+0

+1
260
2

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
+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)   Nov 6, 2018
#2
+1

Thanks!

Nov 7, 2018