SOLUTION: The sides of the right triangular lot has a dimensions of (5x-11), (x) and (3x+4) meters respectively. What is the area of the lot?

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Question 1053024: The sides of the right triangular lot has a dimensions of (5x-11), (x) and (3x+4) meters respectively. What is the area of the lot?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The sides of the right triangular lot has a dimensions of (5x-11), (x) and (3x+4) meters respectively. What is the area of the lot?
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For all sides to be >0, 5x-11>0 --> x>2.2
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3x+4 > x for x>2.2, so x is not the hypotenuse.
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Either (3x+4)^2 + x^2 = (5x-11)^2
or
(3x+4)^2 = (5x-11)^2 + x^2
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Case 1: 5x-11 is the hypotenuse
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(3x+4)^2 + x^2 = (5x-11)^2
10x^2 + 24x + 16 = 25x^2 - 110x + 121
15x^2 - 134x + 105 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 15x%5E2%2B-134x%2B105+=+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=%28-134%29%5E2-4%2A15%2A105=11656.

Discriminant d=11656 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--134%2B-sqrt%28+11656+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-134%29%2Bsqrt%28+11656+%29%29%2F2%5C15+=+8.06543188700421
x%5B2%5D+=+%28-%28-134%29-sqrt%28+11656+%29%29%2F2%5C15+=+0.867901446329125

Quadratic expression 15x%5E2%2B-134x%2B105 can be factored:
15x%5E2%2B-134x%2B105+=+%28x-8.06543188700421%29%2A%28x-0.867901446329125%29
Again, the answer is: 8.06543188700421, 0.867901446329125. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+15%2Ax%5E2%2B-134%2Ax%2B105+%29

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x =~ 8.065
x1 is rejected, it's less than 2.2
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Case 2: 3x+4 is the hypotenuse
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(3x+4)^2 - x^2 = (5x-11)^2
8x^2 + 24x + 16 = 25x^2 - 110x + 121
17x^2 - 134x + 105 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 17x%5E2%2B-134x%2B105+=+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=%28-134%29%5E2-4%2A17%2A105=10816.

Discriminant d=10816 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--134%2B-sqrt%28+10816+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-134%29%2Bsqrt%28+10816+%29%29%2F2%5C17+=+7
x%5B2%5D+=+%28-%28-134%29-sqrt%28+10816+%29%29%2F2%5C17+=+0.882352941176471

Quadratic expression 17x%5E2%2B-134x%2B105 can be factored:
17x%5E2%2B-134x%2B105+=+%28x-7%29%2A%28x-0.882352941176471%29
Again, the answer is: 7, 0.882352941176471. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+17%2Ax%5E2%2B-134%2Ax%2B105+%29

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x = 7 ****** (the other solution is less then 2.2 and rejected)
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Use the 2 values of x to find the lengths of the 2 short sides, a & b
Area = a*b/2