SOLUTION: IN A RIGHT TRIANGLE, THE HYPOTENUSE IS 1 FOOT MORE THAN THE LONG LEG, WHILE THE SHORT LEG IS 11FT. LESS THAN HALF THE LONG LEG. YOUR GOAL IS TO FIND THE PERIMETER OF THIS TRIANGLE.

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Question 952453: IN A RIGHT TRIANGLE, THE HYPOTENUSE IS 1 FOOT MORE THAN THE LONG LEG, WHILE THE SHORT LEG IS 11FT. LESS THAN HALF THE LONG LEG. YOUR GOAL IS TO FIND THE PERIMETER OF THIS TRIANGLE.
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
b=long leg; a=short leg=(0.5b)-11ft; c=hypotenuse=b+1
a%5E2%2Bb%5E2=c%5E2 Substitute for a and c, solve for b
.25b%5E2-11b%2B121%2Bb%5E2=b%5E2%2B2b%2B1
1.25b%5E2-11b%2B121=b%5E2%2B2b%2B1Subtract%28b%5E2%2B2b%2B1%29 from each side.
0.25b%5E2-13b%2B120=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ab%5E2%2Bbb%2Bc=0 (in our case 0.25b%5E2%2B-13b%2B120+=+0) has the following solutons:

b%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-13%29%5E2-4%2A0.25%2A120=49.

Discriminant d=49 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--13%2B-sqrt%28+49+%29%29%2F2%5Ca.

b%5B1%5D+=+%28-%28-13%29%2Bsqrt%28+49+%29%29%2F2%5C0.25+=+40
b%5B2%5D+=+%28-%28-13%29-sqrt%28+49+%29%29%2F2%5C0.25+=+12

Quadratic expression 0.25b%5E2%2B-13b%2B120 can be factored:
0.25b%5E2%2B-13b%2B120+=+0.25%28b-40%29%2A%28b-12%29
Again, the answer is: 40, 12. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+0.25%2Ax%5E2%2B-13%2Ax%2B120+%29

b=40 or b=12 (long leg)
For b=40:
a=0.5b-11ft=0.5(40)-11 ft=9 ft (short leg)
c=b+1=40ft+1 ft=41 ft(hypotenuse)
For b=12
a=.5(12)-11 ft=6-11=-5ft does not work
ANSWER 1: The long leg is 40 feet.
ANSWER 2: The short leg is 9 feet.
ANSWER 3: The hypotenuse is 41 feet.
CHECK:
a%5E2%2Bb%5E2=c%5E2
%289+ft%29%5E2%2B%2840+ft%29%5E2=%2841+ft%29%5E2
81ft%5E2%2B1600ft%5E2=1681ft%5E2
1681ft%5E2=1681ft%5E2