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# In the diagram, points A, B, and C are on the diameter of

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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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Question Stats:

42% (04:08) correct 58% (04:30) wrong based on 406 sessions
Attachment:

Circle.png [ 4.27 KiB | Viewed 7496 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.)

(A) ¾
(B) 5/6
(C) 1
(D) 7/5
(E) 9/7
[Reveal] Spoiler: OA

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MGMAT 3 ---> 610
GMAT ==> 730

Last edited by Bunuel on 14 Jul 2013, 10:01, edited 2 times in total.
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15
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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 8909 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 $$\frac{7}{5}$$.

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Status: Finally Done. Admitted in Kellogg for 2015 intake
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Kudos [?]: 999 [0], given: 217

Re: In the diagram, points A, B, and C are on the diameter of [#permalink]  11 Feb 2012, 11:18
Hi Bunuel - can you please help? How did you get the area of two small semi circles??
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Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Math Expert
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Posts: 27228
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Kudos [?]: 41084 [1] , given: 5665

Re: In the diagram, points A, B, and C are on the diameter of [#permalink]  11 Feb 2012, 12:15
1
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Expert's post
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: $$\frac{\pi{(\frac{r}{2})^2}}{2}=\pi{\frac{r^2}{8}}$$.

Hope it's clear.
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]  23 Feb 2012, 06:51
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]  20 Jun 2013, 05:39
Expert's post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]  22 Jun 2013, 07:27
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Hi 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]  22 Jun 2013, 07:39
Expert's post
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cumulonimbus wrote:
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Hi Bunnel,

Can we have more geometry questions please.
Thanks.

Links to more geometry questions are given in the post you are quoting.
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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"

Thank you
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1
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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"

Thank you

Check the diagram below:
Attachment:

.png [ 5.5 KiB | Viewed 7582 times ]

The same as here:

Hope it's clear.
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]  12 Jun 2014, 01:36
2
KUDOS
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

It 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 7/5

Kudos if you like the post
Attachments

Circle 1.png [ 6.08 KiB | Viewed 4817 times ]

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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 !!
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Please +1 KUDO if my post helps. Thank you.

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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]  16 Jul 2014, 03:34
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]  16 Jul 2014, 21:31
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]  17 Jul 2014, 06:57
Expert's post
himanshujovi wrote:
Bunuel - I am not clear on what is meant by -
Quote:
all arcs pictured are semicircles
?

Red and blue arcs below are semicircles:
Attachment:

Untitled.png [ 4.54 KiB | Viewed 4042 times ]

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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]  02 Aug 2014, 23:04
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*(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 $$\frac{7}{5}$$.

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.

Thanks.
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]  03 Aug 2014, 23:46
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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+1 Kudos please, if you like my post

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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]  22 Aug 2014, 01:32
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.
Re: In the diagram, points A, B, and C are on the diameter of   [#permalink] 22 Aug 2014, 01:32
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