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Triangle ABC is a right triangle with legs AB and AC. Points X and Y lie on legs AB and AC, respectively, so that AX:XB=AY:YC=1:2 . If BY=16 units, and CX=28 units, what is the length of hypotenuse BC? Express your answer in simplest radical form.

 Nov 6, 2018
edited by EmperioDaZe  Nov 6, 2018
 #1
avatar+98172 
+1

B

 

2

X

1

A     1     Y         2               C

 

Note that   BX  = 2AX

And  YC  =  2AY

 

So we have that

 

(3AX)^2 + AY^2  = BY^2   ⇒   9AX^2 + AY^2  = 256  ⇒  AY^2  = 256 - 9AX^2    (1)

And

AX^2  + (3AY)^2  = 28^2  ⇒  AX^2  + 9AY^2  = 784  (2)

 

Subbing (1)  into (2)  we have

 

AX^2 + 9(256 - 9AX^2)  = 784

AX^2 + 2304 - 81AX^2  = 784

-80AX^2  = -1520

AX^2  =  -152 / - 8

AX^2 = 19

AX = sqrt(19)

So....BA  =  3AX  =  3sqrt(19)

 

And

 

AX^2 + 9AY^2  = 784

19 + 9AY^2  = 784

9AY^2  = 765

AY^2 =  85

AY  =  sqrt (85)

So AC  = 3sqrt (85)

 

So  BC  =  sqrt  [ BA^2 + AC^2 ] = 

 

sqrt [ 9 * 19 + 9 * 85 ] =

 

sqrt [ 9 ( 19 + 85)  ]  =

 

3sqrt (104) =

 

3sqrt ( 26 * 4)  =

 

2 * 3 sqrt (26)  =

 

6 sqrt (26)

 

 

 

 

cool cool cool

 Nov 6, 2018
 #2
avatar+47 
+1

Thanks!

 Nov 7, 2018

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