SOLUTION: Please help me with this 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 met
Algebra.Com
Question 1174392: Please help me with this 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 ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The length of a rectangular lot is 14 meters less than twice the lot’s width.
w = the original width
L = (2w - 14); the original length
:
The owner decided to decrease the length and the width by 3 meters each for landscaping purposes.
(w-3) = the new width
and
(2w-14) - 3 = (2w-17), the new length
:
The perimeter of the smaller rectangular lot is 80 meters.
2(2w-17) + 2(w-3) = 80
simplify, divide by 2
2w - 17 + w - 3 = 40
3w - 20 = 40
3w = 40 + 20
3w = 60
w = 60/3
w = 20 meter, the original width
and
2(20) - 14 = 26 meters the original length
:
What is the area of the original lot in square meters?
26 * 20 = 520 sq/meters
:
:
check by finding the new perimeter 23 by 17
2(17) + 2(23) =
34 + 46 = 80
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