In right traingle ABC;
\(AB^2 = BC^2 + AC^2\)
\(AB^2 = 10^2 + 24^2\)
\(AB = \sqrt{676} = 26\)

Also, if CM is the median then AM = MB = \(\frac{26}{2} = 13\)

Approach 1:
We can use the Theorem: In a right triangle, the median drawn to the hypotenuse has the measure half the hypotenuse.
Therefore, CM = \(\frac{26}{2} = 13\)

Approach 2:
Draw AB' parallel to BC with AB' = BC = 10 and B'B parallel to AC with B'B = AC = 24
Notice that we get a rectangle AB'BC

Property: Diagonals of a rectangle are equal and bisect each other
i.e. CM = MB = B'M = AM = 13