Here's the way to solve this
Let E be the foot of the perpendicular drawn from X to AB
By the Pythagorean Theorem
AX^2 = AE^2 + XE^2 and
BX^2 = BE^2 + XE^2
1^2 = AE^2 + XE^2
7^2 = BE^2 + XE^2 subtract the first equation form the second and we have that
48 = BE^2 - AE^2 (1)
Likewise......
Let F be the foot of the perpendicular drawn from X to DC
And note that AE = DF
And BE = CF
So....by the Pythagorean Theorem we have that
DX^2 = DF^2 + XF^2
CX^2 = CF^2 + XF^2
DX^2 = AE^2 + XF^2
CX^2 = BE^2 + XF^2
DX^2 = AE^2 + XF^2
8^2 = BE^2 + XF^2 subtract the first equation from the second
64 - DX^2 = BE^2 - AE^2 (2)
Equating (1) and (2) we have that
64 - DX^2 = 48 rearrange as
64 - 48 = DX^2
16 = DX^2 take the positive root
4 = DX
Just a Dragan found !!!!
