Question 747785
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Drawing the figure as described, we see the square divided into two congruent right triangles and a parallelogram, with all three figures having the same area.<br>
Let x be the side length of the square.<br>
The area of each right triangle is one-third the total area of the square, {{{(1/3)x^2}}}<br>
The length of one leg of the triangle is x; the length of the other leg is such that the area of the triangle -- one-half the product of the lengths of the two legs -- is {{{(1/3)x^2}}}.  That makes the length of the other leg {{{(2/3)x}}}<br>
With legs of lengths {{{x}}} and {{{(2/3)x}}}, the length of the hypotenuse, from the Pythagorean Theorem, is {{{x*sqrt(13)/3}}}<br>
Drawing an altitude of length 1 of the parallelogram produces a right triangle, similar to the two large triangles, with long leg of length 1 and hypotenuse of length {{{x/3}}}.<br>
Using corresponding parts of similar triangles, we then have<br>
{{{(x/3)/1=(x*sqrt(13)/3)/x}}}<br>
{{{x/3=sqrt(13)/3}}}<br>
{{{x=sqrt(13)}}}<br>
The side length of the square in centimeters is {{{sqrt(13)}}}, so the area of the square in square centimeters is 13.<br>
ANSWER: 13<br>