Geometry Question (1 Viewer)

DJel

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Hello,

I'm drawing a blank on this question and some help would be appreciated. The question is;

The area of a rhombus is sqrt(3) cm^2, one of its angles is 120 degrees, find its side length.

Thanks for an help,

DJel.
 

Mark576

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Let A,B,C,D be the vertices of the rhombus, named clockwise so that ∠ABC = ∠CDA = 120°.

Construct diagonal AC:

First we want to find an expression for the area of triangle ABC;

A = (1/2)*BC*AB*sin ∠ABC

As we are dealing with a rhombus, AB=BC=CD=DA. Let the side length be x.

∴ A = (1/2)*x2*sin 120 = (1/2)*x2*(√3)/2, as sin 120 = sin 60 = √3/2.

This simplifies to: (x2√3)/4

We are given the area of the rhombus, √3;

This is equal to twice the area of triangle ABC;

∴ (x2√3)/2 = √3

Solving this we end up with x=√2, as a negative answer is incorrect.
 
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DJel

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Mark576 said:
Let A,B,C,D be the vertices of the rhombus, named clockwise so that *ABC = *CDA = 120°.

Construct diagonal AC:

First we want to find an expression for the area of triangle ABC;

A = (1/2)*BC*AB*sin *ABC

As we are dealing with a rhombus, AB=BC=CD=DA. Let the side length be x.

∴ A = (1/2)*x2*sin 120 = (1/2)*x2*(√3)/2, as sin 120 = sin 60 = √3/2.

This simplifies to: (x2√3)/4

We are given the area of the rhombus, √3;

This is equal to twice the area of triangle ABC;

∴ (x2√3)/2 = √3

Solving this we end up with x=√2, as a negative answer is incorrect.
Thanks for that, really appreciated.

DJel.
 

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