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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.

 Mar 5, 2019
 #1
avatar+98168 
+1

B

2

X

1

A     1    Y     2          C

 

Note that BAY  and XAC are right triangles...so....

 

We have tthis system

 

(3AX)^2 + (AY)^2 =  16^2         and

(AX)^2 +  (3AY)^2 = 28^2

 

Simplify

 

9AX^2 + (AY)^2 =   256   →   -81AX^2 - 9AY^2  = -2304      (1)

AX^2 +  9AY^2 =  784      (2)

 

Add (1)  and (2)

-80AX^2 = -1520

AX^2 =  -1520/ -80  = 19

AX = sqrt (19)

So...3AX = 3sqrt(19) = AB = sqrt (171)

 

And  

AX^2 + 9AY^2 = 784

19 + 9AY^2  784

9AY^2 = 765

AY^2 = 85

AY = sqrt (85)

So ....3AY = 3sqrt(85)  = AC = sqrt (765)

 

So....the hypotenuse BC =   sqrt ( AB^2 + AC^2)    =  sqrt (171 + 765 ) = sqrt (936) = 

sqrt (2^3 * 3^2 * 13)  = 2*3 sqrt (2 * 13) = 6sqrt (26)  units

 

 

cool cool cool

 Mar 5, 2019
edited by CPhill  Mar 5, 2019

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