SOLUTION: A right triangle has a hypotenuse of 41. All of its sides add up to 49. Find the length of the legs of the right triangle. I made these two equations based on the problem: a + b

Algebra ->  Pythagorean-theorem -> SOLUTION: A right triangle has a hypotenuse of 41. All of its sides add up to 49. Find the length of the legs of the right triangle. I made these two equations based on the problem: a + b      Log On


   

Question 933446: A right triangle has a hypotenuse of 41. All of its sides add up to 49. Find the length of the legs of the right triangle.
I made these two equations based on the problem: a + b + 41 = 49, a^2 + b^2 = 41^2
I tried to solve for both a and b on the first equation. I was left with a = 8-b and b = 8-a. I substituted those into the Pythagorean theorem equation and got (8-b)^2 + (8-a)^2 = 1681. I extended the equation and got 64-16b+b^2 + 64-16a+a^2 = 1681. I reduced it to 128-16b+b^2+16a+a^2 = 1681 by adding the coefficients. I then subtracted 128 on both sides and got -16b+b^2+16a+a^2 = 1553. That's where I was stuck. I wasn't sure what to do. What is the correct approach to solve this problem?
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
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