SOLUTION: A rectangle is 5cm. longeer than it is wide.If its length and width are both increased by 3 cm. its area is increased by 60 cm squared.Find the dimensions of the original rectangle
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Question 64020This question is from textbook Algebra Structure and method Book1
: A rectangle is 5cm. longeer than it is wide.If its length and width are both increased by 3 cm. its area is increased by 60 cm squared.Find the dimensions of the original rectangle. This question is from textbook Algebra Structure and method Book1
You can put this solution on YOUR website! A rectangle is 5 cm. longer than it is wide.If its length and width are both increased by 3 cm. it's area is increased by 60 cm squared.Find the dimensions of the original rectangle.
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Let x = width of the original rectangle
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It states; "rectangle is 5 cm. longer than it is wide." so we have:
(x+5) = length of the original rectangle
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Original rectangle is x by (x+5) so it's area = x(x+5) = (x^2 + 5x)
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"length and width are both increased by 3 cm."
New Rectangle:
Width = (x+3)
Length = (x+5) + 3 = (x+8),
New Rectangle area = (x+3)(x+8) = (x^2 + 11x + 24)
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"it's area is increased by 60 sq cm" which means:
New Rectangle area - Old rectangle area = 60
(x^2 + 11x + 24) - (x^ + 5x) = 60
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Remove the brackets, change the signs
x^2 + 11x + 24 - x^2 - 5x = 60
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Group like terms and do the math
x^2 - x^2 + 11x - 5x + 24 = 60
6x + 24 = 60
6x = 60 - 24
x = 36/6
x = 6 cm is the width of the original rectangle
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6 + 5 = 11 cm is the length of the original rectangle
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Check our solutions: old rectangle area 6*11, new rectangle area 9*14
126 - 66 = 60 as stated
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