2008 UNCO Math Contest II Problems/Problem 3
Contents
Problem
A rectangle is inscribed in a square creating four isosceles right triangles. If the total area of these four triangles is , what is the length of the diagonal of the rectangle?
Solution 1
Let the leg of the longer right triangle have length and the shorter one have length , so that the area of the four triangles is . The problem says that this is equal to , so we have
By the Pythagorean theorem, the length of the rectangle is and the width is , so the length of the rectangle's diagonal is . Since , this is simply .
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 , and because the hypotenuse of the 45-45-90 triangle formed by the diagonal is *side length, we see that the diagonal of the square and therefore the diagonal of the rectangle is or .
See Also
2008 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
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 |