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19 Mar 2021, 02:00
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
47% (03:16) correct
53%
(02:43)
wrong
based on 17
sessions
History
Date
Time
Result
Not Attempted Yet
AB is the line joining the points of intersection of a semicircle and a circle as shown in the figure. If AB = m, what is the the area of the circle with radius AB?
A. \(80\pi\)
B. \(90\pi\)
C. \(100\pi\)
D. \(130\pi\)
E. \(160\pi\)
Attachment:
2021-03-08_20-38-14.png [ 44.14 KiB | Viewed 2198 times ]
19 Mar 2021, 22:51
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Bunuel
Logical:
AB has to be less than 10, as the diameter of the smaller circle is 10, and AB is a chord in the same circle.
Thus area < \(\pi 10^2\) or \(Area<100\pi\).
Only A and B are left.
Within 10 seconds we have got the probability of getting the correct answer to \(\frac{1}{2}\) from \(\frac{1}{5}\).
Geometrical approach:
Join the chord AB with the center of each circle.
Also join the centers of the two circles to get line CD. So, CD and AB are perpendicular bisectors.
As A is a point on circumference of smaller circle, the line CA is tangent on smaller circle. Thus, \(\angle CAD = 90\)
So \(CD = \sqrt{10^2+5^2}=5\sqrt{5}\)
As the area of \(\triangle ACD\) will remain same, \(\frac{1}{2}*CD*OA=\frac{1}{2}*AD*AC............5\sqrt{5}*OA=10*5........OA=2\sqrt{5}\)
Since AB=2*OA, AB=\(4\sqrt{5}\)
Area of the circle with radius \(4\sqrt{5}=\pi *(4\sqrt{5})^2=80\pi\)
A
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20 Mar 2021, 17:03
Kudos
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Bunuel
We may solve this with geometry, the equation of a circle is \((x - a)^2 + (y - b)^2 = r^2\), where (a, b) is the center and \(r\) is the radius. Set B as the origin. Thus the bigger circle has an equation of \(x^2 + (y - 10)^2 = 100\), the smaller circle has an equation of \((x - 5)^2 + y^2 = 25\).
Simplifying the equations we have:
\(x^2 + y^2 - 20y = 0\)
\(x^2 - 10x + y^2 = 0\)
Subtract these equations to get \(10x = 20y\) and \(x = 2y\). Plug back in any of the equations to find: \(5y^2 - 20y = 0\) and \(y = 4\), \(x = 8\).
Thus the radius of m would be \(\sqrt{4^2 + 8^2}\) and the area is \(80*\pi\).
Ans: A
