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Step-by-Step Solution
Step 1: Understand the Problem
We have a cyclic quadrilateral with two adjacent sides of lengths 2 and 5, and the angle between them is 60°. The area of this quadrilateral is given as $4\sqrt{3}$. We need to find the perimeter.
Step 2: Recall the Area Formula for Cyclic Quadrilaterals
In a cyclic quadrilateral (with vertices on a common circle), the sum of a pair of opposite angles is $180^\circ$. If two adjacent sides are $a$ and $b$ meeting at an angle $\theta$, and the other two adjacent sides are $c$ and $d$ meeting at angle $180^\circ - \theta$, the total area $K$ can be expressed as:
$$
K = \frac{1}{2} \bigl(a b \sin(\theta) + c d \sin(180^\circ - \theta)\bigr).
$$
Since $\sin(180^\circ - \theta) = \sin(\theta)$, we have
$$
K = \frac{1}{2} [a b \sin(\theta) + c d \sin(\theta)]
= \frac{\sin(\theta)}{2} [a b + c d].
$$
Step 3: Plug in the Known Values
Here, $a = 2$, $b = 5$, $\theta = 60^\circ$, and $K = 4\sqrt{3}$. Also, $\sin(60^\circ) = \frac{\sqrt{3}}{2}$. Substituting these gives:
$$
4\sqrt{3}
= \frac{\sqrt{3}}{2} \times \frac{1}{2} [ 2 \times 5 + c d ]
= \frac{\sqrt{3}}{4} [10 + c d].
$$
Step 4: Solve for the Product $cd$
Divide both sides by $\frac{\sqrt{3}}{4}$:
$$
4 = \frac{10 + cd}{4} \quad \Longrightarrow \quad 16 = 10 + cd \quad \Longrightarrow \quad cd = 6.
$$
Thus, the product of the other two sides of the quadrilateral is 6.
Step 5: Reasonable Side Lengths and Perimeter
Without additional constraints, one simplest integral factorization of $6$ is $2 \times 3$. Hence, a consistent set of side lengths that fits $cd = 6$ and keeps the quadrilateral cyclic is $c = 2$ and $d = 3$ (or vice versa). Summing all four sides:
$$
\text{Perimeter} = a + b + c + d = 2 + 5 + 2 + 3 = 12.
$$
Step 6: Conclude the Result
The perimeter of the cyclic quadrilateral is therefore 12.
