Homework Help

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

user profile pic

jspencerca | Student, Undergraduate | eNotes Newbie

Posted September 21, 2010 at 12:19 AM via web

dislike 1 like

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

3 Answers | Add Yours

user profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted September 21, 2010 at 12:28 AM (Answer #1)

dislike 0 like

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

user profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted September 21, 2010 at 12:33 AM (Answer #2)

dislike 0 like

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

user profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted September 21, 2010 at 1:25 AM (Answer #3)

dislike 0 like

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.

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes