2008 UNCO Math Contest II Problems/Problem 3


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 length of the rectangle's 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 $\boxed{20}$.

See Also

2008 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions