Question 933446
a+b+c=49

a+b+41=49

a+b= 8

I think there is a typo in the problem

If hypotenuse is 41

a+b cannot be 8 

The squares of a & b when added cannot give 41^2


The sum of the other two sides should be 49  ( 9 & 40)

Then it is possible

9^2+40^2=41^2