# Rectangles.The perimeter of a rectangle is 7 times its width. What are its sides if the area is 40 ?

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Let the length and width of the rectangle be L and W.

The perimeter of a rectangle is 7 times its width.

=> 2L + 2W = 7W

=> 2L = 5W

=> L = (5/2)W

Area = 40

=> L*W = 40

=> (5/2)W*W = 40

=> W^2 = 16

=> W = 4

L = 10

**The length of the rectangle is 10 and width is 4**

We'll apply the most common formula of the area of a rectangle:

A = x*y (1), where x is the width and y is the length.

P = 7x

But the perimeter of a rectangle is:

P = 2(x+y)

7x = 2x + 2y

We'll subtract 2x both sides:

5x = 2y

y = 5x/2 (2)

The value of the area is 40:

40 = x*y (3)

We'll substitute (2) in (3):

40 = x*5x/2

80 = 5x^2

We'll divide by 5:

x^2 = 16

width: x = 4 units

We'll keep only the positive solution since the value of a side cannot be negative.

The length is y = 40/x units

y = 40/4

length: y = 10 units