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# Length of the side of a square given its area is 60 more than its perimeter

Summary:

The length of the side of a square, given that its area is 60 more than its perimeter, can be found by solving the equation \( s^2 = 4s + 60 \). Solving this quadratic equation, we get \( s = 10 \). Thus, the side length of the square is 10 units.

Square problem.Find the length of the side of a square if it's area is 60 more than its perimeter.

Area and perimeter have different units, so I'll consider only their numeric values. Let the side of the square be S.

The perimeter is 4S and the area is S^2

As the area is 60 more than the perimeter

S^2 - 4S = 60

=> S^2 - 4S - 60 = 0

=> S^2 - 10S + 6S - 60 = 0

=> S(S - 10) + 6(S - 10) = 0

=> (S + 6)(S - 10) = 0

S = -6 and S = 10

As length is positive we eliminate S = -6

The side of the square is 10

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Square problemFind the length of the side of a square if it's area is 60 more than its perimeter.

Area and perimeter have different units, so I'll consider only their numeric values. Let the side of the square be S.

The perimeter is 4S and the area is S^2

As the area is 60 more than the perimeter

S^2 - 4S = 60

=> S^2 - 4S - 60 = 0

=> S^2 - 10S + 6S - 60 = 0

=> S(S - 10) + 6(S - 10) = 0

=> (S + 6)(S - 10) = 0

S = -6 and S = 10

As length is positive we eliminate S = -6

The side of the square is 10