Bunuel wrote:
In the diagram below, P and Q are the centers of the two circles. What is the area of triangle XYZ?
(1) Radius of each circle is 6.
(2) XY = 2.
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It should be
C.
Statement 1: InsufficientKnowing the radius tells us nothing about the points of intersection.
Statement 2: InsufficientIt tells us nothing about the position of points \(Z\) and \(X\) relative to the centers of the circles and points \(Y\) and \(X\).
Statement 1 + 2: Sufficient - Bear with me as it may not be easy to understand.
Circle P:- We know that arc \(ZQV\) is \(120\) degrees since circle \(P\) is being carved by a circle \(Q\) of equal radius.
- Angle that point \(Y\) creates with points \(Z\) and \(V\) is half of the central angle \(ZPV\)
(central angle theorem).
- This makes angle \(Y = 120*\frac{1}{2} = 60\) degrees.
See figure 1.Circle Q:- We know that point \(Q\) splits angle \(Y\) exactly in half (project line \(YZ\) further into the circle Q where it hits the opposite end internally to understand. You can shift Y as well on the circumference of circle P).
- Thus radius \(QX\) meets the chord \(YV\) exactly where \(XY = YZ\) (\(Z\) is already on radius of \(Q\)).
See figure 2 to understand- We now have two sides \(XY\) and \(XZ\) equal and their central angle \(60\) degrees.
- This is all we need since this makes it an equilateral triangle (draw it to understand better). We can now apply the formula for area of equilateral triangle \(XY^2 * \sqrt{3}/4\). - This results in \(\sqrt{3}\)
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