Question 401653
If you recall the 8-15-17 Pythagorean triple, you can conclude the base is 15 and the height is 8.


Or, we can turn this into a quadratic by letting the height be h and the base h+7. Then,


{{{h^2 + (h+7)^2 = 17^2 = 289}}}


{{{2h^2 + 14h + 49 = 289}}}


{{{2h^2 + 14h - 240 = 0}}}


{{{h^2 + 7h - 120 = 0}}} (dividing by 2)


{{{(h-8)(h+15) = 0}}}


h = 8 (taking the positive solution).