SOLUTION: hello can you please help me solve," rectangular garden has a perimeter 66 ft and area 216 ftsquared. how can i find the dimensions of the garden." I know that P=2L +2W and Area= L

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: hello can you please help me solve," rectangular garden has a perimeter 66 ft and area 216 ftsquared. how can i find the dimensions of the garden." I know that P=2L +2W and Area= L      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 661498: hello can you please help me solve," rectangular garden has a perimeter 66 ft and area 216 ftsquared. how can i find the dimensions of the garden." I know that P=2L +2W and Area= LxW can i do L times W = 216 ft squared and is L=2w so is it 2w times w = 216 AND THEN I AM STUCK...please help!
Found 3 solutions by ReadingBoosters, stanbon, josmiceli:
Answer by ReadingBoosters(3246) About Me  (Show Source):
You can put this solution on YOUR website!
p = 2l+2w=66 or l = (66-2w)/2
a = lw = 216
using area
[(66 - 2w)/2](w) = 216
Multiply both sides by 2
(66-2w)(w) = 216(2)
66w - 2w^2 = 432
-2w^2 + 66w - 432 = 0
-2(w^2 - 33w + 216)=0
divide both sides by -2
w^2 - 33w + 216 = 0
(w - )(w - ) mixed signs due to + 216, but - 33
Look for multiple of 216 that add to make 33: 24,9
(w-24)(w- 9)
w = 24, w = 9
width or length is 24ft
width or length is 9ft

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
rectangular garden has a perimeter 66 ft and area 216 ftsquared
Equations:
L + W = 33 ft
L*W = 216 ft^2
------
Substitute for L and solve for "W:
(33-W)*W = 216
-------
-W^2 + 33W - 216 = 0
-----
W^2 - 33W + 216 = 0
---
Factor:
(W-9)(W-24) = 0
------
If W = 9 ft, L = 24 ft
If W = 24 ft, L = 9 ft
==========================
Cheers,
Stan H.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+P+=+2L+%2B+2W+
+66+=+2L+%2B+2W+
(1) +33+=+L+%2B+W+ ft
and
+A+=+216+ ft2
(2) +L%2AW+=+216+
------------------
(1) +W+=+33+-+L+
Substitute (1) into (2)
(2) +L%2A%28+33+-+L+%29+=+216+
(2) +33L+-+L%5E2+=+216+
(2) +-L%5E2+%2B+33L+-+216+
Solve using the quadratic equation
+L+=+%28-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29+
+a+=+-1+
+b+=+33+
+c+=+-216+
+L+=+%28-33+%2B-+sqrt%28+33%5E2+-+4%2A%28-1%29%2A%28-216%29+%29%29+%2F+%282%2A%28-1%29%29+
+L+=+%28-33+%2B-+sqrt%28+1089+-+864+%29%29+%2F+%28-2%29+
+L+=+%28-33+%2B-+sqrt%28+225+%29%29+%2F+%28-2%29+
+L+=+%28-33+%2B+15%29+%2F+%28-2%29+
+L+=+%28-18%29+%2F+%28-2%29+
+L+=+9+
and
+L+=+%28-33+-+15%29+%2F+%28-2%29+
+L+=+%28-48%29+%2F+%28-2%29+
+L+=+24+
Note that +9%2A24+=+216+
I'll say +L+=+24+, and
(1) +33+=+L+%2B+W+
(1) +33+=+24+%2B+W+
(1) +W+=+9+
The width is 9' and the length is 24'