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## Isoperimetric Geometry

The isoperimetric theorem states that: “Among all shapes with an equal area, the circle will be characterized by the smallest perimeter” which is equivalent to “Among all shapes with equal perimeter, the circle will be characterized by the largest area.” The theorem’s name derives from three Greek words: ‘isos’ meaning ‘same’, ‘peri’ meaning ‘around’ and ‘metron’ meaning ‘measure’. A perimeter (= ‘peri’ + ‘metron’) is the arc length along the boundary of a closed two-dimensional region (= a planar shape). So, the theorem deals with shapes that have equal perimeters.

**Grade Level:** High School (9-12)

**Curriculum Topic Benchmarks:**
M1.4.8, M2.4.9, M5.4.2, M5.4.10, M5.4.11

**Subject Keywords:** Geometry, Shape, Area, Perimeter

**Author(s):** Jan Bogaert

**PUMAS ID:** 01_22_03_1

**Date Received:** 2003-01-22

**Date Revised:** 2003-03-24

**Date Accepted:** 2003-03-26

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### Comments

**Comment by ***Ismail Kocayusufoglu* on May 24, 2004

"There is an easy way proving the main part of this example. Here it is.

Theorem:

Let Area(Rectangle) = Area(Circle) (Denote A(R)=A(C) for convenience). Show that Perimeter(Circle) < Perimeter(Rectangle). (Denote as P(C)

Proof. A(R)=A(C)

ab=ƒÎr2

r= ã[(ab)/ƒÎ]

Need to show that P(C)< 2(a+b) .

To show:

2ƒÎr<2(a+b) ƒÎr<(a+b)

ƒÎã[(ab)/ƒÎ] < (a+b)

ƒÎab<(a+b)2

So, it is enough to show that ƒÎab<(a+b)2.

Lemma : Given a,b, a>b, then ƒÎab<(a+b).

Proof of Lemma : Since (a-b)>0:

(a-b)2 = a2+b2-2ab > 0

a2+b2 > 2ab

a2+b2 > (ƒÎ-2)ab ;

since 2>(ƒÎ-2)

a2+b2 > ƒÎab-2ab

a2+b2+2ab > ƒÎab

(a+b) 2 > ƒÎab

as claimed.

This proves the theorem."

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