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geom circle

by maihuna » Sun Jan 16, 2011 3:58 am
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
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by jaxis » Sun Jan 16, 2011 4:39 am
[spoiler]Ans:D - 7/5[/spoiler]


Image

Angle ABY = 360 - Angle (YXA+ XAB+XYB) --- (1)

Since both tringle XAB and triangle YXB are isoceles.

We have:
Angle BXA=Angle BAX
Angle BXY =Angle BYX

Now, BXY + BXA = Angle YXA = 105 (given) = BYX+BAX

Substituting this in (1)..we get Angle ABY = 360-(105+105) = 150

------------------------------------------------------------------------------------------
Image
Let the radius of the Big circle be .2R
Area of the BIG circle = � (2R)² = 4�R²
Area of BIG semicircle = 2�R²
Are of the Sector AYB = (150/180) * 2�R² = (5/3) �R²
Area of sector YCB = (30/180)* 2�R² = (1/3) �R²
------------------------------------------------------------------------------------------
Raius of the semicircles inside the BIG circle= R
Area of the semicircle = (�R²)/2
-------------------------------------------------------------------------------------------
Area of shaded portion above YB = (5/3) �R² - (�R²)/2
Are of shaded region below YB =(1/3) �R² + (�R²)/2


Ratio = ((5/3)-(1/2))/((1/3)+(1/2)) =[spoiler] 7/5.[/spoiler]

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Jaxis.
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by Rahul@gurome » Sun Jan 16, 2011 5:21 am
Image

Refer to the image above.

So in quadrilateral XYAB we have,
=> (x + y + x + y + angle B) = 360°
=> angle B = 360° - 2(x+y) = 150°

Suppose radius of the circle = r

angle YBA = 150°
angle YBC = 180° - 150° = 30°

Area of the sector YBC = πr²*30/360 = πr²/12 ----------------------(1)

Area of the sector YBA = πr²*150/360 = 5πr²/12 --------------------(2)

Area of the small semicircle (diameter x) = π(r/2)²/2 = πr²/8 -------(3)

Area of the shaded region above red line = (2)-(3) = 7πr²/24 -----(4)

Area of the shaded regions below red line = (1)+(3) = 5πr²/24 -----(5)

Required ratio (4)/(5) = 7/5

The correct answer is D.
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by maihuna » Sun Jan 16, 2011 8:42 am
i think angle a B can be very easily calculated, compared to cumbersome tchniques above, using circle formulae of angle is twice at center than at any other point, since B is minor angle it will be : 360-(105*2)
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