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BC

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Source: — Data Sufficiency |
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by mehravikas » Sun Oct 11, 2009 8:19 pm
I am not able to understand why 'D' is the answer.

From statement 1 - Can we assume the sides to be in the ratio 1:sqrt(3):2 ?
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by rohan_vus » Mon Oct 12, 2009 1:44 am
Yes stmnt (1) is also sufficient . consider dropping a perpendicular from B onto AD .. Take M a point on line AD and draw a simple vertical line to connect to B . thus now u have tow right anngle traingles ..One is BMC ( right angle at M ) and other is BMD ( again right angle at M) ..

Now u can derive easily CM = DM - CD .. ==> 3.. coz DM is given by 30-60-90 degree rule and you already know BD , so DM = 9

Similarly BM = 3*(sqrt of 3).. Consider triangle BMC now..( given BM and CM.. you know its 60degree at angle BCA ...So easy to BC then now.. BC = 6 = CD
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by mehravikas » Mon Oct 12, 2009 11:40 am
Can you draw the figure and add it as an attachment please?
rohan_vus wrote:Yes stmnt (1) is also sufficient . consider dropping a perpendicular from B onto AD .. Take M a point on line AD and draw a simple vertical line to connect to B . thus now u have tow right anngle traingles ..One is BMC ( right angle at M ) and other is BMD ( again right angle at M) ..

Now u can derive easily CM = DM - CD .. ==> 3.. coz DM is given by 30-60-90 degree rule and you already know BD , so DM = 9

Similarly BM = 3*(sqrt of 3).. Consider triangle BMC now..( given BM and CM.. you know its 60degree at angle BCA ...So easy to BC then now.. BC = 6 = CD
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by 2010gmat » Thu Nov 05, 2009 6:41 am
stmt 1 is suff...no need to draw the figure...simply drop a prependicular from B on AD...say it meets AD at E...

BE/ BD = Sin 30 --calclate BE

(dC+EC = DE) --> DE/BD = Cos 30 we know ec and bd calculate dc...

now in triangle BEC we know both BE and BC

BE/BC = Tanx -- x comes out as 60

x = exterior angle = sum of two int angles

=> <cbd = <cdb => CD = BC = 6

2 is simple...both angles equal hence sides equal

answer shud be D
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