If L₁ and L₂ are two parallel lines and ΔABC is an equilateral triangle, then the area of triangle ABC is

The correct answer is 21√3.

Step-by-Step Explanation:
Let us analyze the details from the provided image:
1. We are given two parallel lines, L₁ (top line) and L₂ (bottom line).
2. Vertex A of the equilateral triangle ΔABC lies on line L₁, and vertex B lies on line L₂.
3. The vertical distance from line L₁ (and thus from vertex A) to vertex C is labeled as 6.
4. The vertical distance from vertex C to line L₂ is labeled as 3.
5. Therefore, the total vertical distance between line L₁ and line L₂ is:
$6+3=9$

Let $a$ be the side length of the equilateral triangle ΔABC.
Let $\theta$ be the angle that the side AB makes with the horizontal line L₂.
Since A lies on L₁ and B lies on L₂, the vertical height of A relative to B is 9. Thus, we have:
$a\sin \theta =9\Rightarrow \sin \theta =\frac{9}{a}$

Similarly, let $\alpha$ be the angle that the side AC makes with the horizontal.
Since the vertical distance from A to C is 6, we have:
$a\sin \alpha =6\Rightarrow \sin \alpha =\frac{6}{a}$

Since ΔABC is an equilateral triangle, the angle between the sides AB and AC is 60°.
Therefore, the difference between the angles they make with the horizontal is 60°:
$\theta -\alpha =60°$

Taking the cosine of both sides:
$\cos (\theta -\alpha )=\cos (60°)=\frac{1}{2}$

Using the trigonometric identity:
$\cos (\theta -\alpha )=\cos \theta \cos \alpha +\sin \theta \sin \alpha$

We can express $\cos \theta$ and $\cos \alpha$ in terms of $a$:
$\cos \theta =\sqrt{1-{\sin }^2\theta }=\sqrt{1-\frac{81}{a^2}}=\frac{\sqrt{a^2-81}}{a}$

$\cos \alpha =\sqrt{1-{\sin }^2\alpha }=\sqrt{1-\frac{36}{a^2}}=\frac{\sqrt{a^2-36}}{a}$

Substituting these values into our identity:
$\frac{1}{2}=\frac{\sqrt{a^2-81}}{a}\frac{\sqrt{a^2-36}}{a}+\left(\frac{9}{a}\right)\left(\frac{6}{a}\right)$

$\frac{1}{2}=\frac{\sqrt{a^2-81}\sqrt{a^2-36}+54}{a^2}$

Multiplying both sides by $a^2$:
$\frac{a^2}{2}-54=\sqrt{(a^2-81)(a^2-36)}$

Squaring both sides:
${\left(\frac{a^2}{2}-54\right)}^2=(a^2-81)(a^2-36)$

$\frac{a^4}{4}-54a^2+2916=a^4-117a^2+2916$

Subtracting 2916 from both sides:
$\frac{a^4}{4}-54a^2=a^4-117a^2$

Multiplying by 4:
$a^4-216a^2=4a^4-468a^2$

Rearranging the terms:
$3a^4=252a^2$

Since $a\ne 0$, dividing both sides by $3a^2$ gives:
$a^2=84$

The area of an equilateral triangle with side length $a$ is given by:
$\text{Area}=\frac{\sqrt{3}}{4}{a}^2$

Substituting $a^2=84$ into the area formula:
$\text{Area}=\frac{\sqrt{3}}{4}\times 84=21\sqrt{3}$