SOLUTION: Perimeter is 116 area is 825 what are the dimensions

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Question 846742: Perimeter is 116 area is 825 what are the dimensions
Answer by josh_jordan(263) About Me  (Show Source):
You can put this solution on YOUR website!
To find the dimensions of this rectangle, we need to know the formulas for the perimeter and area of rectangles:

Perimeter = 2L + 2W

Area = L x W

We are told the perimeter is 116 and the area is 825. So:

2L + 2W = 116
L x W = 825

Now that we have our equations, we can solve by rewriting our second equation in terms of either L or W and substituting that result into our first equation. Let's rewrite equation 2 in terms of L:

L = 825/W

Now, let's substitute 825/b for a in our first equation:

2(825/W) + 2W = 116

Multiply 2 by 825/W, which will give us

1650/W + 2W = 116

Now, we need to add 1650/W to 2W. To do this, multiply 1650/W by 1 and multiply 2W by W. add those results together and place in the numerator, and place "W" in the denominator:

%281650%2FW%29%2B%28%282W%2AW%29%2FW%29=116 ----->

%281650%2B2W%5E2%29%2FW=116

Next, multiply both sides of the equation by W, giving us:

1650%2B2W%5E2=116W

Now, we need to subtract 116W from both sides, giving us:

1650%2B2W%5E2-116W=0

Rewrite in standard quadratic form:

2W%5E2-116W%2B1650=0

Now, we will use the quadratic formula to solve for W:

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aW%5E2%2BbW%2Bc=0 (in our case 2W%5E2%2B-116W%2B1650+=+0) has the following solutons:

W%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=%28-116%29%5E2-4%2A2%2A1650=256.

Discriminant d=256 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--116%2B-sqrt%28+256+%29%29%2F2%5Ca.

W%5B1%5D+=+%28-%28-116%29%2Bsqrt%28+256+%29%29%2F2%5C2+=+33
W%5B2%5D+=+%28-%28-116%29-sqrt%28+256+%29%29%2F2%5C2+=+25

Quadratic expression 2W%5E2%2B-116W%2B1650 can be factored:
2W%5E2%2B-116W%2B1650+=+2%28W-33%29%2A%28W-25%29
Again, the answer is: 33, 25. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-116%2Ax%2B1650+%29



Therefore, our width = 25 or 33

To find the length of our rectangle, we need to substitute 25 and 33 one at a time into one of our original equations. Let's use our original second equation, L x W = 825:

L x 25 = 825 ----->

L = 825/25 ----->

L = 33
So, one of the dimensions of our rectangle is: 33 x 25

Now, we need to find the other possible dimensions of our rectangle by substituting 33 for our width in L x W = 825:

L x 33 = 825 ----->

L = 825/33 ----->

L = 25

So the other possible dimensions of our rectangle are: 25 x 33

Therefore, our two possible sets of dimensions are: 33 x 25 OR 25 x 33