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In the diagram, points A, B, and C are on the diameter of [#permalink]
08 Feb 2012, 17:27

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Difficulty:

85% (hard)

Question Stats:

38% (03:23) correct
61% (02:42) wrong based on 202 sessions

Attachment:

Circle.png [ 4.27 KiB | Viewed 3275 times ]

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.)

Re: Area of shaded region [#permalink]
08 Feb 2012, 17:59

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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:

untitled.PNG [ 4.59 KiB | Viewed 4703 times ]

According to the central angle theorem <ABY=2*(180-105)=150 (for more on this check Circles chapter of Math Book: math-circles-87957.html). Hence <CBY=180-150=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}.

Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
11 Feb 2012, 12:15

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enigma123 wrote:

Hi Bunuel - can you please help? How did you get the area of two small semi circles??

The radius of the small semicircles is r/2, where r is the radius of the large circle. Thus the area of each is half of the area of the circle with the radius of r/2: \frac{\pi{(\frac{r}{2})^2}}{2}=\pi{\frac{r^2}{8}}.

Re: Area of shaded region [#permalink]
30 Jun 2013, 04:09

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*(180-105)=150 (for more on this check Circles chapter of Math Book: math-circles-87957.html). Hence <CBY=180-150=30"

Re: Area of shaded region [#permalink]
30 Jun 2013, 07:58

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Genfi wrote:

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*(180-105)=150 (for more on this check Circles chapter of Math Book: math-circles-87957.html). Hence <CBY=180-150=30"

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

Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
25 Jun 2014, 23:20

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 !!

gmatclubot

Re: In the diagram, points A, B, and C are on the diameter of
[#permalink]
25 Jun 2014, 23:20