SOLUTION: there is a similar question on here but it doesnt show how to solve it.
The perimeter of a rectangle yard is 270 feet. If its length is 25 feet greater than its width what are th
Algebra.Com
Question 235396: there is a similar question on here but it doesnt show how to solve it.
The perimeter of a rectangle yard is 270 feet. If its length is 25 feet greater than its width what are the dimensions of the yard?
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
L=W+25
2L+2W=270
2(W+25)+2W=270
2W+50+2W=270
4W=270-50
4W=220
W=220/4
W=55 THE WIDTH.
L=55+25=80 THE LENGTH.
PROOF:
2*80+2^55=270
160+110=270
270=270
RELATED QUESTIONS
Here is the question: "A rectangle has an area of 25 square feet. A similar rectangle has (answered by LinnW)
a rectangle yard measures 50 ft. long and 40 ft wide, it contains a rectangular swimming... (answered by fractalier)
the perimeter of a rectangular yard is 270 feet. if its length is 25 feet greater than... (answered by CharlesG2)
I would greatly appreciate some assistance in trying to solve a problem. I cannot draw... (answered by josgarithmetic)
i am not sure if this is the right category but here it goes. I am in grade 9 math and i (answered by ankor@dixie-net.com)
If the rectangle's width is tripled and its length is doubled, the perimeter of the new... (answered by Photonjohn)
this is part of my summer math work and i dont know how to solve it ...
{{{6x - 18y =... (answered by homeworkhelpanytime,jim_thompson5910)
I've been working on this problem for 5 days and can't figure it out. This is the... (answered by Fombitz,solver91311)
A problem on my math homework really gave me a hard time trying to understand it because... (answered by josgarithmetic,stanbon)