Bunuel wrote:

PQ, QR and PR are diameters of the three circles shown above. If QR = 4, and PQ = 2QR, what is the area of the shaded region?

(A) 6π

(B) 8π

(C) 10π

(D) 12π

(E) 16π

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Find areas of circles, then divide by 2 (shaded and unshaded regions are in semicircles)

To find shaded area: Use just the bottom half + \(\frac{1}{2}\) Circle QR

Circle QR diameter = 4, r = 2

Circle QR area =

\(4\pi\)

\(\frac{1}{2} * 4\pi = 2\pi\) = \(\frac{1}{2}\) Area of Circle QR

Circle PQ diameter: (2*4) = 8, r = 4

Circle PQ area =

\(16\pi\)

\(\frac{1}{2} * 16\pi = 8\pi\) = \(\frac{1}{2}\) Area of Circle PQ

Circle PR diameter: (4 + 8 = 12), r = 6

Circle PR area =

\(36\pi\)

\(\frac{1}{2} * 36\pi = 18\pi\) = \(\frac{1}{2}\) Area of Circle PR

Shaded area = Use just the bottom half (\(\frac{1}{2}\) PR + \(\frac{1}{2}\) QR) - \(\frac{1}{2}\)PQ

\(\frac{1}{2}\) QR is the little shaded area in the top half

Shaded area = Areas of

(\(\frac{1}{2}\) of shaded large Circle PR + \(\frac{1}{2}\) shaded little Circle QR)* - \(\frac{1}{2}\) unshaded medium Circle PQ

\((18\pi + 2\pi) - 8\pi\)

\(20\pi - 8\pi = 12\pi\)Answer D

No need to account for the \(\frac{1}{2}\) shaded little Circle QR in the bottom half; it is included in \(\frac{1}{2}\) big Circle QR in the bottom half
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