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In the drawing attached, the shaded region is bound by the

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In the drawing attached, the shaded region is bound by the [#permalink] New post 16 Jul 2006, 15:07
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In the drawing attached, the shaded region is bound by the arc of a circle of radius 50 yards and a line that is parallel to the thick line tangent to the arc. If |BC| =25 yards, the area of the shaded region in square yards is between

(A) 850 and 1000 (B) 1000 and 1150 (C) 1150 and 1300 (D) 1300 and 1450 (E) 1450 and 1600
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Senior Manager
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 [#permalink] New post 16 Jul 2006, 16:22
toughie..

Given BC = 25, we can calculate AB, comes out to arnd ~43.5 (drawing a line from center of the circle to A and midpt of AB and then using pythagoras).. Given AB, we can calculate shaded area = ~ 1/2 (43.5 * 25) ~ 1100

(B)

could find a better way..
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 [#permalink] New post 16 Jul 2006, 17:23
Since we have the radius of the circle we can find out the length of the cord(AB).

Key: BC is the half of the radius.

AB = 2*sqrt(50^2-25^2)
or
AB = 2*cos30*50.

So, AB = 50sqrt(3)

let O as the center of the circle
then the area of triangle AOB = 50sqrt(3)*25*(1/2) = 1082.5

(1/3)*50^2*pi - 1082.5 = 15xx.xxxx

So, ans should be EEEEEEEEEEE
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Last edited by freetheking on 16 Jul 2006, 18:03, edited 1 time in total.
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 [#permalink] New post 16 Jul 2006, 17:58
sgrover wrote:
freetheking wrote:
Since we have the radius of the circle we can find out the length of the cord(AB).

Key: BC is the half of the radius.

AB = 2*sqrt(50^2+25^2)

Shdnt this be: 2*(50^2 - 25^2)


shoot,,, you are right.. it's typo.. thanks
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 [#permalink] New post 21 Jul 2006, 15:41
We should be able to solve this without using sin/cos/tan
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 [#permalink] New post 21 Jul 2006, 15:47
You're quite right! As long as you know your 45-45-90 and 30-60-90 triangles
  [#permalink] 21 Jul 2006, 15:47
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In the drawing attached, the shaded region is bound by the

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