contestada

A square sheet of paper has area $15\text{ cm}^2$. the front side is white and the back side is black. a corner of the sheet is lifted and placed so that the crease is at a $45^\circ$ angle. if the fold is such that the visible black area is equal to the visible white area, how many centimeters long is the crease? express your answer in simplest radical form. [asy] unitsize(2 cm); pair translate = (1.5,0); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw(shift(translate)*((1, 1 - sqrt(2/3))--(1,0)--(0,0)--(0,1)--(1 - sqrt(2/3), 1))); draw((1, 1 - sqrt(2/3))--(1 - sqrt(2/3),1),dashed); filldraw(shift(translate)*((1 - sqrt(2/3), 1 - sqrt(2/3))--(1, 1 - sqrt(2/3))--(1 - sqrt(2/3), 1)--cycle),black); [/asy]

Respuesta :

Hagrid

The best answer for this solution would be:

 

2 square root of 5

 

The first thing you do is to use the pythagorean theorem, and solve for the x. After that, you must square both sides. And keep continuing to find the x.

 

Then for the second method use the Pythagorean theorem in terms of the fold of the sheet.

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