In the diagram, points A, b and C are on the diameter...

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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 dregrees. What is the ratio of the area og 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

The OA is D.

Is there a strategic approach to this PS question? I don't have it clear. I appreciate if any expert explain it for me. Thanks!

[GMATGuruNY]

RULE:
In a quadrilateral with all four vertices on a circle, opposite angles SUM TO 180.

Let the radius of the circle = 12, implying that the area of the circle = πr² = π(12²) = 144π.


Since quadrilateral XAYC has all four vertices on the circle, ∠YXA and ∠YCA must sum to 180.
Since ∠YXA = 105, ∠YCA = 180-105 = 75.

In ∆YBC, sides BY and BC are both radii and thus are equal.
Since angles opposite equal sides of a triangle must also be equal, ∠BYC = ∠YCA = 75, with the result that ∠YBC = 180-75-75 = 30.

Yellow sector:
(central angle YBC)/360 = 30/360 = 1/12.
Since (sector area)/(circle area) = (central angle)/360, the yellow sector is 1/12 of the circle:
(1/12)(144Ï€) = 12Ï€.

Red semicircle:
Since r=6 -- implying a circle with an area of 36Ï€-- the area of the red semicircle = (1/2)(36Ï€) = 18Ï€.

Shaded area below YB:
(yellow sector) + (red semicircle) = 12Ï€ + 18Ï€ = 30Ï€.

Green sector:
(semicircle CYXA) - (yellow sector) = (1/2)(144π) - 12π = 72π - 12π = 60π.

Shaded area above YB:
Since semicircle ADB and the red semicircle have the same radius, they have equal areas (18Ï€).
Thus:
Shaded area above YB = (green sector) - (semicircle ADB) = 60π - 18π = 42π.

Resulting ratio:
(shaded area above YB)/(shaded area below YB) = (42Ï€)/(30Ï€) = 7/5.

The correct answer is D..