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 $200$ , 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 1

Let the leg of the longer right triangle have length $a$ and the shorter one have length $b$ , so that the area of the four triangles is $\frac{1}{2}a\cdot a+\frac{1}{2}b\cdot b+\frac{1}{2}a\cdot a+\frac{1}{2}b\cdot b = a^2+b^2$ . The problem says that this is equal to $200$ , so we have $a^2+b^2=200$

By the Pythagorean theorem, the length of the rectangle is $\sqrt{a^2+a^2}=a\sqrt{2}$ and the width is $\sqrt{b^2+b^2}=b\sqrt{2}$ , so the side length of the diagonal is $\sqrt{(\sqrt{2}a)^2+(\sqrt{2}b)^2} = \sqrt{2(a^2+b^2)}$ . Since $a^2+b^2=200$ , this is simply $\sqrt{2\cdot 200} = \boxed{20}$ .

Solution 2

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 $\sqrt{200}$ , and because the hypotenuse of the 45-45-90 triangle formed by the diagonal is $\sqrt{2}$ *side length, we see that the diagonal of the square and therefore the diagonal of the rectangle is $\sqrt{400}$ or $\boxed{20}$ .

2008 UNCO Math Contest II (Problems • Answer Key • Resources)
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2008 UNCO Math Contest II Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

See Also