SOLUTION: If the length of one side of a square is tripled and the length of an adjacent side is increased by 10, the resulting rectangle has an area that is 6 times the area of the original
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Question 726749: If the length of one side of a square is tripled and the length of an adjacent side is increased by 10, the resulting rectangle has an area that is 6 times the area of the original square. Find the length of a side of the original square. Answer by DrBeeee(684) (Show Source):
You can put this solution on YOUR website! Let s = the length of the side of the square.
Then the rectangle is 3*s by s+10 giving the area
(1) (3s)*(s+10) = 6s^2 or
(2) 3s^2 + 30s = 6s^2 or
(3) 3s^2 - 30s = 0 or
(4) 3s*(s - 10) = 0 which has two solutions
(5) s = 0,10
Obviously we choose
(6) s = 10
Let's check this.
Is (3*10*(10+10) = 6*10^2)?
Is (30*20 = 6*100)?
Is (600 = 600)? Yes
Answer: The length of the side of the original square is 10 units.
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