Geometry !

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neelgandham: Community Manager; Posts: 1060; Joined: Fri May 13, 2011 6:46 amFollowed by: 52 members; Location: Utrecht, The Netherlands; Thanked: 318 times

by neelgandham » Fri Nov 25, 2011 7:50 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?

(A) 3/4 (B) 5/6 (C) 1 (D) 7/5 (E) 9/7

p.s:I don't know the OA, but solved it for the correct answer !

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Anil Gandham
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Source: Beat The GMAT — Problem Solving | Fri Nov 25, 2011 7:50 am

user123321: Master | Next Rank: 500 Posts; Posts: 385; Joined: Fri Sep 23, 2011 9:02 pmFollowed by: 6 members; Thanked: 62 times

by user123321 » Fri Nov 25, 2011 10:49 am

is it D?

adding some hints as to how I arrived at this. We can connect Y & C and YCAX becomes a cyclic quadrilateral. And then you can get angle YCA(since sum of opp angles in a cyclic quad is 180).
since YB = BC, We have BYC & YCB angles same. using these two conditions you can get angle YBC & hence angle YBA.
now using sector formula with additional condition where a part of semi circle with radius r/2 adding and subtracting to individual sector areas, we can find the ratios.

user123321

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shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Fri Nov 25, 2011 1:26 pm

Complete the quadrilateral ABYX. We have (BY=BX and XB=AB - all radius)

Therefore the base angles are equal, i.e BXY=BYX and BAX=BXA.

2*(105) + ABY = 360

Angle ABY = 150. (Therefore YBC = 30)

Assume radius of circle = 2 (Hence radius of smaller semicircles will be =1)

So area of sector AYB(150) = (150/360)* 3(2^2) = 5 (pi~3)

Subtracting area of semicircle with radius = 1 (3*1/2 = 1.5)

Effective area = 5-1.5 = 3.5

Similarly area of sector BYC will be 1/5th of AYB = 5/5 = 1 (Since 150 = 5*(30) )

Area of the sector will be 1 unit and adding area of semicircle(1.5 as calculated before) in this case we have 1+1.5 = 2.5

Required ratio is 3.5/2.5 = 7/5 D IMO

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pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

by pemdas » Fri Nov 25, 2011 1:31 pm

initially i put an inscribed angle instead of central angle (correction in diff. color)
i think it's 7/5 and d

interesting question, intercepted arc AY=2*angle(YXA)=2*105=210`. Hence arc YA=360`-210`=150` and angle(YCB) as inscribed angle of arc YA must be 150/2=75`. Triangle YBC is isosceles and angle(YBC)=180-2*75=30. Angle(YBC) is a central angle (not an inscribed angle) of arc YC, therefore arc YC=30`

From drawing semicircles AB and BC are equal as share the same radius. Semicircles are deducted and added in the areas above and below the line YB respectively. Hence the squares sought will be Pi*(r^2)*(150/360) - [Pi(r/2)^2]/2 and (Pi*r^2)*(30/360) + [Pi(r/2)^2]/2.
To rewrite this will be Pi*r^2(5/12 - 1/8) and Pi*r^2(1/12 + 1/8)
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The required ratio will be Pi*r^2(5/12 - 1/8) divided by Pi*r^2(1/12 + 1/8)
(10-3)/24 : (2+3)/24 = 7/24 * 24/5 = 7/5

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neelgandham: Community Manager; Posts: 1060; Joined: Fri May 13, 2011 6:46 amFollowed by: 52 members; Location: Utrecht, The Netherlands; Thanked: 318 times

by neelgandham » Fri Nov 25, 2011 3:29 pm

I got a D !

Anil Gandham
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