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In the figure above, A and B are the centers of the two circles. If th
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20 Jun 2023, 01:02
20 Jun 2023, 01:02
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47% (02:52) correct
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In the figure above, A and B are the centers of the two circles. If the distance between A and B is 2 units, then what is the area, in square units, of the shaded region ?
A. \(\frac{4\pi}{3}-2 \sqrt{3}
\)
B. \(\frac{2\pi}{3}-\sqrt{3}\)
C. \(\frac{4\pi}{3}-\sqrt{3}\)
D. \(\frac{2\pi}{3}-2 \sqrt{3}\)
E. \(\frac{4\pi}{3}-4 \sqrt{3}\)
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Re: In the figure above, A and B are the centers of the two circles. If th
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09 Jul 2023, 14:25
09 Jul 2023, 14:25
Any explanations? I am completely lost here.
Posted from my mobile device
Posted from my mobile device
In the figure above, A and B are the centers of the two circles. If th
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09 Jul 2023, 23:22
09 Jul 2023, 23:22
2
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OE
As A and B are centers of the circle, AB becomes the radius of both the circles. AC is the radius of circle with center A, and BC is the radius of circle with center B. So, we can say AB, BC and AC are equal in length.
Hence ⊿ ABC is an equilateral triangle.
Similarly, ⊿ ABD is also an equilateral triangle.
Now, Area of shaded region = Area of sector BAD + Area of sector ABD - Area of Triangle ABD. We know, Area of sector \(= \frac{\pi r^2 \times 0 }{360}\) where, r = radius of the circle, ϴ = angle subtended by the arc at centre and area of an equilateral triangle
\(= \frac{\sqrt{3}}{4}\) \((side )^2\)
Area of shaded region \(= \frac{60}{360} \times \pi \times 2^2+\frac{60}{360} \times \pi \times 2^2-\frac{\sqrt{3}}{4} \times 2^2=\frac{4 \pi}{3} - \sqrt{3}\)
C is the answer
As A and B are centers of the circle, AB becomes the radius of both the circles. AC is the radius of circle with center A, and BC is the radius of circle with center B. So, we can say AB, BC and AC are equal in length.
Hence ⊿ ABC is an equilateral triangle.
Similarly, ⊿ ABD is also an equilateral triangle.
Now, Area of shaded region = Area of sector BAD + Area of sector ABD - Area of Triangle ABD. We know, Area of sector \(= \frac{\pi r^2 \times 0 }{360}\) where, r = radius of the circle, ϴ = angle subtended by the arc at centre and area of an equilateral triangle
\(= \frac{\sqrt{3}}{4}\) \((side )^2\)
Area of shaded region \(= \frac{60}{360} \times \pi \times 2^2+\frac{60}{360} \times \pi \times 2^2-\frac{\sqrt{3}}{4} \times 2^2=\frac{4 \pi}{3} - \sqrt{3}\)
C is the answer
