+47 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.
+129852 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)
