

This statement tells us that the inscribed triangle is a right angled triangle.

Right angled triangle have three sides;

Two smaller sides and one hypotenuse.

Area = 1/2*(smaller side)*(bigger side)=40 if we assume that it is not an isosceles right triangle.

We know the radius = 1/2*(Diameter) = 1/2*(Hypotenuse) [Hypotenuse is referred as "BASE" in this statement]

Not Sufficient.

(2)

The measure of one of the angles in the triangle is 30.

We can make plenty such triangles with one angle as 30.

Not sufficient.

Combining both;

We know that the triangle is a right angled triangle with 30-90-60 as its angles.

The opposite sides will be in ratio:

x:2x:\sqrt{3}xArea = \frac{1}{2}*x*\sqrt{3}x=40x^2=\frac{80}{\sqrt{3}}x=\sqrt{\frac{80}{\sqrt{3}}}Hypotenuse = Diameter = 2x= 2*\sqrt{\frac{80}{\sqrt{3}}}Area = \pi*(\frac{Diameter}{2})^2=\pi*(\sqrt{\frac{80}{\sqrt{3}}})^2=\pi*\frac{80}{\sqrt{3}}Sufficient.

Ans: "C"

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~fluke