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In the figure above, point B has an xcoordinate of 0.5 and a ycoordi
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27 Nov 2019, 01:56
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In the figure above, point B has an xcoordinate of 0.5 and a ycoordinate that is higher than any other point on the semicircle. If AB lies on the line y=2x+1, what is the area of the shaded region, in square units? A. \(\frac{2\pi}{3}\) B. \(\pi\frac{\sqrt{3}}{8}\) C. \(\pi1\) D. \(2\pi\frac{3}{2}\) E. \(\frac{3\pi}{8}\) Are You Up For the Challenge: 700 Level QuestionsAttachment:
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Re: In the figure above, point B has an xcoordinate of 0.5 and a ycoordi
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27 Nov 2019, 04:41
Bunuel wrote: In the figure above, point B has an xcoordinate of 0.5 and a ycoordinate that is higher than any other point on the semicircle. If AB lies on the line y=2x+1, what is the area of the shaded region, in square units? A. \(\frac{2\pi}{3}\) B. \(\pi\frac{\sqrt{3}}{8}\) C. \(\pi1\) D. \(2\pi\frac{3}{2}\) E. \(\frac{3\pi}{8}\) Let a perpendicular dropped from Point B meets xaxis at point C > ACB is a right angle triangle with ∠C = 90 deg > Area of Shaded region = 1/4th the area of circle  Area of the triangle ABCLet the point B = (0.5, b) & A = (a, 0) > y = 2x + 1 > b = 2*0.5 + 1 = 1 + 1 = 2 > Point B = (0.5, 2) For point A, 0 = 2*a + 1 > 2a = 1 > a = 0.5 > Point A = (0.5, 0) So, Point C = (0.5, 0) > Radius of the circle = BC = 2  0 = 2 units AC = 0.5  (0.5) = 1 Area of Shaded region = 1/4*π*\(2^2\)  1/2*AC*BC = π  1/2*1*2 = π  1 IMO Option C



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Re: In the figure above, point B has an xcoordinate of 0.5 and a ycoordi
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08 Dec 2019, 22:25
Quote: Let a perpendicular dropped from Point B meets xaxis at point C > ACB is a right angle triangle with ∠C = 90 deg
> Area of Shaded region = 1/4th the area of circle  Area of the triangle ABC
Shouldn't it be Area of shaded region = Area of semi circle  [Area of 1/4th circle +Area of triangle]?? I did not understand how you got 1/4th area of circle  Area of traingle??



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Re: In the figure above, point B has an xcoordinate of 0.5 and a ycoordi
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08 Dec 2019, 22:26
Dillesh4096Never mind understood my mistake Thanks



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Re: In the figure above, point B has an xcoordinate of 0.5 and a ycoordi
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08 Dec 2019, 23:31
devavrat wrote: Quote: Let a perpendicular dropped from Point B meets xaxis at point C > ACB is a right angle triangle with ∠C = 90 deg
> Area of Shaded region = 1/4th the area of circle  Area of the triangle ABC
Shouldn't it be Area of shaded region = Area of semi circle  [Area of 1/4th circle +Area of triangle]?? I did not understand how you got 1/4th area of circle  Area of traingle?? Hi devavrat, See the highlighted part, Area of shaded region = Area of semi circle  [Area of 1/4th circle +Area of triangle] We know, Area of semicircle = 1/2 of area of circle!! So, Area of shaded region = Area of semi circle  [Area of 1/4th circle +Area of triangle] = 1/2 of area of circle  [Area of 1/4th circle +Area of triangle] = (1/2  1/4) of area of circle  Area of triangle = 1/4 of area of circle  Area of triangle !!! Which is what I have mentioned in the formula. Hope it’s clear! Posted from my mobile device



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Re: In the figure above, point B has an xcoordinate of 0.5 and a ycoordi
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09 Dec 2019, 04:43
Bunuel wrote: In the figure above, point B has an xcoordinate of 0.5 and a ycoordinate that is higher than any other point on the semicircle. If AB lies on the line y=2x+1, what is the area of the shaded region, in square units? A. \(\frac{2\pi}{3}\) B. \(\pi\frac{\sqrt{3}}{8}\) C. \(\pi1\) D. \(2\pi\frac{3}{2}\) E. \(\frac{3\pi}{8}\) Are You Up For the Challenge: 700 Level QuestionsAttachment: 102Geometry.jpg "point B has an xcoordinate of 0.5" Also, it lies on y = 2x + 1. So y = 2. Coordinates of B are (0.5, 2) and it is the radius of the circle from the centre, say C (0.5, 0) lying on the x axis. Coordinates of point A: 0 = 2x + 1 (because at A, y coordinate is 0). So A (0.5, 0) So area of shaded region is: Area of quarter of a circle  Area of triangle ABC = \((1/4)*4*\pi  (1/2)*1*2 = \pi  1\) Answer (C)
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Re: In the figure above, point B has an xcoordinate of 0.5 and a ycoordi
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10 Dec 2019, 23:41
How did we get Area of 1/4th of circle ? Can somebody please explain this.



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In the figure above, point B has an xcoordinate of 0.5 and a ycoordi
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10 Dec 2019, 23:47
aarushisingla wrote: How did we get Area of 1/4th of circle ? Can somebody please explain this. Dear aarushisingla , As Given in data, we have semicircle. And Point B is at the highest level of semicircle. Suppose, a perpendicular dropped from Point B meets xaxis at point C By the Coordinates available, We can find BC which is Radius of the circle = 2 units Area of 1/4 of circle = 1/4 * π * r^2 = π For finding the area of the shaded region, We are subtracting area of Triangle with 1/4 of circle. I hope you will get it. Do PM if you didn't understand Regards, Rajat Chopra




In the figure above, point B has an xcoordinate of 0.5 and a ycoordi
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10 Dec 2019, 23:47






