SOLUTION: one leg of a right is 2 millimeters shorter than the longer leg and the hypotenuse is 2 millimeters longer than the longer leg. find the lengths of the sides of the triangle. find

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Question 949117: one leg of a right is 2 millimeters shorter than the longer leg and the hypotenuse is 2 millimeters longer than the longer leg. find the lengths of the sides of the triangle. find the lengths of the sides of the triangle
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
a=shorter leg; b=longer leg=a+2; c=hypotenuse=a+4
a%5E2%2Bb%5E2=c%5E2 Substitute for b and c
a%5E2%2B%28a%2B2%29%5E2=%28a%2B4%29%5E2
a%5E2%2Ba%5E2%2B4a%2B4=a%5E2%2B8a%2B16 Subtract a^2 from each side.
a%5E2%2B4a%2B4=8a%2B16 Subtract 8a from each side.
a%5E2-4a%2B4=16 Subtract 16 from each side.
a%5E2-4a-12=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aa%5E2%2Bba%2Bc=0 (in our case 1a%5E2%2B-4a%2B-12+=+0) has the following solutons:

a%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=%28-4%29%5E2-4%2A1%2A-12=64.

Discriminant d=64 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--4%2B-sqrt%28+64+%29%29%2F2%5Ca.

a%5B1%5D+=+%28-%28-4%29%2Bsqrt%28+64+%29%29%2F2%5C1+=+6
a%5B2%5D+=+%28-%28-4%29-sqrt%28+64+%29%29%2F2%5C1+=+-2

Quadratic expression 1a%5E2%2B-4a%2B-12 can be factored:
1a%5E2%2B-4a%2B-12+=+1%28a-6%29%2A%28a--2%29
Again, the answer is: 6, -2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-4%2Ax%2B-12+%29

So the solution we want is 6
The length of the shorter leg is 6 mm.
b=a+2 mm=6 mm+2 mm=8 mm The length of the longer leg is 8 mm.
c=a+4 mm=6 mm+4 mm=10 mm The length of the hypotenuse is 10 mm, so we have a 3-4-5 right triangle.