The width of the rectangle is 13 inches and its length is 12 inches or the width of the rectangle is 12 inches and its length is 13 inches.
Let
The area of a rectangle, A = LW. Since A = 156 square inches,
LW = 156 (1)
Also, the perimeter of the rectangle, P = 2(L + W). Since P = 50 inches,
2(L + W) = 50
L + W = 50/2
L + W = 25 (2)
From (2), L = 25 - W
Substituting L into (1), we have
LW = 156
(25 - W)W = 156
25W - W² = 156
Re-arranging, we have
W² - 25W + 156 = 0
Using the quadratic formula, we find W.
For a quadratic equation ax² + bx + c = 0, the roots of the equation,
[tex]x = \frac{-b +/- \sqrt{b^{2} - 4ac} }{2a}[/tex]
So,comparing both equations, a = 1, b = -25 and c = 156.
So,
[tex]W = \frac{-(-25) +/- \sqrt{(-25)^{2} - 4X1X156} }{2X1}\\W = \frac{25 +/- \sqrt{625 - 624} }{2}\\W = \frac{25 +/- \sqrt{1} }{2}\\W = \frac{25 +/- 1}{2}\\W = \frac{25 + 1}{2} or W = \frac{25 - 1}{2}\\W = \frac{26}{2} or W = \frac{24}{2}\\W = 13 or 12[/tex]
Since L = 25 - W
Substituting W = 13 into the equation, we have
L = 25 - 13
= 12 inches
Substituting W = 12 into the equation, we have
L = 25 - 12
= 13 inches.
So, the width of the rectangle is 13 inches and its length is 12 inches or the width of the rectangle is 12 inches and its length is 13 inches.
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