Question 1096449
<br>GH is the same length as GI:
{{{x^2 = -2x+15}}}
{{{x^2+2x-15 = 0}}}
{{{(x+5)(x-3) = 0}}}<br>
The potential solutions are x=-5 and x=3; but we need to check to see if they make sense in the actual problem.<br>
x = -5:  GH = x^2 = (-5)^2 = 25; GI = -2x+15 = 10+15 = 25; HI is -x+4 = 5+4 = 9.
That works; the side lengths are 25, 25, and 9.<br>
x = 3:  GH = x^2 = 3^2 = 9; GI = -2x+15 = -6+15 = 9; HI = -x+4 = 1.
That also works: the side lengths are 9, 9, and 1.<br>
So there are two different isosceles triangles that satisfy the given conditions.