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In quadrilateral \(BCED\), we have \(BD = 11\), \(BC = 9\), and \(CE=2\). Sides \(\overline{BD}\) and \(\overline{CE}\) are extended past \(B\) and \(C\), respectively, to meet at point \(A\). If \(AC = 8\) and \(AB = 5\), then what is \( DE\)?

 Mar 23, 2020
 #1
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See  the following image

 

 

We can find angle BAC  with the Law of Cosines

 

We have

 

BC^2  = AB^2 + AC^2  - 2 ( AB * AC)  cos(BAC)

 

9^2  = 5^2  + 8^2  - 2 (5*8)cos(BAC)

 

[ 9^2  - 5^2 - 8^2]

______________  = cos BAC =   1/10

[ -2 * 5 * 8 ]

 

So using the Law of Cosines, again we  have  that

 

DE  = √[ AE^2 + AD^2  - 2 (AE * AD) cos (BAC)]

 

DE  =   √[ 10^2  + 16^2  - 2 (10*16)(1/10)  ] 

 

DE  = √ [ 100 + 256  - 32]

 

DE = √324  =  18

 

 

cool cool cool

 Mar 23, 2020

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