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?
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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?
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}
