Bunuel wrote:
In the figure above, A and B are the centers of the two circles. If each circle has radius X, what is the area of the shaded region?
A. \(\frac{(2\pi - \sqrt{3}) x^2}{6}\)
B. \(\frac{(4\pi - 3 \sqrt{3}) x^2}{12}\)
C. \(\frac{(4\pi - 3 \sqrt{3}) x^2}{6}\)
D. \(\frac{(4\pi - \sqrt{3}) x^2}{6}\)
E. \(\frac{(6\pi - 1) x^2}{6}\)
Please refer the enclosed diagram,
It is given that C1C2=AC1=AC2=C1D=C2D=radius=x,
So, AC1C2 is an equilateral triangle. You know, In an equilateral triangle, each of the three angles are equal. So, \(\angle{AC1C2}=60^{\circ}\)
Similarly, DC1C2 is an equilateral triangle. So, \(\angle{ADC1C2}=60^{\circ}\)
Hence, So, \(\angle{AC1D}=\angle{AC1C2}+\angle{ADC1C2}=120^{\circ}\), which is the central angle of circle with center C1.
Similarly, \(\angle{AC2D}=120^{\circ}\), which is the central angle of circle with center C2.
Since both the circles subtend an equal angle at their centers, hence the sector formed are of equal area.
So area of any of the sectors under discussion= \(\pi\)\(x^2*(\frac{120}{360})\)=\(\pi\)\(\frac{x^2}{3}\)
Now Area of shaded region=Area of sector of 1st circle+Area of sector of 2nd circle-4*Area of \(\triangle\)AC1B-----(1)
(We are subtracting the area of triangles since total area of both the sectors contains additional area of these 4 triangles whose areas is equal)
From the figure, each of the triangles is a 30-60-90 triangle with sides in the the ratio \(1:\sqrt{3}:2\)
In the \(\triangle\)AC1B, \(C1B=\frac{x}{2}\), hence \(AB=\sqrt{3}C1B=\sqrt{3}\frac{x}{2}\)
So, Area of \(\triangle\)AC1B=\(\frac{1}{2}*\frac{x}{2}*\sqrt{3}\frac{x}{2}\)=\(\sqrt{3}\frac{x^2}{8}\)
From(1), we have AREA OF SHADED PORTION=\(2*\pi\)\(\frac{x^2}{3}\) - \(4*\sqrt{3}\frac{x^2}{8}\)=\(\frac{(4\pi - 3 \sqrt{3}) x^2}{6}\)
Ans. (C)
P.S:- I have changed the nomenclature from 'A' 'B' of original figure to 'C1' 'C2' for easy understanding.
kanthaliya,
Sandy56,
rraman,
Srija221 Hope it helps.
Open to further queries(if any).
Attachments
Area of shaded.JPG [ 27.62 KiB | Viewed 8719 times ]
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Regards,
PKN
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