In rhombus DEFG, if m<DEF=60, and the shorter diagonal DF has a length of 7, find the length of the side of a rhombus.

In a rhombus, the diagonals right-bisect each other.

I recognize the 30-60-90 triangle in the diagram.
then
x/2 = 3.5/1
x = 7
(sure enough, triangle DEF is equilateral)

So one side of the rhombus is equal to seven? I'm still confused as to how you got your answer...

Ok, I will use a different approach this time.

Hope you make a sketch.
Since one angle is 60º, then its opposite angle is also 60º, properties of a parallelogram.
Which means the other angle pair is 120º each, (angle D = angle F = 120)
Now look at your diagram, don't you see two equilateral triangles,
triange DEF and also triangle DGF.
So all sides in those two triangles must be equal,
since DF = 7 , the sides of the rhomubus are 7 each.

I see it now,because a diagonal divides a parallelogram into two congruent triangles! Thank you!!

To find the length of the side of a rhombus, we need to use the properties of a rhombus and the given information.

A rhombus is a quadrilateral with all four sides of equal length. It also has two pairs of opposite congruent angles.

Given that angle DEF is 60 degrees, we know that angle DEG (the adjacent angle) is also 60 degrees since opposite angles in a rhombus are congruent.

Since the diagonals of a rhombus intersect at a right angle, we can use triangle DEF to find the length of side EF. In triangle DEF, we have an angle of 60 degrees and the length of the shorter diagonal DF, which is 7 units.

To find the length of side EF, we can use the trigonometric relationship of sine. The sine of an angle is equal to the ratio of the length of the side opposite the angle to the length of the hypotenuse.

In this case, the side opposite angle DEF is EF, and the hypotenuse is DF. Using the equation sin(60) = EF/7, we can solve for EF.

sin(60) = EF/7
√3/2 = EF/7
EF = 7 * √3/2 = 7√3/2

So, the length of the side of rhombus DEFG is 7√3/2 units.