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Perimeter

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by parallel_chase » Sun Aug 10, 2008 9:52 am
We know that CD is parallel to the diameter.

Adjacent angles will be same. Therefore x=30

X+X = 60

Therefore the central angle would be = 120

arc CAE = 120/360 * 2pi5 = 10/3 pi

we have to find the length of EB & CB.

Triangles CAB & ABE are right triangle, and one of the angle is 30.

The diameter is the hypotenuese therefore

EB = 5 sqrt 3
CB = 5 sqrt 3

The total perimeter = 10/3 pi + 10sqrt3

The answer is D.
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by raunekk » Sun Aug 10, 2008 10:18 am
shaded region's part of the circle can be found by

formula of the length of arc= @/360 * 2*pie*r

i.e x=30 (CD II AB)

thus @=x+x= 60

Lenght of the shaded arc= 60/360 *2 * pie * 5 = 10/6 *pie..(1)

Angle CBA = 30 , Angle ACB = 90, angle BAC=60

thus a right angle triangle will have sides in ratio : 1:sqroot 3 : 2

Thus,CB=5 sqroot3 , AB=10 and CA = 5

CB=EB=5sqroot3

Perimeter: 5sqroot3+5sqroot3+10/6 *pie (frm 1)

= 5/3 * pie + 10 sqroot3

IMO:B
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by sibbineni » Sun Aug 10, 2008 10:35 am
i agree with chase
IMO D
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by sibbineni » Sun Aug 10, 2008 10:39 am
shaded region's part of the circle can be found by

formula of the length of arc= @/360 * 2*pie*r
(@ is central angle)
i.e x=30 (CD II AB) (Inscribed angle)



thus @=x+x= 60 (Central angle would be 2 times the inscribed angle)



Lenght of the shaded arc= 60/360 *2 * pie * 5 = 10/6 *pie..(1)

Angle CBA = 30 , Angle ACB = 90, angle BAC=60

thus a right angle triangle will have sides in ratio : 1:sqroot 3 : 2

Thus,CB=5 sqroot3 , AB=10 and CA = 5

CB=EB=5sqroot3

Perimeter: 5sqroot3+5sqroot3+10/6 *pie (frm 1)

= 5/3 * pie + 10 sqroot3

IMO:B
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by sudhir3127 » Sun Aug 10, 2008 10:44 am
OA D.

chase got it right !!!

thanks all..
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by raunekk » Sun Aug 10, 2008 11:14 am
opps ... took the wrong angle..!!

IT would be 120..ie @=120

thanks all!!!
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