All the side lengths of a rhombus are 6, and one of the angles is 120 degrees. Find the area of the rhombus.

Step-by-step explanation:

Let the rhombus be $ABCD$, where $\angle DAB = 120^\circ$. Then $\angle ABC = 180^\circ - \angle DAB = 180^\circ - 120^\circ = 60^\circ$.

[asy] unitsize(1 cm); pair A, B, C, D; A = (0,1); B = (sqrt(3),0); C = (0,-1); D = (-sqrt(3),0); draw(A--B--C--D--cycle); draw(A--C); label("$A$", A, N); label("$B$", B, E); label("$C$", C, S); label("$D$", D, W); label("$6$", (A + D)/2, NW); [/asy]

Since $AB = BC$, triangle $ABC$ is equilateral. By the same argument, triangle $ACD$ is also equilateral. Each triangle has area

\[\frac{\sqrt{3}}{4} \cdot 6^2 = 9 \sqrt{3},\]so the area of the rhombus is $2 \cdot 9 \sqrt{3} = \boxed{18 \sqrt{3}}$.

idk

step-by-step explanation:

payback