The median CD to the hypotenuse AB cuts a right triangle ABC into two

The median CD to the hypotenuse AB cuts a right triangle ABC into two smaller triangles (isosceles) with the perimeters of 36 and 50 cm, respectively.
We need to find value of AC + BC =?
Property of median of right angle triangle, \(CD = AD = BD = \frac{AB}{2} i.e. AB=2*CD \)
\(AC+CD+AD = 36\)
\(AC + AB = 36\).....(1)
\(BC+CD+BD = 50\)
\(BC + AB = 50\) ........(2)
Subtract equation 1 from 2 we get \(BC - AC = 14.\).....(3)

Also from Pythagoras theorem, \(AC^2 + BC^2 = AB^2\)
subtract \(2*AC*BC\) from both sides and re write the equation as
\(AB^2 - 2* AC*BC = (AC - BC)^2 \)
substitute the value of AC and BC from equation 1 and 2 and value AC-BC from equation 3 in above equation
\(AB^2 - 2 (36-AB)* (50-AB) = (-14)^2\)
solving the equation we get
\(AB^2 - 172*AB + 3796 = 0\)
=> \((AB-26)*(AB-146) = 0\)
AB = 26 or 146 (Very high value so can be ignored)
Now substitute the value of AB in equation 1 and 2 to get value of AC = 10 and BC = 24
Therefore, \(AC + BC = 34\)
Answer: D

PS: At 0030 hrs with a drained mind I could workout this solution which is pretty lengthy. Will request the experts to post a short and crisp solution