SOLUTION: The length of a side of one square is 3 inches greater than the length of a side of a second square. If the area of the first square exceeds the area of the second by 51 squares i
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Question 31069: The length of a side of one square is 3 inches greater than the length of a side of a second square. If the area of the first square exceeds the area of the second by 51 squares inches, find the lengths of the sides of each square. Answer by mukhopadhyay(490) (Show Source):
You can put this solution on YOUR website! Assume that the length of smaller square is x inch
The length of the larger square is then (x+3)
Area of smaller Square = x^2
Area of larger Square=(x+3)^2
Based on the question,
(x+3)^2 = x^2 + 51
=>x^2 + 6x + 9 = x^2 + 51
=>6x = 42
=>x = 7
Length of the sides of the smaller square is 7 inch and the length of the side of the larger square is 10 inch.
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