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Question 18: In the figure above, A and b are the centers of
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19 May 2016, 16:21
19 May 2016, 16:21
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45% (02:12) correct
54% (03:58) wrong based on 11 sessions
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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. \(\frac{(6\pi - 1) x^2}{6}\)
Re: Question 18: In the figure above, A and b are the centers of
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23 May 2016, 14:51
23 May 2016, 14:51
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Expert Reply
Solution
Here we have two circles with centers A and B they intersect on points say X and Y. Now to find the area of the shaded region.
Let us connect the points X, A, B to form a triangle and Y, A, B to form another triangle. Since the radius of circle A and B are same. We can say that XAB is an equilateral triangle. With XA=AB= XB= radius of circle. Similarly YAB is also a triangle.
So now we can clearly see that the area of the two triangles = \(2*\sqrt{3}*\frac{x^2}{4}\). ........(i)
Now the segment, i.e. the area between the triangle and a circle.
Area of segment = \(\frac{x^2}{2}(\frac{pi*theta}{180}- sin(theta))\).
Where theta is the central angle.
Or in our case angle XAB = 60 degrees= theta.
Putting theta = 60 we have
Area of segment= \(\frac{pi*x^2}{6} - \sqrt{3}\frac{x^2}{2}\).
Now we have 4 such segments. Now adding this to (i) and simplifying we get
Area of the shaded region as = \(\frac{4*pi}{6}*x^2 - 3*\sqrt{3}\frac{1}{2}*x^2\)
Here we have two circles with centers A and B they intersect on points say X and Y. Now to find the area of the shaded region.
Let us connect the points X, A, B to form a triangle and Y, A, B to form another triangle. Since the radius of circle A and B are same. We can say that XAB is an equilateral triangle. With XA=AB= XB= radius of circle. Similarly YAB is also a triangle.
So now we can clearly see that the area of the two triangles = \(2*\sqrt{3}*\frac{x^2}{4}\). ........(i)
Now the segment, i.e. the area between the triangle and a circle.
Area of segment = \(\frac{x^2}{2}(\frac{pi*theta}{180}- sin(theta))\).
Where theta is the central angle.
Or in our case angle XAB = 60 degrees= theta.
Putting theta = 60 we have
Area of segment= \(\frac{pi*x^2}{6} - \sqrt{3}\frac{x^2}{2}\).
Now we have 4 such segments. Now adding this to (i) and simplifying we get
Area of the shaded region as = \(\frac{4*pi}{6}*x^2 - 3*\sqrt{3}\frac{1}{2}*x^2\)
Re: Question 18: In the figure above, A and b are the centers of
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23 May 2016, 20:24
23 May 2016, 20:24
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Let us assume other two points of intersection of both circles are C and D. The distance between A and B is x. The line CD intersects AB at mid point and let us assume the point of intersection is E.
Now triangle AEC is 30, 60, 90 triangle. (sides x, x/2 and x*Square root of 3/2). Hence the arc CADB makes 120 degrees, so the area of arc CADB is 120/360*pi*x^2 = pi*x^2 / 3
Area of the triangle CAD is 1/2 * x * x*(Square root 3)/2.
The answer is two times (pi*x^2 / 3 - 1/2 * x * x*(Square root 3)/2). = (4 pi - 3*(Square root 3))*x^2/6. Hence C.
Now triangle AEC is 30, 60, 90 triangle. (sides x, x/2 and x*Square root of 3/2). Hence the arc CADB makes 120 degrees, so the area of arc CADB is 120/360*pi*x^2 = pi*x^2 / 3
Area of the triangle CAD is 1/2 * x * x*(Square root 3)/2.
The answer is two times (pi*x^2 / 3 - 1/2 * x * x*(Square root 3)/2). = (4 pi - 3*(Square root 3))*x^2/6. Hence C.
