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

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It is given that the perimeter of the rectangle is 7 times its width. Let the sides of the rectangle be denoted by W and L.

Perimeter = 2(W + L) = 7W

2W + 2L = 7W

=> 2L = 5W

=> L = (5/2)*W

The area of the rectangle is given as 40

W*L = 40

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

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

=> W^2 = 40*2/5

=> W^2 = 16

=> W = 4

We don’t consider the negative root as it is meaningless here.

So the width is 4, the length is (5/2)*4 = 10.

**The sides of the rectangle are 4 and 10.**

If p is the perimeter, A the area, l and w the sides of a rectangle , then,

P = 2(l+w).

A = lw.

Given that p = 7w and A = 40.

Therefore , the perimmeter equation could be written as:

7w = 2(l+w)...(1) and area equation as: 40 = lw...(2).

From (1) , we get 7w = 2l+2w. Or 2l = 7-2w = 5w. So l = 5w/2.

We put l = 5w/2 in eq (2):

40 = (5w/2)w = 5w^2/2.

Multiply both sides by 2.

80 = 5w^2.

Divide both sides by 5:

16 = w^2.

Take square root:

4 = w.

Therefore from (2) 40 lw, Or 40 =l*4. So we get l= 40/4 = 10.

Therefore the length and width of the given rectangle are 10 and 4 .

We'll calculate the area of a rectangle as the product of the lengths and the width.

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 reject the negative solution because the value of a side cannot be negative.

The length is y = 40/x units

y = 40/4

**length: y = 10 units**