62. Replace x with 5
Okay...anyways, I was able to get the first answer (61) and I'm on 62 right now kind of confused even though I think it's done in the exact same way. I may be wrong.
A) The length of the diagonal to GD is 5 square root of 2 (if anyone knows a place or program where I can type my stuff in and have it converted into latex code that would be great because I've had to copy and paste the coding and it just takes...forever) because in a 45-45-90 triangle the hypotenuse would be t square root 2 and t would be 5 because it says 5. That's how the first problem had it as a 45-45-90 triangle and I would assume it's the same because they're back to back once you slice them up but if it's not...then I worked out a 30-60-90 triangle solution for the B part as well and the answer would be that the hypotenuse is 10 in that case.
Answer: 5 square root 2...or 10 (if 30-60-90 triangle)
B) I'm suppose to find (GB)^2 = (GD)^2+(BD)^2.
(GB)^2 = 50 + 25
GB = Square root of 75
(10)^2+((5) square root (3))^2
(GB)^2 = 115
GB = Square root of 115