# A rectangular piece of paper has an area of 100sq rt2 cm^2. The piece is such that, when it is folded in half along a dashed line, the new rectangle is similar (of the same shape) to the...

A rectangular piece of paper has an area of 100sq rt2 cm^2. The piece is such that, when it is folded in half along a dashed line, the new rectangle is similar (of the same shape) to the original rectangle. What are the dimensions of the piece of paper?

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It does not matter which variable you choose. I made x the width and y the length.

The dimensions of the original rectangle are approximately 10 by 14.14

The fold will be parallel to the width (so the length of the fold is 10) located about 7.07 units from the end. (This folds the rectangle in half.)

The ratio of length to width in the original rectangle is `(10sqrt(2))/10=sqrt(2)`

The ratio of length to width in the folded rectangle is `10/(5sqrt(2))` . This simplifies: `10/(5sqrt(2))*sqrt(2)/sqrt(2)=(10sqrt(2))/10=sqrt(2)` so the rectangles are similar. (All angles are 90 degrees, and corresponding sides are in proportion.)

Let the dimensions of the rectangle be x and y. After folding, the dimensions are x and `y/2` .

The area of the original rectangle is `100sqrt(2)"cm"^2` . Thus `xy=100sqrt(2)` .

Since the two rectangles are similar we have `x/y=((y/2))/x`

Then `x^2=y^2/2 ==> y^2=2x^2 ==> y=xsqrt(2)` .

Substituting into the area formula we get:

`x(xsqrt(2))=100sqrt(2) ==> x=10` so `y=10sqrt(2)` .

The original rectangle had dimensions `10 "x" 10sqrt(2)` and the folded rectangle has dimensions `5sqrt(2)"x"10` .

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The dimensions of the original piece of paper are

`10"cm"` by `10sqrt(2)"cm"`

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