SOLUTION: Here is my problem: The hypotenuse of a right triangle is 1cm more than twice the shorter leg, and the longer leg is 9cm less than three times the shorter leg. Find the lengths of

Algebra ->  Triangles -> SOLUTION: Here is my problem: The hypotenuse of a right triangle is 1cm more than twice the shorter leg, and the longer leg is 9cm less than three times the shorter leg. Find the lengths of       Log On


   

Question 82008: Here is my problem: The hypotenuse of a right triangle is 1cm more than twice the shorter leg, and the longer leg is 9cm less than three times the shorter leg. Find the lengths of the three sides of the triangle.
I know that a%5E2+%2B+b%5E2+=+c%5E2 (pythagorean theorem), so if a is the shorter leg, I would get: a%5E2+%2B+%283a-9%29%5E2+=+%282a%2B1%29%5E2 So if I follow what I think I know about solving equations, I wind up with 10a%5E2+-+54a+%2B+81+=+4a%5E2+%2B+4a+%2B1 which I can simplify down to 6a%5E2+-+58a+=+-80. This is where I get stuck. I'm not sure how to solve for a with a%5E2 in the equation, and if I try to factor out the a I am getting a negitive, never ending number for a.
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
A^2+(3A-9)^2=(2A+1)^2
A^2+9A^2-54A+81=4A^2+4A+1
10A^2-4A^2-54A-4A+81-1=0
6A^2-58A+80=0 HERE YOU HAVE TO FIND FACTORS OF 6 & 80 WHEN MULTIPLIED & THEN ADDED WILL EQUAL -58.
IF YOU CAN'T DO THAT THEN USE THE QUADRATIC EQUATION (aX^2+bX+c=0:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
(2A-16)(3A-5)=0
2A-16=0
2A=16
A=16/2
A=8 ANSWER.
3A-5=0
3X=5
A=5/3
A=1 2/3 ANSWER.