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In right triangle $ABC$, $BD=CD+9$.  If $AB=10$ and $BC=25$, what is $AD$?

 

 Sep 10, 2023
 #1
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AC = sqrt (25^2  - 10^2)  = sqrt (525) 

 

AD = sqrt [ ( CD + 9 )^2 + 10^2 )    =  sqrt [ CD^2 + 18CD + 181 ]

 

AD  + DC =  AC =  sqrt (525)

 

sqrt [ CD^2 + 18CD + 181 ] + CD  = sqrt (525)

 

sqrt [ CD^2 + 18CD + 181 ]  =  sqrt [ 525 ] - CD        square both sides

 

CD^2 + 18CD + 181  =  525 - 2 CD*sqrt (525)  + CD^2

 

18CD  + 181  =  525 - 2CD * sqrt (525)

 

18CD + 2CD * sqrt (525)  = 525 - 181

 

CD ( 18  + 2sqrt (525))  = 344

 

CD  =  344 /  [ 18 + 2sqrt (525) ] =  172 / [ 9 + sqrt 525 ]  

 

AC - CD =  AD

 

sqrt (525)  -  172 / [ 9 + sqrt 525] =  AD

 

(sqrt (525) [ 9 + sqrt (525) ]  - 172) / [ 9 + sqrt (525) ] =  AD

 

[9sqrt (525) + 525 - 172] / [9 + sqrt 525)  = AD

 

[9 sqrt (525)  + 353 ] / [ 9 + sqrt (525) ] = AD  ≈  17.523

 

cool cool cool

 Sep 10, 2023

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