SOLUTION: For a certain right triangle it is known that the hypotenuse is 1cm longer than one of the legs with 18 cm longer than the other leg. How long is each side?

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Question 667161: For a certain right triangle it is known that the hypotenuse is 1cm longer than one of the legs with 18 cm longer than the other leg. How long is each side?
Found 2 solutions by vleith, MathLover1:
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the shorter leg. You are told the longer leg is x%2B18 and the hypotenuse is x+%2B+19
Use the Pythagorean theorem
c%5E2+=+a%5E2+%2B+b%5E2
%28x%2B19%29%5E2+=+x%5E2+%2B+%28x%2B18%29%5E2
x%5E2+%2B+38x+%2B+361+=+x%5E2+%2B+%28x%5E2+%2B+36x+%2B+324%29
0+=+x%5E2+-+2x+-+37
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-2x%2B-37+=+0) has the following solutons:

x%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=%28-2%29%5E2-4%2A1%2A-37=152.

Discriminant d=152 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--2%2B-sqrt%28+152+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-2%29%2Bsqrt%28+152+%29%29%2F2%5C1+=+7.16441400296898
x%5B2%5D+=+%28-%28-2%29-sqrt%28+152+%29%29%2F2%5C1+=+-5.16441400296898

Quadratic expression 1x%5E2%2B-2x%2B-37 can be factored:
1x%5E2%2B-2x%2B-37+=+1%28x-7.16441400296898%29%2A%28x--5.16441400296898%29
Again, the answer is: 7.16441400296898, -5.16441400296898. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B-37+%29


Since the triangle legs must be positive, then the sides are 7.16cm, 25.16cm and 26.16cm

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

let x be hypotenuse
the hypotenuse is 1cm longer than one of the legs,
x%2B1cm........1
with 18cm longer than the other leg
x%2B18cm..........2
x%5E2=%28x-1cm%29%5E2%2B%28x-18cm%29%5E2.
x%5E2=x%5E2-2xcm%2B1cm%5E2%2Bx%5E2-36cm%2B324cm%5E2.
cross%28x%5E2%29=cross%28x%5E2%29%2Bx%5E2-38xcm%2B325cm%5E2
0=x%5E2-38xcm%2B325cm%5E2..use quadratic formula
Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve x%5E2-38%2Ax%2B325=0 ( notice a=1, b=-38, and c=325)





x+=+%28--38+%2B-+sqrt%28+%28-38%29%5E2-4%2A1%2A325+%29%29%2F%282%2A1%29 Plug in a=1, b=-38, and c=325




x+=+%2838+%2B-+sqrt%28+%28-38%29%5E2-4%2A1%2A325+%29%29%2F%282%2A1%29 Negate -38 to get 38




x+=+%2838+%2B-+sqrt%28+1444-4%2A1%2A325+%29%29%2F%282%2A1%29 Square -38 to get 1444 (note: remember when you square -38, you must square the negative as well. This is because %28-38%29%5E2=-38%2A-38=1444.)




x+=+%2838+%2B-+sqrt%28+1444%2B-1300+%29%29%2F%282%2A1%29 Multiply -4%2A325%2A1 to get -1300




x+=+%2838+%2B-+sqrt%28+144+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)




x+=+%2838+%2B-+12%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%2838+%2B-+12%29%2F2 Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


x+=+%2838+%2B+12%29%2F2 or x+=+%2838+-+12%29%2F2


Lets look at the first part:


x=%2838+%2B+12%29%2F2


x=50%2F2 Add the terms in the numerator

x=25 Divide


So one answer is

x=25




Now lets look at the second part:


x=%2838+-+12%29%2F2


x=26%2F2 Subtract the terms in the numerator

x=13 Divide


So another answer is

x=13


So our solutions are:

x=25 or x=13



So our solutions are:
x=25 or x=13
then we have one triangle with sides:the hypotenuse is 25cm, one of the legs 24cm, and the other leg is 7cm.....solution
and other triangle with sides:the hypotenuse is 13cm, one of the legs 12cm, and the other leg is 13cm-18cm...this cannot be solution because the length cannot be negative