+0

# Geometry

0
1
1
+388

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.

Jun 7, 2024

#1
+129401
+1

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

Jun 8, 2024
edited by CPhill  Jun 8, 2024