SOLUTION: Find the dimensions of a rectangle whose area is 187 cm2 and whose perimeter is 56 cm. (Enter your answers as a comma-separated list.)

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Question 988671: Find the dimensions of a rectangle whose area is 187 cm2 and whose perimeter is 56 cm. (Enter your answers as a comma-separated list.)
Found 2 solutions by Boreal, solver91311:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
2L+2W=56;L+W=28
LW=187
L=187/W
187/W+W=28
187+W^2=28W, multiplying by W
W^2-28W+187=0
W=(1/2)(28+/-) sqrt (784-748);sqrt term =6
W=17 or 11
L=11 or 17
By convention, L is longer, so it is 17*11
17,11,17,11 is perimeter
17*11 is area.
Notice that the roots are interchangeable.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


The sum of the length and the width of a rectangle is half of the perimeter. Proof of this assertion is left as an exercise for the student.

For this rectangle, length plus width equals 28, so we can say .

Then we know that the area, which is the product of the length times the width, is 187. Substituting from above:



Solve the quadratic. One of the roots is the width, the other is the length.

John

My calculator said it, I believe it, that settles it