SOLUTION: The hypotenuse of a right triangle is 8 ft. longer than one leg and 4 ft. longer than the other leg. What are the dimensions of this triangle? Show the equation you used to solve t

Algebra ->  Triangles -> SOLUTION: The hypotenuse of a right triangle is 8 ft. longer than one leg and 4 ft. longer than the other leg. What are the dimensions of this triangle? Show the equation you used to solve t      Log On


   

Question 128479: The hypotenuse of a right triangle is 8 ft. longer than one leg and 4 ft. longer than the other leg. What are the dimensions of this triangle? Show the equation you used to solve this and then your answer.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let c = the hypotenuse and a and b the other two legs.
From the problem description, you can write:
c+=+a%2B8 and c+=+b%2B4 Rewrite these in terms of a and b to get:
a+=+c-8 and b+=+c-4 Now you can use the Pythagorean theorem to find c, the hypotenuse:
c%5E2+=+a%5E2+%2B+b%5E2 Substitute the two equations above for a and b:
c%5E2+=+%28c-8%29%5E2%2B%28c-4%29%5E2 Simplify.
c%5E2+=+%28c%5E2-16c%2B64%29%2B%28c%5E2-8c%2B16%29 Combine like-terms.
c%5E2+=+2c%5E2-24c%2B80 Subtract c%5E2 from both sides.
c%5E2-24c%2B80+=+0 Solve this quadratic by factoring:
%28c-4%29%28c-20%29+=+0, so then...
c+=+4 or c+=+20 Discard the first solution c+=+4 as c must be greater than 8 or else you have one leg of the triangle equal to zero!
So, the hypotenuse is 20 and the other two legs are a=12 and b = 16.
Check:
c%5E2+=+a%5E2%2Bb%5E2 Substitute c = 20, a = 12, and b = 16.
20%5E2+=+12%5E2%2B16%5E2
400+=+144%2B256
400+=+400 OK!