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area of shaded region

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area of shaded region [#permalink] New post 10 Dec 2009, 20:08
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pls provide a short method to solve the attached problem
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Re: area of shaded region [#permalink] New post 10 Dec 2009, 20:21
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xcusemeplz2009 wrote:
pls provide a short method to solve the attached problem

answer is c

bc = 6 and bd = 6sqrt 3 so dc = 12

now draw a triangle adc..d will = 90 degree
so short leg = 12, ad = 12 sqrt 3 and ac = 24 but bc = 6 so ab = 18

get the area of all semicircles

big semicircle: radius = 12 so area = 144 but divide by 2 so 72
smaller area: radius = 9 so area = 81/2
smallest area: radius = 3 so area = 9/2
72 - 81/2-9/2 = 27
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Re: area of shaded region [#permalink] New post 10 Dec 2009, 20:58
lagomez wrote:
xcusemeplz2009 wrote:

now draw a triangle adc..d will = 90 degree
so short leg = 12, ad = 12 sqrt 3 and ac = 24


thanks lagomez for the solution ,
cud u pls explain the above(in red) funda , i am not able to recollect this rule.
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Re: area of shaded region [#permalink] New post 10 Dec 2009, 21:05
xcusemeplz2009 wrote:
lagomez wrote:
xcusemeplz2009 wrote:

now draw a triangle adc..d will = 90 degree
so short leg = 12, ad = 12 sqrt 3 and ac = 24


thanks lagomez for the solution ,
cud u pls explain the above(in red) funda , i am not able to recollect this rule.


a triangle inscribed in a semicircle that has one side as the diameter is a right triangle
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Re: area of shaded region [#permalink] New post 10 Dec 2009, 21:17
lagomez wrote:
xcusemeplz2009 wrote:
so short leg = 12, ad = 12 sqrt 3 and ac = 24 but bc = 6 so ab = 18



i am intrested to know how AD=12 sqrt 3 and AC=24
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Re: area of shaded region [#permalink] New post 10 Dec 2009, 21:33
xcusemeplz2009 wrote:
lagomez wrote:
xcusemeplz2009 wrote:
so short leg = 12, ad = 12 sqrt 3 and ac = 24 but bc = 6 so ab = 18



i am intrested to know how AD=12 sqrt 3 and AC=24


bdc = 30-60-90 triangle..

somebody correct me if i'm wrong here, but if I recall I think there is a rule that that if a 30-60-90 is divided into 90 degree angles then the triangle also has the same degrees as the larger angles

a 30-60-90 divided into two triangles that have 90 degrees then the other angles are 30-60..don't quote me on this part but I think that's the rule

so if the smaller triangle bdc = 30-60-90 then the larger triangle must be 30-60-90
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Re: area of shaded region [#permalink] New post 13 Dec 2009, 08:08
consider triangle BCD
sqr(DC)=sqr(BC) sqr(BD)
substituting values we get DC=12
Let AB=x and AD = y
In triangle ADB
sqr(y)=sqr(x)+ sqr(DB)
sqr(y) =sqr(x) + 3*36 ------ 1

In triangle ADC
angle ADC = 90

sqr(AC)= sqr(AD)+sqr(DC)
sqr(x+6)= sqr(y)+ sqr(12) ---- 2
solving 1 and 2
AB=x=18

hence shaded portion = 0.5*pi(144-81-9)
= 27*pi

Ans C
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Re: area of shaded region [#permalink] New post 14 Dec 2009, 08:10
Angle ADC is a right angle because it touches the circle and each ends of the diameter. We also already know that angle ACD is 60 deg, so ADC will be 30.
Re: area of shaded region   [#permalink] 14 Dec 2009, 08:10
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