# Answer to Question #16646 in Geometry for lola

Question #16646

use folding to construct a square with exactly 3/4 the area of the original square. Write an explanation of your methods. explain how you know the area is exactly 3/4 of the original area

Expert's answer

Let the initial square be square 1, and the resulting square be square 2.

S2 = ¾ * S1 => for sides of the square the following holds: a2 = a1*sqrt(3)/2.

Fold the square in half, you will get the segment of the length a1/2. Straightenit out.

Now choose the vertex of the square, and draw a circle of radius a1 / 2with the center in it.

Choose the vertex neighbor to the first vertex and draw a tangent from thepoint of this vertex to the circle from the previous step.

The distance between the length of this tangent will be a1* sqrt(3)/2,because the square triangle was formed with cathetus of a length a1 / 2 and

hypotenuse of length a1.

So, we get the segment of length a1* sqrt(3)/2. Now choose some vertex, measure this lengthalong two sizes of the square, and fold the square in the points of ends of

these segments along sizes of the square.

S2 = ¾ * S1 => for sides of the square the following holds: a2 = a1*sqrt(3)/2.

Fold the square in half, you will get the segment of the length a1/2. Straightenit out.

Now choose the vertex of the square, and draw a circle of radius a1 / 2with the center in it.

Choose the vertex neighbor to the first vertex and draw a tangent from thepoint of this vertex to the circle from the previous step.

The distance between the length of this tangent will be a1* sqrt(3)/2,because the square triangle was formed with cathetus of a length a1 / 2 and

hypotenuse of length a1.

So, we get the segment of length a1* sqrt(3)/2. Now choose some vertex, measure this lengthalong two sizes of the square, and fold the square in the points of ends of

these segments along sizes of the square.

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