When is the area of a rectangle equal to the perimeter?

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Suppose the sides of the rectangle are l and b.

Then the area A of threctangle A = lb.

Perimeter p = 2(l+b).

When A = p, lb = 2(l+b)

lb = 2l+2b

lb-2l = 2b

l (b-2) = 2b

Or

b = 2l/(l-2)

l = 2b/(b-2) is the condition when area of the rectangle = perimeter.

Example

l= 12. then b = 2l/((l-2) = 2*12/(12-2) = 24/10 = 2.4

verification:

p = 2(12+2.40 = 28.8

A= lb = 12*2.4 = 28.8

Area of the rectangle is the product of the length and the width.

A = l*w

On the other hand, the perimeter of the rectangle is:

P = 2(l+w)

Now, we'll put the area and the perimeter in the relation of equality:

l*w = 2(l+w)

Now, we'll form the second degree equation, when knowing the product and the sum of the length and width.

x^2 - Sx + P = 0

We'll use Viete's relations:

l + w = S

l*w = P

But, l*w = 2(l+w)

P = 2S

x^2 - Sx + 2S = 0

delta = S^2 - 8S

S^2 - 8S = 0

S(S-8) = 0

S = 0 impossible

S = 8

l+w = 8 => l = 8-w

l*w = 16

(8-w)*w - 16 = 0

w^2 - 8w + 16 = 0

w1 = [8+sqrt(64-64)]/2

w = 4

l = 4

For a rectangle let the length be equal to L and the width be equal to W. The perimeter of the rectangle is the sum of the lengths of all the sides which is 2*( L+W). Area of a rectangle is L*W.

If the area of a rectangle is equal to the perimeter it means

2*(L+W)= L*W

- This does not have a unique solution.

For example take L as 10

2*(10+W)= 10W

=> 20 + 2W = 10W

=> 20 = 8W

W= 2.5

Similarly let L= 18

2*(18+W)= 18W

=>36+ 2W = 18W

=> 36 = 16W

=> W= 2.25

**As it is not specified that the length and width have to be whole numbers there can be innumerable such rectangles with the area equal to the perimeter.**

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