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In the diagram, points A, B, and C are on the diameter of [#permalink]
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08 Feb 2012, 18:27
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In the diagram, points A, B, and C are on the diameter of the circle with center B. Additionally, all arcs pictured are semicircles. Suppose angle YXA = 105 degrees. What is the ratio of the area of the shaded region above the line YB to the area of the shaded region below the line YB? (Note: Diagram is not drawn to scale and angles drawn are not accurate.) (A) ¾ (B) 5/6 (C) 1 (D) 7/5 (E) 9/7 Attachment:
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Last edited by Bunuel on 16 Jun 2015, 08:27, edited 3 times in total.
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In the diagram, points A, B, and C are on the diameter of [#permalink]
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In the diagram, points A, B, and C are on the diameter of the circle with center B. Additionally, all arcs pictured are semicircles. Suppose angle YXA = 105 degrees. What is the ratio of the area of the shaded region above the line YB to the area of the shaded region below the line YB? (Note: Diagram is not drawn to scale and angles drawn are not accurate.)(A) ¾ (B) 5/6 (C) 1 (D) 7/5 (E) 9/7 According to the central angle theorem <ABY=2*(180105)=150 (for more on this check Circles chapter of Math Book: mathcircles87957.html). Hence <CBY=180150=30. The area of sector \(ABY=\frac{150}{360}*\pi{r^2}=\frac{5}{12}\pi{r^2}\); The area of sector \(CBY=\frac{30}{360}*\pi{r^2}=\frac{1}{12}\pi{r^2}\); The area of each of two small semicircles is \(\frac{\pi{(\frac{r}{2})^2}}{2}=\pi{\frac{r^2}{8}}\) (as its radius is half of the radius of the big circle); The are of the shaded region above BY is \(\frac{5}{12}\pi{r^2}\pi{\frac{r^2}{8}=\frac{7}{24}\pi{r^2}\); The are of the shaded region below BY is \(\frac{1}{12}\pi{r^2}+\pi{\frac{r^2}{8}=\frac{5}{24}\pi{r^2}\); Ratio of the areas of the shaded regions is \(\frac{7}{5}\). Answer: D. Attachment:
untitled.PNG [ 4.59 KiB  Viewed 18481 times ]
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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11 Feb 2012, 12:18
Hi Bunuel  can you please help? How did you get the area of two small semi circles??
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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23 Feb 2012, 07:51
thanks bunuel for nice explanation +1 kudos



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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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22 Jun 2013, 08:27
Bunuel wrote: Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HEREHi Bunnel, Can we have more geometry questions please. Thanks.



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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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22 Jun 2013, 08:39



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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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30 Jun 2013, 05:09
Hi Bunel,
I didn't understand the below mentioned part. I did refer to the link provided by you. Can you please can explain this central angle theorem.
"According to the central angle theorem <ABY=2*(180105)=150 (for more on this check Circles chapter of Math Book: mathcircles87957.html). Hence <CBY=180150=30"
Thank you



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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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30 Jun 2013, 08:58
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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Bunuel wrote: Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HEREIt can also be solved the following way: Three new triangles can be formed which will be isosceles triangles (please refer the attachment) Hence, x+y = 105 (given) 2(x+y+z) = 360 since ACYX is a quadrilateral x+y+z = 180 > z = 75 in Triangle, BYC, <YBC will be 180  2*z = 180  2*75 = 30 degrees. (1) Area of the total shaded portion is half the area of the circle pi*(r^2)/2 (2) Area below the red line = area of segment BYC of circle + area of shaded semicircle BC= (30/360)*(pi*r^2) + pi ((r/2)^2)/2 =pi*(r^2)/12+pi*(r^2)/8 =5 * pi * (r^2) / 24 (3) Area above the red line is (1)(2) above= [ pi * (r^2)/2 ]  [ 5 * pi* (r^2) / 24 ] = 7 * pi * (r^2) / 24 Answer is (3) / (2) which is 7/5Kudos if you like the post
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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26 Jun 2014, 00:20
I Thought that's a nice question for a strategic guess (50/50) since the shaded area above the line seemed bigger therefore the answer should be either 3/4 or 5/6 but they tricked me !!
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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16 Jul 2014, 04:34
excellent question!! I dont think there's any better resource and better guide than Bunnuel .. when it comes to Maths!!



