SOLUTION: The length of a square is increased by a 5th so that its new area is 44cm^2 more than the original value. Find the difference in perimeter of two shapes.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: The length of a square is increased by a 5th so that its new area is 44cm^2 more than the original value. Find the difference in perimeter of two shapes.      Log On


   

Question 1078906: The length of a square is increased by a 5th so that its new area is 44cm^2 more than the original value. Find the difference in perimeter of two shapes.
Answer by Alan3354(69443) About Me  (Show Source):
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The length of a square is increased by a 5th so that its new area is 44cm^2 more than the original value. Find the difference in perimeter of two shapes.
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10 cm square --> area of 100 sq cm
12 cm square --> area of 144 sq cm
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4*(12-10) = 8 cm difference in perimeter