SOLUTION: A rectangular garden which is 12ft by 16ft is surrounding by a walkway which is a uniform width. If the walkway has an area of 165 ft^2, what is the width of the walkway?

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Question 902225: A rectangular garden which is 12ft by 16ft is surrounding by a walkway which is a uniform width. If the walkway has an area of 165 ft^2, what is the width of the walkway?
Answer by Stitch(470) About Me  (Show Source):
You can put this solution on YOUR website!
The area of a rectangle is given by the equation A = L*W, where L is the length and W is the width.
We can find the total area of the larger rectangle (The area of the garden + the area of the walkway) by adding the two areas together.
The area of the garden is :12+%2A+16+=+192ft^2
Now we will added the two given areas together to find the total area of the garden and the walkway.
192ft^2 + 165ft^2 = 357ft^2
Now we need to write the equation of the area of the large rectangle.
Let X be the width of the walkway.
The new length: L+=+X+%2B+12+%2B+X
The new width: W+=+X+%2B+16+%2B+X
The new area equation is:
357+=+%28X+%2B+12+%2B+X%29+%2A+%28X+%2B+16+%2B+X%29
Combine like terms
357+=+%282X+%2B+12%29+%2A+%282X+%2B+16%29
Use FOIL to simplify the right hand side of the equation.
ERROR Algebra::Solver::Engine::invoke_solver_noengine: solver not defined for name 'foil'.
Error occurred executing solver 'foil' .

Rewrite the equation
357+=+4X%5E2+%2B+56X+%2B+192
Subtract 357 from both sides
+0+=+4X%5E2+%2B+56X+-+165
Now use the quadratic equation to solve for X.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aX%5E2%2BbX%2Bc=0 (in our case 4X%5E2%2B56X%2B-165+=+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=%2856%29%5E2-4%2A4%2A-165=5776.

Discriminant d=5776 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-56%2B-sqrt%28+5776+%29%29%2F2%5Ca.

X%5B1%5D+=+%28-%2856%29%2Bsqrt%28+5776+%29%29%2F2%5C4+=+2.5
X%5B2%5D+=+%28-%2856%29-sqrt%28+5776+%29%29%2F2%5C4+=+-16.5

Quadratic expression 4X%5E2%2B56X%2B-165 can be factored:
4X%5E2%2B56X%2B-165+=+4%28X-2.5%29%2A%28X--16.5%29
Again, the answer is: 2.5, -16.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+4%2Ax%5E2%2B56%2Ax%2B-165+%29

X = 2.5 & -16.5
Now since we can not have a negative distance -16.5ft will not work for this problem.
So the width of the walkway is 2.5ft