SOLUTION: I'm stuck on this question. A rect. has a perimeter of 68 cm, and it's area is 273 sq.cm. Find the dimensions of this rect. (For purposes of this problem only, the width is shorte

Question 272992: I'm stuck on this question.
A rect. has a perimeter of 68 cm, and it's area is 273 sq.cm. Find the dimensions of this rect. (For purposes of this problem only, the width is shorter than the length.)
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Write the equation for each statement
:
A rect. has a perimeter of 68 cm,
2L + 2W = 68
Simplify, divide by 2
L + W = 34
or
L = (34-W); use this form for substitution
:
it's area is 273 sq.cm.
L * W = 273
Substitute (34-W) for L
(34-W)*W = 273
-W^2 + 34W = 273
Arrange as a quadratic equation
-W^2 + 34W - 273 = 0
multiply by -1, it's easier to factor
W^2 - 34W + 273
Factors to
(W - 13)(W - 21) = 0
Two solutions
W = 13
and
W = 21
:
W = 13 is the width and L = 21 is the length
;
:
You can check this by finding the perimeter and area with these values.

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