Question 268929: If I know the perimeter of a rectangle is 34 and the area is 60 how do I find the sides? Found 2 solutions by CharlesG2, unlockmath:Answer by CharlesG2(834) (Show Source):
You can put this solution on YOUR website! If I know the perimeter of a rectangle is 34 and the area is 60 how do I find the sides?
perimeter is 2 * (length + width) or 2L + 2W
area is length * width or L*W
2L + 2W = 34 (divide both sides by 2)
L + W = 17 (you can solve this for L to have L in terms of W, or solve for W to have W in terms of L)
L * W = 60 (so now we know the sum is 17 and the product is 60)
should subsitute in to the product equation (L*W=60) either L in terms of W and solve for W then solve for L or W in terms of L and solve for L then solve for W
L * W = 60
L * (17 - L) = 60 (substituted in)
17L - L^2 = 60 (rearrange)
-L^2 + 17L - 60 = 0 (divide both sides by -1)
L^2 - 17L + 60 = 0 (solve for L)
(L-12)(L-5) = 0 (foil: L^2 - 5L - 12L + 60)
L=5 or L=12
since L+W=17 then W=12 or W=5
L*W = 12 * 5 = 60
2 * (L+W) = 2 * 17 = 34
You can put this solution on YOUR website! Hello,
We can set up two equations and solve it. Let x be one side and y be the other side as follows:
2x+2y=34
x(y)=60 Rewrite this as: x=60/y then plug this into the first equation as:
2(60/y)+2y=34 Now multiply by y to get:
120+2y^2=34y Subtract 34y from both sides to get:
2y^2-34y+120=0 Rewrite it as:
2(y^2-17y+60)=0 factor this to get:
2(y-12)(y-5)=0 Solve for y:
y=12
y=5
Now solve for x by plugging in 12 for y:
2x+2(12)=34
x=5
Plug in 5 to get:
x=12
So we see one side is 12 and the other side is 5.
There we go!
RJ
There is a book I wrote that might help:
www.math-unlock.com
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