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Re: area of shaded region [#permalink]
10 Dec 2009, 20:21

1

This post received KUDOS

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

Re: area of shaded region [#permalink]
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

Re: area of shaded region [#permalink]
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

Re: area of shaded region [#permalink]
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.

gmatclubot

Re: area of shaded region
[#permalink]
14 Dec 2009, 08:10

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