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# Special Right Triangles

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A square sheet of paper has area $$6 \text{cm}^2$$ The front is white and the back is black. When the sheet is folded so that point A rests on the diagonal as shown, the visible black area is equal to the visible white area. How many centimeters is A from its original position? Express your answer in simplest radical form.

Oct 6, 2018

#1
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The sides of the square must be √6 cm  each

Let x be the distance between the legs of the black triangle and the side of the square

The white area is composed of a square of sides of  x  and two rectangles with dimensions of x and [ √6 - x ]

So.....the white area  =    x^2 + 2x ( √6 - x )

And the area of the  black triangle  =  ( √6 - x )^2 / 2  = [ 6 - [2√6]x  + x^2] / 2

Since these areas are equal  we have that

x^2 + 2x ( √6 - x )  = [ 6  - [2√6]x + x^2] / 2

2 [ x^2  + 2x (√6 - x) ] = 6 - [2√6]x + x^2

2x^2 + 4x(√6 - x ]   = 6 - [2√6]x  + x^2

x^2 + [4√6] x - 4x^2  = 6 - [2√6]x

-3x^2 + [ 6√6]x   = 6

3x^2  - (6√6]x  = -6

x^2 - [2√6]x = -2     complete the square

x^2  - [2√6]x  +  6  =  -2 + 6

( x - √6)^2  = 4    take both roots

x - √6  = 2        or     x - √6  = -2

x = √6 + 2       or     x  =  √6  - 2

(reject)                           (accept)

So  ...let the original position of A   = (0, 0)

And the new position  of A  =  ( √6  - ( √6  - 2) ,  √6 - ( √6 - 2) )  = ( 2 , 2)

And its distance from its original position is

√ [ 2^2  +  2^2  ]  =

√8  =

2√2

Oct 6, 2018
edited by CPhill  Oct 6, 2018
#2
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Thank you CPhill

Oct 8, 2018