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E
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Two circles with equal radius pass through centre of each other as shown in figure. If radius = r, the area of the overlapping region would be:
A. \(2\pi r^2 - \frac{\sqrt{3}}{2}\)
B. \(2\pi r^2 - \frac{\pi}{8}\)
C. \(r^2(\frac{4\pi}{3} - \frac{\sqrt{3}}{4})\)
D. \(r^2(\frac{\pi}{3} - \sqrt{2})\)
E. \(r^2(\frac{2\pi}{3} - \frac{\sqrt{3}}{2})\)
Attachment:
1 (1).jpg [ 9.44 KiB | Viewed 4533 times ]
Are You Up For the Challenge: 700 Level Questions
10 Sep 2022, 22:30
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We can find the area of each sector and subtract the area that's counted twice to obtain the area of the overlap region.
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Screenshot 2022-09-11 105902.png [ 249.42 KiB | Viewed 3958 times ]
10 Sep 2022, 23:00
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Let the center of the first circle be C and the center of the second circle be C′.
Let the points of intersection of the two circles be A and B.
we see that the lengths AC = AC′ = BC = BC′ = CC′ = r (radius of the circles).
draw line segment AC, AC', BC' and BC
Clearly ACBC' forms a rhombus with angle CAC' = 60degree (as triangle ACC' and BCC' are equilateral triangles) and angle ACB = 120 degree = \(\frac{2π}{3}\)
from center C, we can determine the circular segment (shaded region in the diagram attached below) subtended with angle \(\frac{2π}{3}\), and radius r,
the area of circular segment =\(\frac{ r^2}{2} * (\frac{2π}{3} - sin(\frac{2π}{3})) = \frac{r^2}{2} * (\frac{2π}{3}−\frac{√3}{2})\)
the area of desired region is twice the area of circular segment
therefore area of overlapping region = \(r^2 * (\frac{2π}{3}−\frac{√3}{2})\)
Answer: E
Let the points of intersection of the two circles be A and B.
we see that the lengths AC = AC′ = BC = BC′ = CC′ = r (radius of the circles).
draw line segment AC, AC', BC' and BC
Clearly ACBC' forms a rhombus with angle CAC' = 60degree (as triangle ACC' and BCC' are equilateral triangles) and angle ACB = 120 degree = \(\frac{2π}{3}\)
from center C, we can determine the circular segment (shaded region in the diagram attached below) subtended with angle \(\frac{2π}{3}\), and radius r,
the area of circular segment =\(\frac{ r^2}{2} * (\frac{2π}{3} - sin(\frac{2π}{3})) = \frac{r^2}{2} * (\frac{2π}{3}−\frac{√3}{2})\)
the area of desired region is twice the area of circular segment
therefore area of overlapping region = \(r^2 * (\frac{2π}{3}−\frac{√3}{2})\)
Answer: E
Attachments
Picture3.png [ 115.75 KiB | Viewed 3813 times ]
