Question

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:

• A. \( 7\sqrt{3} \)
• B. \( 4\sqrt{3} \)
• C. \( 21\sqrt{3} \)
• D. \( 84 \)

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}

Answer 1

The Correct Option is C

Step 1: From the figure, point \( C \) lies on the median of the equilateral triangle. The distances are: \[ AC = 6, CB = 3 \]
Step 2: In an equilateral triangle, the centroid divides the median in the ratio \( 2:1 \). Hence, total height \( h \) of the triangle is: \[ h = AC + CB = 6 + 3 = 9 \]
Step 3: Height of an equilateral triangle is given by: \[ h = \frac{\sqrt{3}}{2}a \] \[ \Rightarrow a = \frac{2h}{\sqrt{3}} = \frac{2 \times 9}{\sqrt{3}} = 6\sqrt{3} \]
Step 4: Area of an equilateral triangle: \[ \text{Area} = \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4}(6\sqrt{3})^2 = 21\sqrt{3} \]