A rectangular box has a length of 8 ft and a width of 2 ft. The length of the three dimentional diagonal is 10 ft. What is the height of the box?

I think answer is 12.96, but not sure.

any help is appreciated
thanks in advance

Obviously the diagonal of 10 must be greater than either the length, width or height, so the height cannot be greater than 10 and you are incorrect

Look at the diagonal of the base, call it d
d^2 = 8^2 + 2^2 = 68
d = √68

now the diagonal we want is the hypotenuse of the right-angled triangle with base √68 and height of x
x^2 + √68^2 = 10^2
x^2 = 32
x = √32 = appr 5.66