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In figure 5, rectangle ABCD is inscribed in a circle. If the radius of the circle is 1 and AB = 1, what is the area of the shaded region?

A) 0.091
B) 0.285
C) 0.614
D) 0.705
E) 0.732

Draw a diagonal of the rectangle. That’s going to be the diameter of the circle, too, so you know it’ll have a length of 2. That makes a right triangle with a leg of 1 and a hypotenuse of 2. Do you know your special right triangles? If you do, you recognize that that’s a 30º-60º-90º triangle, so you can save yourself a tiny bit of time: the long side of the rectangle is \sqrt{3}. Nice, right?

OK, now draw the other diagonal.

What you’ve got there, now, is a central angle, which tells you that a 120º angle cuts out the only part of the circle you care about. If you don’t know why that’s important, read this.

The area of the whole circle is π, which you know because the radius is 1. The sector you care about is \dfrac{120}{360} = \dfrac{1}{3} of that, so \dfrac{\pi}{3}. The triangle you’re taking away from it has an area of \dfrac{1}{2}(\sqrt{3})\left(\dfrac{1}{2}\right)=\dfrac{\sqrt{3}}{4}.

To find the area of the shaded region, subtract!

\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{4}=0.614...

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