SOLUTION: Not quite sure where to put this problem but please help me solve.
The perimeter of the den is 88ft. If the area is 480 square feet then what are the dimensions of the rectangul
Algebra.Com
Question 83471: Not quite sure where to put this problem but please help me solve.
The perimeter of the den is 88ft. If the area is 480 square feet then what are the dimensions of the rectangular room?
Answer by dolly(163) (Show Source): You can put this solution on YOUR website!
Let the length be x ft and the width be y ft.
The perimeter is 88ft
==> 2(l + w) = 88
==> l + w = 88/2
==> l + w = 44
==> w = 44 - l -------------------(1)
Given area = 480
==> l x w = 480
==> l (44-l) = 480
==> 44l - l^2 = 480
==> - l^2 + 44l - 480 = 0
==> l^2 - 44l + 480 = 0
==> l^2 - 20l - 24l + 480 = 0
==> l(l-20) - 24(l - 20) = 0
==> (l - 20) ( l- 24) = 0
==> l-20 = 0 or l - 24 = 0
==> l = 20 or l = 24
==> w = 24 0r w = 20
Thus the dimensions of the retangular room are 20 ft and 24 ft.
RELATED QUESTIONS
Please help me solve the equation:
sq root of 28x^2y
I am not sure how to type the... (answered by MathLover1)
Sorry i didnt know which category to place this word problem in but anyways here's the... (answered by scott8148)
Please help me solve this. I do not even know where to start!! :(
The area of a... (answered by jim_thompson5910)
Please Help, I'm not quite sure if I answered this right or not but the question is,... (answered by Earlsdon,MathLover1)
problem, a garden area is 30ft long and 20ft wide. a path of uniform width is set around... (answered by ankor@dixie-net.com)
I need help solving this problem. I'm not quite sure that I understand how to do it:... (answered by lwsshak3)
I'm not sure where to put this question. If I should be posting it in under another topic (answered by richard1234)
Hi i was looking at the Ken's yard problem. Ken's area of his rectangular yard is 480... (answered by Rajendrakar)
I am not quite sure the name of this kind of problem because I am still new at this, but... (answered by Fombitz)