SOLUTION: Find the perimeter of a rectangle whose area =63ft^2 and whose length is 2ft longer than its width? If you can solve it normally or using the Pythagorean Theorem? Thank You

Question 433574: Find the perimeter of a rectangle whose area =63ft^2 and whose length is 2ft longer than its width? If you can solve it normally or using the Pythagorean Theorem?

Thank You
Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
"Find the perimeter of a rectangle whose area =63ft^2 and whose length is 2ft longer than its width?"

x = width
x + 2 = length {length is 2 more than width}

Area of a rectangle = width x length

x(x + 2) = 63 {area is width x length}
x^2 + 2x = 63 {used distributive property}
x^2 + 2x - 63 = 0 {subtracted 63 from both sides}
x + 9)(x - 7) = 0 {factored into two binomials}
x + 9 = 0 or x - 7 = 0 {set each factor equal to 0}
x = - 9 or x = 7 {subtracted 9 and added 7, respectively}
x = 7 {width cannot be negative}
x + 2 = 9 {substituted 7, in for x, into x + 2}
width = 7ft and length = 9ft

Perimeter of a rectangle is 2(width) + 2(length)
= 2(7) + 2(9) {substituted width and length into perimeter formula}
= 14 + 18 {multiplied}
= 32 ft {added}

Perimeter of the rectangle is 32 ft.
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