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

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MrJglass
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In the diagram, points A, B, and C are on the diameter of [#permalink] New post   17 Jul 2018, 17:32
Bunuel wrote:

Hope it's clear.

I solved this question in less than 10 seconds, I was surprised to find out that it is 95% hard. I probably did something wrong and was lucky to get the correct answer. Please look at what I did and correct me, where I'm wrong.

Since the question stem indicates that the diagram is not drawn to scale, using the provided info, I drew a proper diagram with decent semi-circles.

Instead of using angle YXA as 105°, I turned it upside down and used angle ABY as 105°. Since we are only looking for the ratio of the regions, irrespective of which position the degree is coming from.

Thus,I had the upper shaded region to be 105°, and the lower region to be therefore (360-(180+105)) = 75°.

I went ahead to compute 105/75 = 7/5.

It might seem pretty long but it was pretty fast. Although it seems like a ludicrous luck!
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In the diagram, points A, B, and C are on the diameter of [#permalink] New post   23 Aug 2020, 19:19
without the addition of a picture,


when an Inscribed Angle like Angle YXA touches the circumference of the circle on the SAME SIDE of the Circle that it is Inscribed upon, then I always like to think of the "Central Angle" as the Reflex Angle.

What I mean is if you draw a Radius from B to A, then you have Angle YBA "Touching" the same 2 Points (Point A and Point Y) as Inscribed Angle YXA does.

The Central Angle that measures the Major Arc from A all the way around the circle to C and then Y will be measured by the Reflex Angle on the Opposite Side of Angle YBA. That Reflex Angle will Equal 2 * (Inscribed Angle of 105 degrees).

If that Reflex Angle = 210 degrees, then the Angle YBA will = 150 degrees because together they are 360 degrees around Point B-Center.

I don't know if this helped anyone. I'll try to draw a picture and get it attached if someone is interested.
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Vivek1707
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink] New post   02 Jan 2021, 04:51
arichinna wrote:
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 :)



Hello, could you please explain further how the area above the line is 1-2
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink] New post   16 Aug 2021, 13:32
enigma123 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

Show Spoiler
Attachment:
Circle.png


Simple way using property of cyclic quadrilateral.
Make cyclic quadrilateral AXYC.Join AB-BC-YC
In AXYC:
AXY = 105
So ACY = 180-105 = 75 (Sum of opposite angles of cyclic quadrilateral is 180)

Make triangle YBC.
In YBC,YB = CB = radius
Hence BYC = BCY
But ACY = BCY (A-B-C)
So BYC = BCY = 75

YBC = 180-(BYC+BCY) = 180-150 = 30
So ABY = 150 (A-B-C is straight line)

Area of circular region AYB = 150/360 * Pi * r^2
Area of circular region BYC = 30/360 * Pi * r^2

Let radius be r,then area of semicircular region with radius as r/2 is (Pi*r^2)/8

Required ratio:

\(\frac{5}{12} *Pi*r^2 - Pi*r^2*\frac{ 1}{8}\)

Upon

\(\frac{1}{12} *Pi*r^2 + Pi*r^2*\frac{ 1}{8}\)

Which gives

\(\frac{7}{5}\)
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink] New post   13 Oct 2022, 07:25
Bunuel wrote:
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


The same as here:


Hope it's clear.



Hi Bunuel,
Why is the angle below the central angle alpha and not its (alpha's) adjacent one? Shouldn't the adjacent angle be alpha as it subtends the arc ?
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Re: In the diagram, points A, B, and C are on the diameter of [#permalink] New post   20 Oct 2023, 05:20
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In the diagram, points A, B, and C are on the diameter of [#permalink]
20 Oct 2023, 05:20
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