geometry

First and foremost, I would draw a diagram. This is a good start with any geometry problem. I have constructed a mock version of a sketch and presented it below. This way, both of us can visualize the problem together.

To be honest, I either never learned the formula for the area of a rhombus, or I forgot the formula. Either way, I will have to resort to some prior knowledge that I have in order to figure out the area of this rhombus. By definition, rhombuses are special parallelograms, so the area formula for a parallelogram will hold true for a rhombus. 

\(A_{parallelogram}=bh\) where \(b \) is the length of one of the bases and \(h\) is the perpendicular height of the parallelogram. I have labeled these lengths in the diagram above as well. 

\(b=AB=1\) as the length of any side of a rhombus is 1, as given by the original problem.

Finding \(h\) requires a little bit more work, but not too much either. \(\triangle EDA\) is a right triangle where \(DA=1\) because that is a length of a side length of a rhombus. \(m\angle EAB = 120^\circ\), which is, once again, given information. It is also possible to find the measures of the angles of the triangles, which will also reveal the length of the side of \(\triangle EDA\)

\(m\angle EAD + m\angle DAB = m\angle EAB\\ m\angle EAD + 90^\circ= 120^\circ\\ m\angle EAD= 30^\circ\)

\(m\angle DEA + m\angle EAD + m\angle ADE = 180^\circ \\ m\angle DEA + 30^\circ + 90^\circ = 180^\circ \\ m\angle DEA = 60^\circ\)

Now, we have determined that \(\triangle EDA\) is a 30-60-90 triangle, which also means that its side lengths have a consistent ratio of \(1:\sqrt{3}:2\). Let's set up a proportion then to find the length of the perpendicular height of the triangle.

\(\frac{\sqrt{3}}{2}=\frac{h}{1}\\ h=\frac{\sqrt{3}}{2}\)

Now that we have both b and h, we can now solve for the area of this mystery rhombus.

\(A_{\text{rhombus}}=bh\\ A_{\text{rhombus}}=1*\frac{\sqrt{3}}{2}\\ A_{\text{rhombus}}=\frac{\sqrt{3}}{2}\)

I hope this explanation helped you!