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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]

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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

[Reveal] Spoiler:
Attachment:
Circle.png
Circle.png [ 4.27 KiB | Viewed 22328 times ]
[Reveal] Spoiler: OA

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Originally posted by enigma123 on 08 Feb 2012, 18:27.
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

Image

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 \(\frac{7}{5}\).

Answer: D.

[Reveal] Spoiler:
Attachment:
untitled.PNG
untitled.PNG [ 4.59 KiB | Viewed 23437 times ]

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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]

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New post 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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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]

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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]

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

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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]

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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
.png [ 5.5 KiB | Viewed 20317 times ]


The same as here:
Image

Hope it's clear.
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]

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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 :)
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]

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New post 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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New post 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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New post 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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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]

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New post 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*(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 \(\frac{7}{5}\).

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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New post 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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New post 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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In the diagram, points A, B, and C are on the diameter of [#permalink]

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New post 13 Jul 2016, 20:48
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE


Attachment:
Circle.png
Circle.png [ 7.13 KiB | Viewed 2725 times ]

Here is the another approach for finding the angle of arc YBC
angle AXC=90(angle in semicircle is 90)
so angle YXC =angle YXA-angle AXC= 105-90=15
as angle YBC=2*angle YXC =2*15=30deg.
as we find the area of sector YBC ..rest can be find with the solution suggested above by bunuel..

thanks
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In the diagram, points A, B, and C are on the diameter of [#permalink]

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New post 25 Jan 2017, 00:38
I used a visual approach to solve this question because I couldn't think of the central angle theorem.

If you joined the points A and C, you will see two smaller semicircles: one on the right side which is shaded, and one on the left which is not shaded. These will have the same area. For convenience's sake, redraw the the circle without the yin-yang portion and only consider the semicircle on the left side of diameter AC.

Now you need the ratio of the areas above and below the line BY. Clearly, the upper portion is larger than the lower portion, hence the ratio has to be greater than 1.
That allows us to eliminate options A,B and C.

Now between D and E, I chose D because sum of the ratios of D and E are 7+5 = 12 and 9+7=16 respectively and only 12 is a multiple of 360. My reasoning: If the ratio was 9:7 the calculations for the respective areas would be more complex if one opted for the conventional method of solving this questions
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]

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New post 21 Dec 2017, 13:14
Bunuel wrote:

Check the diagram below:
Attachment:
.png


The same as here:
Image

Hope it's clear.


Dear Bunuel,

I will grateful to you if you can explain how the central angle theorem help you.

I have reviewed the lesson you stated. I understood the 2 clear cases of the central angle but the third angle is not sharing same arc like the other 2. IN the question the angle is formed by two intersecting lines Ax & Yx. How come they share same arc with ABY?

Thanks
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]

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New post 21 Dec 2017, 20:08
Mo2men wrote:
Bunuel wrote:

Check the diagram below:
Attachment:
.png


The same as here:
Image

Hope it's clear.


Dear Bunuel,

I will grateful to you if you can explain how the central angle theorem help you.

I have reviewed the lesson you stated. I understood the 2 clear cases of the central angle but the third angle is not sharing same arc like the other 2. IN the question the angle is formed by two intersecting lines Ax & Yx. How come they share same arc with ABY?

Thanks


The following link should help: https://www.mathopenref.com/arccentralangletheorem.html
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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