### Instruction: Two adjacent sides of a rhombus form a $60$-degree angle. If each side of the rhombus measures $2$ cm, what is the area of the rhombus, in square centimeters? Express your answer in simplest radical form.### Response: I know that a rhombus is a quadrilateral with four congruent sides, and that the opposite angles of a rhombus are also congruent.
So if one angle is $60$ degrees, then the opposite angle is also $60$ degrees, and the other two angles are $120$ degrees each.
I also know that the diagonals of a rhombus are perpendicular bisectors of each other, and that they form four right triangles inside the rhombus.
So if I draw the diagonals of the rhombus, I can use the Pythagorean theorem to find their lengths, and then use the formula for the area of a rhombus, which is half the product of the diagonals.
Let me label the vertices of the rhombus as $A, B, C, D$, and the intersection of the diagonals as $E$.
Then $AB = BC = CD = DA = 2$ cm, and $\angle A = \angle C = 60^\circ$, and $\angle B = \angle D = 120^\circ$.
By the Pythagorean theorem, $AE^2 + BE^2 = AB^2$, or $AE^2 + BE^2 = 4$.
But I also notice that $\triangle ABE$ is a $30$-$60$-$90$ triangle, so I can use the special ratio of the sides to find $AE$ and $BE$.
Since $\angle A = 60^\circ$, then $AE$ is the longer leg, and $BE$ is the shorter leg.
So $AE = \sqrt{3} \cdot BE$.
Substituting this into the equation, I get $(\sqrt{3} \cdot BE)^2 + BE^2 = 4$, or $3BE^2 + BE^2 = 4$, or $4BE^2 = 4$.
Dividing both sides by $4$, I get $BE^2 = 1$, or $BE = 1$ cm.
Then $AE = \sqrt{3} \cdot BE = \sqrt{3}$ cm.
Similarly, I can find that $CE = \sqrt{3}$ cm and $DE = 1$ cm, using the same logic.
So the diagonals of the rhombus are $AE + CE = 2\sqrt{3}$ cm and $BE + DE = 2$ cm.
The area of the rhombus is half the product of the diagonals, so $A = \frac{1}{2} \cdot (2\sqrt{3})(2)$.
Simplifying, I get $A = 2\sqrt{3}$ square centimeters.