Question 773066
Let ABC be a right triangle , right angled at B. So AC is the hypotenuse.

Therefore by Pythagorus theorem AB^2 + BC^2 = AC^2 ----------(1)

Let D be the middle point of AC i.e. AD = CD ----------------(2)

BD is the the median from the right angle to the hypotenuse.
So BD = 5 in.

Also by Apollonius theorem we know that 'In any triangle , the sum of the squares on two sides is equal to twice the square on half the third side together with twice the square on the median that bisects the third side'. 
Therefore in our case we can state that AB^2 + BC^2 = 2(AC/2)^2 + 2(BD)^2 --(3)

But AC/2 = AD -------(4)
or  AC = 2AD --------(5)

Using equation 1 and equation 4 in equation 3 we have
                 AC^2 = 2(AD)^2 + 2(BD)^2 --------------(6)
Using equation 5 we can write equation 6 as
                  (2AD)^2 = 2(AD)^2 + 2(BD)^2
                or 4(AD)^2 = 2(AD)^2 + 2(BD)^2
                or 4(AD)^2 - 2(AD)^2 = 2(BD)^2
                or 2(AD)^2 = 2(BD)^2
                or AD^2 = BD^2
Putting the value of BD in the above equation we get
                   AD^2 = 5^2 = 25
Therefore          AD = 5
From equation 5 we have AC = 2AD = 2*5 = 10
So the length of the hypotenuse is 10 in.