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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)
ƒÎã[(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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