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Argh

Guest

Area of a Rhombus

I was asked a question today and I cant figure it out. A square with side A has two equilateral triangles inscribed within it also with side A. The triangles cross to form a rhombus. What is the area of the Rhombus in terms of A?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Area of a Rhombus

Hi;

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Bob

Administrator

Registered: 2010-06-20

Posts: 10,643

Re: Area of a Rhombus

hi Argh

Welcome to the forum.

I cannot make a diagram with this information.  I made a square and called one side A . Then I made one equilateral triangle inside the square with one side A.  I cannot make another.  Did you mean that the second triangle has side B where B is opposite to A ?

EDIT:  Ah ha!  Just read the second post.  Ok here we go:

#Using trig. you can work out EJ and GI and as you know IJ you can work out EG.

Angle HGE = 30 so you can work out HF. 

The rhombus is 4 right angled triangles so its area is half EG x HF.

Bob

Last edited by Bob (2015-06-23 20:24:14)

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phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,895

Re: Area of a Rhombus

Hi;

Deducting the area of the square from the sum of the areas of the four triangles that together fill the square gives the area of the only overlapped region: the rhombus.

The pair of isosceles triangles is congruent (as is the pair of equilateral triangles), and, using the area formulas for these triangles, I got the following in terms of A:

Last edited by phrontister (2017-02-27 00:34:43)

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