Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. A circle has a radius of 6 in. The inscribed equilateral triangle will have an area of:

There are several ways you can go at this. One is to recognize that the triangle formed by two vertices and the circle center is an isosceles triangle with apex angle 120°. Its area will be

A = (1/2)·r²·sin(120°) = (1/2)·(6 in)²·(√3)/2 = 9√3 in²

The total area of the inscribed triangle is the area of 3 of these, so is ...

A = 3·9√3 in² = 27√3 in²

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Alternate solution

If we call the center of the circle point O, and the vertices of the inscribed equilateral triangle A, B, and C, we can define point X on segment AB such that OX⊥AB.

Triangle AOX is a 30°-60°-90° triangle, so has sides in the ratios 1 : √3 : 2. That is, AO is twice the length of OX, and AX is √3 times the length OX.

This makes OX = 3, AX = XB = 3√3, and XC = OX+OC = 3+6 = 9.

Now we have the dimensions of triangle ABC: the base AB is 6√3, and the height (XC) is 9. The area can be found in the usual way:

A = 1/2·bh = 1/2·(6√3 in)·(9 in) = 27√3 in²