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.
+129847 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
