The problem describes a parallelogram $ABCD$ with the following details:

1. The perimeter of the parallelogram is $60\sqrt{3}$.
2. The ratio of the lengths of the sides is $2:3$.
3. A perpendicular altitude $BK$ is dropped from $B$ to the longer side.
4. The angle between the adjacent sides is $120^\circ$.
5. We need to calculate the area of the quadrilateral $BCDK$.

Step-by-Step Solution

Step 1: Determine the side lengths of the parallelogram

The perimeter of a parallelogram is given by: $2(a + b) = 60\sqrt{3},$ where $a$ and $b$ are the side lengths. Simplifying: $a + b = 30\sqrt{3}.$ Since the ratio of the side lengths is $2:3$, let: $a = 2x, \quad b = 3x.$ Substitute into $a + b = 30\sqrt{3}$: $2x + 3x = 30\sqrt{3},$ $5x = 30\sqrt{3}.$ Solve for $x$: $x = 6\sqrt{3}.$ Thus: $a = 2x = 12\sqrt{3}, \quad b = 3x = 18\sqrt{3}.$

Step 2: Find the height ($h$)

The area of the parallelogram can also be expressed as: $\text{Area} = b \cdot h,$ where $b$ is the longer side ($18\sqrt{3}$) and $h$ is the altitude ($BK$). The parallelogram has an angle of $120^\circ$ between its adjacent sides. Using the trigonometric formula for the area of a parallelogram: $\text{Area} = a \cdot b \cdot \sin(120^\circ).$ The sine of $120^\circ$ is $\sin(120^\circ) = \frac{\sqrt{3}}{2}$, so: $\text{Area} = (12\sqrt{3})(18\sqrt{3})\left(\frac{\sqrt{3}}{2}\right).$ Simplify: $\text{Area} = 12 \cdot 18 \cdot 3 \cdot \frac{\sqrt{3}}{2} = 324\sqrt{3}.$ Using $\text{Area} = b \cdot h$, we solve for $h$: $h = \frac{\text{Area}}{b} = \frac{324\sqrt{3}}{18\sqrt{3}} = 18.$

Thus, the height $BK = h = 18$.

Step 3: Calculate the area of quadrilateral $BCDK$

The quadrilateral $BCDK$ is a trapezoid, with one parallel side $CD = b = 18\sqrt{3}$, the other side $BK = h = 18$, and the height $BK$ between them.

The area of a trapezoid is: $\text{Area} = \frac{1}{2} \cdot (b_1 + b_2) \cdot h,$ where $b_1 = BK = 18$, $b_2 = CD = 18\sqrt{3}$, and $h = BK = 18$. Substituting: $\text{Area} = \frac{1}{2} \cdot (18 + 18\sqrt{3}) \cdot 18.$ Simplify: $\text{Area} = 9 \cdot (18 + 18\sqrt{3}),$ $\text{Area} = 162 + 162\sqrt{3}.$

Thus, the area of the quadrilateral $BCDK$ is: $\boxed{162 + 162\sqrt{3}}.$

Do you want further clarification or a breakdown of any specific part?

Relative Questions:

1. How is the area of a parallelogram related to its height and base?
2. What is the significance of the $120^\circ$ angle in the calculation?
3. How do trigonometric functions simplify calculations in geometry?
4. What happens to the trapezoid's area if the angle changes?
5. Can the quadrilateral $BCDK$ be decomposed into simpler shapes?

Tip:

Always confirm whether the dimensions and trigonometric relationships are consistent before applying formulas in geometry problems.