SOLUTION: The perimeter of a rectangle is 52 yard, and the rectangle is 168yd^2. find the dimensions of the rectangle

Question 91865: The perimeter of a rectangle is 52 yard, and the rectangle is 168yd^2. find the dimensions of the rectangle
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The perimeter of a rectangle is 52 yard, and the rectangle is 168yd^2. find the dimensions of the rectangle
:
Let x = length and y = width
Perimeter:
2x + 2y = 52
Simplify, divide by 2:
x + y = 26
y = (26-x): can use for substitution
:
Area:
x * y = 168
Substitute (26-x) for y
x(26-x) = 168
26x - x^2 = 168
0 = x^2 - 26x + 168; arrange as a quadratic equation, solve for x
Factors to:
(x - 12)(x - 14) = 0
:
x = +12 and x = +14
:
Length 14 and width 12, would be the answer
:
:
Check using the perimeter
2(14) + 2(12) = 52
:
Did this make sense to you? Any questions?
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