prove-that-among-all-closed-figures-having-same-perimeter-circle-has-maximum-area-

Question Number 3417 by RasheedSindhi last updated on 13/Dec/15

p r o v e t h a t a m o n g a l l c l o s e d f i g u r e s

h a v i n g s a m e p e r i m e t e r c i r c l e

h a s m a x i m u m a r e a .

Commented by Filup last updated on 13/Dec/15

I dont understand the question

Commented by RasheedSindhi last updated on 13/Dec/15

C o n s i d e r c l o s e d f i g u r e s t r i a n g l e,

q u a d i l a t e r a l, p e n t a g o n \dots o r o t h e r

c l o s e d c u r v e s o r m i x t u r e o f c u r v e s

a n d s t r a i g h t l i n e s e g m e n t s, a l l

h a v i n g s a m e p e r i m e t e r . A m o n g

a l l t h e a r e a s t h e s e f i g u r e s h a v e

t h e l a r g e s t i s t h e a r e a o f c i r c l e .

Commented by 123456 last updated on 13/Dec/15

this is isoperemtric inequality

4 π A ⩽ L^{2}

Commented by Filup last updated on 13/Dec/15

Circle is essentially an \infty - s i d e d

p o l y g o n

Commented by Filup last updated on 13/Dec/15

3 - gon (triangle) < 4 - gon (square)

4 - gon < 5 - gon

etc

∴ n - g o n < \infty - g o n (c i r c l e)

Commented by Rasheed Soomro last updated on 13/Dec/15

W h a t i s m e a n t b y isoperemtric inequality ?

Commented by RasheedSindhi last updated on 13/Dec/15

T h \propto n k S

Commented by prakash jain last updated on 13/Dec/15

isoperimetric inequality for all closed

plane figure

4 π A ⩽ L^{2}

A area

L perimeter

equality holding true only for circle .