Question

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The length of a rectangle is four times the width. The perimeter of the rectangle is 45 inches. Write a system of equations that represents this problem. What is the area of the rectangle? show all work

Answer 1

L = 4W

P = 2W + 2L = 45 inches.

Find W and L. Then find the Area: A = W*L.

P = 2W + 2(4W) = 10W=45 inches. Thus, W = 4.5 inches.

L = 4L = 18 inches

A= area = (4.5 inches)(18 inches) = 81 square inches

Answer 2

x= length
w=width

x = 4w (length=4 times the width)
2x+2w=45 (perimeter is 45)

since we already have what x is equal to (4w), we can plug it into our second equation and solve for w.

2(4w)+2w=45
8w+2w=45
10w=45
w=4.5
4.5in

now that we have the value for w, we can plug in w in either equation and solve for x

x=4(4.5)
x=18
18in

since we know the length and the width, we multiply the two to find the area

18*4.5=81
Area = 81 inches squared

hope this helped!