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Given two parallelograms $P1$ and $P2$ sharing a diagonal, such that area of $P1$ is greater than area of $P2$, can we say that the perimeter of $P1$ is greater than the perimeter of $P2$ ?

Actually I was trying to prove another question : Proof of a geometric statement

I thought of using this argument as shown in the following picture . Here $1$ represents parallelogram with sides $AB$ and $AC$ and $2$ represents area of parallelogram with sides $BD$ and $CD$.My attempt :

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  • $\begingroup$ I'm only looking at parallelograms where one is inside the other and they share a doagonal. Can you give an answer keeping the diagram in mind ? I need this only as a way to prove another statement. $\endgroup$ – Srinivas K Mar 1 '15 at 7:15
  • $\begingroup$ Ok, in that case, I will have to think. Can we use trigonometry? $\endgroup$ – Sawarnik Mar 1 '15 at 8:59
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Try the following two parallelograms:

(-2,0),(0,1.01),(2,0),(0,-1.01)

(-2,0),(-2,1),(2,0),(2,-1)

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Since I can not comment - Your question boils down to a simpler claim: The perimeter of a triangle created by connecting a point in a scalene and acute triangle to any two vertices, is smaller than the original triangle perimeter.

I tested this with Geogebra - and it seems correct. The comment made by Sawarnik is not relevant to your claim - which is true but require proof.

I provided a proof on your first post, that the parallelogram that contains the smaller one has a larger perimeter.

You may accept the answers on both posts. Thanks.

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