**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**

**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**

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