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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, 17:27

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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.)

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

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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"

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]

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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 !!
_________________

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

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

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.

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

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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]

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16 Jun 2015, 05:34

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.

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..

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

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24 Jan 2017, 23:38

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

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?

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?