Question 464769
If the 5-12-13 triangle sticks out at you, and you recognize 5+12 = 17, times 2 is 34, you're done.


To set it as a quadratic, we can let L and 17-L be the length and width (17-L instead of 34-L because the perimeter is 2(L+W) = 2(L + 17-L) = 34). The triangle with sides L, 17-L, and the diagonal of length 13 must satisfy the Pythagorean theorem, i.e.


*[tex \LARGE L^2 + (17-L)^2 = 169]


*[tex \LARGE 2L^2 - 34L + 289 = 169]


*[tex \LARGE 2L^2 - 34L + 120 = 0]


*[tex \LARGE 2(L-5)(L-12) = 0 \Rightarrow L = 5, 12]


This implies the dimensions are 5 cm by 12 cm (note that both dimensions are under L because we can choose either length to be L).