+595 In rectangle WXYZ, A is on side \overline{WX}, B is on side \overline{YZ}, and C is on side \overline{XY}. If AX = 2, BY = 4, \angle ACB = 90^\circ, and CY = CX, then find AB.
+129847 W y A 2 X
x
C
x
Z B 4 Y
angle ACX + angle BCY = 90
Let
sin ACX = 2 / sqrt [2^2 + x^2]
cos ACX = x/ sqrt ;2^2 + x^2]
sin BCY = 4 /sqrt [4^2 + x^2]
cos BCY = x /sqrt [ 4^2 + x^2]
sin ( ACX + BCY) = sin 90
sin ACX cos BCY + cos ACX sin BCY
2 / sqrt [ 2^2 + x^2[ * x / sqrt [ 4^2 + x^2] + x/ /sqrt [ 2^2 + x^2 ] * 4 /sqrt [ 4^2 + x^2] =1
2x + 4x = sqrt [ 4 + x^2] sqrt [ 16 + x^2]
6x = sqrt [ (4 + x^2) (16 + x^2)] square both sides
36x^2 = (4 + x^2) (16 + x^2)
36x^2 = x^4 + 20x^2 + 64
x^4 - 16x^2 + 64 = 0
(x^2 - 8)^2 = 0
x^2 - 8 = 0
x = sqrt (8)
2x = 2sqrt 8
Draw yB perpendicular to ZY
yB = 2sqrt 8
yA = BY - AX = 4 - 2 = 2
By the Pythagorean Theorem
yA^2 + yB^2 = AB^2
2^2 + (2sqrt 8)^2 = AB^2
4 + 32 = AB^2
36 = AB^2
AB = sqrt 36 = 6
