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Mathegocart

Member

Registered: 2012-04-29

Posts: 2,226

The Perplexing Dimensions of The Rhombus.

I have been trying to solve and develop a comprehensive solution, but I haven't. I would profoundly appreciate the help from the jovial people of MIF.
HCEG is a rhombus formed by connecting the midpoints of the rectangle ABDF. Note that O is the center of the circle circumscribing ABDF, and that it is the intersection of the diagonals of the rhombus HCEG. Some dimensions: OG = 10, CK = 8.
Find a length of the side of the rhombus HCEG.

My diagram is currently quite convoluted and looks more unsystematic than planned out.

I did some EM work( was intuitive) and found the solution, though can't find a geometric explanation

concluded after a quick peruse of the problem/

Last edited by Mathegocart (2017-01-16 00:58:26)

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bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: The Perplexing Dimensions of The Rhombus.

Hi;

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,640

Re: The Perplexing Dimensions of The Rhombus.

hi Mathegocart

HE is a line of symmetry for the rectangle, the rhombus and the circle.  So proceed like this:

Find the radius OL = OF.

The diagonals of a rhombus bisect at right angles, so EOG = 90 and also OE // GF and  OG // EF.  (// = parallel)

=> Triangles GOF and FGE and congruent, so GE = OF = radius.

Bob

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