Question 1174393: Please help me with this word problem:
The length of a rectangular lot is 14 meters less than twice the lot’s width. The owner decided to decrease
the length and the width by 3 meters each for landscaping purposes. The perimeter of the smaller
rectangular lot is 80 meters. What is the area of the ORIGINAL lot in square meters?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let L = the length
let W = the width.
the perimeter of the lot is equal to 2 * (L + W)
the area of the lot is equal to L * W
you are given that the length of a rectangular lot is 14 meters less than twice the lot’s width.
the equation for that is:
L = 2W - 14
The owner decided to decrease the length and the width by 3 meters each for landscaping purposes. The perimeter of the smaller rectangular lot is 80 meters.
since perimeter = 2 * (L - 3 + W - 3), then your equation for this is:
80 = 2 * (L - 3 + W - 3)
combine like terms in this to get:
80 = 2 * (L + W - 6)
simplify further to get:
80 = 2L + 2W - 12
you were given that L = 2W - 14, therefore replace L with 2W - 14 in the equation of 80 = 2L + 2W - 12 to get:
80 = 2 * (2W - 14) + 2W - 12
simplify to get:
80 = 4W - 28 + 2W - 12
combine like terms to get:
80 = 6W - 40
add 40 to both sides of this equation to get:
120 = 6W
solve for W to get:
W = 120 / 6 = 20
since L = 2W - 14, then L = 40 - 14 = 26
you have L = 26 and W = 20
taking 3 off of each, you have:
L - 3 = 23
W - 3 = 17
the formula for the perimeter becomes 80 = 2 * (23 + 17) = 46 + 34 = 80.
this confirms the values for L and W are correct.
you want to know the area of the original lot.
that would be area = L * W = 26 * 20 = 520 square meters.
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