View source for 2008 UNCO Math Contest II Problems/Problem 3

== Problem ==

A rectangle is inscribed in a square
creating four isosceles right triangles.
If the total area of these four triangles
is <math>200</math>, what is the length of the diagonal
of the rectangle?

<asy>
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle,black);
draw((0,2/3)--(2/3,0)--(1,1/3)--(1/3,1)--cycle,black);
</asy>

== Solution ==
Without loss of generality, squeeze the rectangle into a line that becomes the diagonal of the square. 2 of the triangles approach 0 area as the rectangle approaches a line and the diagonal of the rectangle approaches the line, so we can treat this as a question of "What is the length of the diagonal of a square of area 200?" 
We see that the side of the square must be <math>\sqrt{200}</math>, and because the hypotenuse of the 45-45-90 triangle formed by the diagonal is <math>\sqrt{2}</math>*side length, we see that the diagonal of the square and therefore the diagonal of the rectangle is <math>\sqrt{400}</math> or <math>\boxed{20}</math>.

== See Also ==
{{UNCO Math Contest box|n=II|year=2008|num-b=2|num-a=4}}

[[Category:Introductory Geometry Problems]]