Question 350758: The length of a rectangle is five times its width.
If the area of the rectangle is 405ft, find its perimeter.
Found 2 solutions by stanbon, haileytucki:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The length of a rectangle is five times its width.
If the area of the rectangle is 405ft, find its perimeter.
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L = 5W
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Equation:
Area = LW
405 = (5W)W
81= W^2
W = 9
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L = 5W = 45
======================
Perimeter = 2(L+W)
P = 2(45+9)
P = 2(54)
P = 108
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Cheers,
Stan H.
Answer by haileytucki(390)  (Show Source):
You can put this solution on YOUR website! The length of a rectangle is five times its width.
If the area of the rectangle is 405ft, find its perimeter.
l = 2w
We know the area is 405 so,
lw = w(w) = 2w^2 = 405
5w^(2)=405
Divide each term in the equation by 5.
(5w^(2))/(5)=(405)/(5)
Simplify the left-hand side of the equation by canceling the common factors.
w^(2)=(405)/(5)
Now, The perimeter is the sum of the lengths of all four sides.
2L + 2W=
18 +90= 108 perimeter
Simplify the right-hand side of the equation by simplifying each term.
w^(2)=81
Take the square root of both sides of the equation to eliminate the exponent on the left-hand side.
w=\~(81)
Pull all perfect square roots out from under the radical. In this case, remove the 9 because it is a perfect square.
w=\9
First, substitute in the + portion of the \ to find the first solution.
w=9
So, simply,
9 ft is the width
and 45 is the length (45 is the length because it is 5 times the width, which is 9)
Proof:
45*9
Multiply 45 by 9 to get 405.
405
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