If \( L_1 \) and \( L_2 \) are two parallel lines and \( \triangle ABC \) is an equilateral triangle as shown in the figure, then the area of triangle \( ABC \) is:
Show Hint
For equilateral triangles:
\begin{itemize}
\item Height \( = \frac{\sqrt{3}}{2}a \)
\item Area \( = \frac{\sqrt{3}}{4}a^2 \)
\item Centroid divides the median in the ratio \( 2:1 \)
\end{itemize}
To find the area of the equilateral triangle \( \triangle ABC \), let's use the height of the triangle provided in the diagram.
The height of an equilateral triangle can be expressed using the formula:
\( h = \frac{\sqrt{3}}{2} \times a \)
where \( a \) is the side length of the equilateral triangle.
From the diagram, the total height \( AC \) is calculated as:
\( AC = 6 + 3 = 9 \)
We equate this to the formula for the height:
\( \frac{\sqrt{3}}{2} \times a = 9 \)
Solve for \( a \):
\( a = \frac{2 \times 9}{\sqrt{3}} = \frac{18}{\sqrt{3}} \)
Rationalizing the denominator:
\( a = \frac{18 \times \sqrt{3}}{3} = 6\sqrt{3} \)
The area \( A \) of an equilateral triangle is given by the formula:
\( A = \frac{\sqrt{3}}{4} \times a^2 \)
Substitute the value of \( a \):
\( A = \frac{\sqrt{3}}{4} \times (6\sqrt{3})^2 \)
\( A = \frac{\sqrt{3}}{4} \times 108 \)
\( A = 27\sqrt{3} \)
Since there's an inconsistency in the question's given correct answer \( 21\sqrt{3} \), let's verify again:
The correct calculation confirms the area as:
\( A = 27\sqrt{3} \)
Therefore, based on calculations, the correct area of \( \triangle ABC \) should be revised.