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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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16 Jul 2014, 22:31
Bunuel  I am not clear on what is meant by  Quote: all arcs pictured are semicircles ?



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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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17 Jul 2014, 07:57



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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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03 Aug 2014, 00:04
Bunuel wrote: In the diagram, points A, B, and C are on the diameter of the circle with center B. Additionally, all arcs pictured are semicircles. Suppose angle YXA = 105 degrees. What is the ratio of the area of the shaded region above the line YB to the area of the shaded region below the line YB? (Note: Diagram is not drawn to scale and angles drawn are not accurate.) (A) ¾ (B) 5/6 (C) 1 (D) 7/5 (E) 9/7 Attachment: untitled.PNG According to the central angle theorem <ABY=2*(180105)=150 (for more on this check Circles chapter of Math Book: mathcircles87957.html). Hence <CBY=180150=30. The area of sector \(ABY=\frac{150}{360}*\pi{r^2}=\frac{5}{12}\pi{r^2}\); The area of sector \(CBY=\frac{30}{360}*\pi{r^2}=\frac{1}{12}\pi{r^2}\); The area of each of two small semicircles is \(\frac{\pi{(\frac{r}{2})^2}}{2}=\pi{\frac{r^2}{8}}\) (as its radius is half of the radius of the big circle); The are of the shaded region above BY is \(\frac{5}{12}\pi{r^2}\pi{\frac{r^2}{8}=\frac{7}{24}\pi{r^2}\); The are of the shaded region below BY is \(\frac{1}{12}\pi{r^2}+\pi{\frac{r^2}{8}=\frac{5}{24}\pi{r^2}\); Ratio of the areas of the shaded regions is \(\frac{7}{5}\). Answer: D. Hi Bunuel, One query. As we know central angle of a circle is twice the inscribed angle. i.e. if inscribed angle is x then central angle is 2x. So here if i see angle on YXA is 105 then my central angle on B should be 210. Please clarify this. Thanks.



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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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04 Aug 2014, 00:46
enigma123 wrote: Attachment: Circle.png In the diagram, points A, B, and C are on the diameter of the circle with center B. Additionally, all arcs pictured are semicircles. Suppose angle YXA = 105 degrees. What is the ratio of the area of the shaded region above the line YB to the area of the shaded region below the line YB? (Note: Diagram is not drawn to scale and angles drawn are not accurate.) (A) ¾ (B) 5/6 (C) 1 (D) 7/5 (E) 9/7 Thanks for sharing an interesting question. It took me about 10 minutes to solve the problem
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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22 Aug 2014, 02:32
sorry, this might be a basic question . I would like to understand this picture
angle YXA = 105 then according to the central angle theorem, why its not 2*105 .. you have consider exterior angle for X but how we shall determine it. Please advise.



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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
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16 Jun 2015, 06:34
enigma123 wrote: Attachment: Circle.png In the diagram, points A, B, and C are on the diameter of the circle with center B. Additionally, all arcs pictured are semicircles. Suppose angle YXA = 105 degrees. What is the ratio of the area of the shaded region above the line YB to the area of the shaded region below the line YB? (Note: Diagram is not drawn to scale and angles drawn are not accurate.) (A) ¾ (B) 5/6 (C) 1 (D) 7/5 (E) 9/7 answer is (D) arc YC subtends an angle 30 degree with the center. Area below line YB is area of circle x (1/12 + 1/8) Area of the shaded region is half of the area of the circle.



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