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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
avatar+129401 
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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

 

cool cool cool

 Jun 8, 2024
edited by CPhill  Jun 8, 2024

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