SOLUTION: The perimeter of a rectangle is 22, and its diagonal is \sqrt{61}. Find its dimensions and area, accurate to two decimal places.

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Question 947304: The perimeter of a rectangle is 22, and its diagonal is \sqrt{61}. Find its dimensions and area, accurate to two decimal places.
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
Let L=length; Let W=Width; Perimeter=P=2(L+W); Diagonal=D=hypotenuse
L and W are two legs of a right triangle with the diagonal as hypotenuse, so:
P=2(L+W)
22=2(L+W) Divide each side by 2
11=L+W Solve for W, subtract L from each side
11-L=W
L%5E2%2BW%5E2=D%5E2
L%5E2%2BW%5E2=%28sqrt%2861%29%29%5E2
L%5E2%2BW%5E2=61 Substitute for W
L%5E2%2B%2811-L%29%5E2=61
L%5E2%2B%28121-22L%2BL%5E2%29=61
2L%5E2-22L%2B121=61 Subtract 61 from each side
2L%5E2-22L%2B60=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aL%5E2%2BbL%2Bc=0 (in our case 2L%5E2%2B-22L%2B60+=+0) has the following solutons:

L%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=%28-22%29%5E2-4%2A2%2A60=4.

Discriminant d=4 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--22%2B-sqrt%28+4+%29%29%2F2%5Ca.

L%5B1%5D+=+%28-%28-22%29%2Bsqrt%28+4+%29%29%2F2%5C2+=+6
L%5B2%5D+=+%28-%28-22%29-sqrt%28+4+%29%29%2F2%5C2+=+5

Quadratic expression 2L%5E2%2B-22L%2B60 can be factored:
2L%5E2%2B-22L%2B60+=+2%28L-6%29%2A%28L-5%29
Again, the answer is: 6, 5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-22%2Ax%2B60+%29

ANSWER:Length is 6 or 5
If L=6 units, then W=11-L=11-6=5 units
If L=5 units, then W=11-L=11-5=6 units
ANSWER: The dimensions are 6 units x 5 units (or 5 units x 6 units)
The area is the same either way: Area =L x W=6 units x 5 units= 30 square units