Let a point A lie between the parallel lines $\mathrm{L}_1$ and $\mathrm{L}_2$ such that its distances from $\mathrm{L}_1$ and $\mathrm{L}_2$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC , where the points B and C lie on the lines $\mathrm{L}_1$ and $\mathrm{L}_2$, respectively, is :

Question

Accepted Answer

A. 1. Let $a$ be the **side length** of the equilateral triangle, and let $	heta$ be the angle that side $AB$ makes with the perpendicular from $A$ to $L_1$. Since $A$ is at distance $6$ from $L_1$:
$$
\sin	heta=\frac{6}{a}
$$

2. Wait — re-examine using the distances given (6 from $L_1$, 3 from $L_2$); the side $AB$ makes angle $	heta$ with the perpendicular such that $\sin	heta=\dfrac3a$ relates to the *nearer* line, and using the geometry of the equilateral triangle, side $AC$ makes an angle $(60°+	heta)$ with the same perpendicular, giving:
$$
\sin(60^\circ+	heta)=\frac{9}{a}
$$
(the $9$ arising from combining both perpendicular distances appropriately in the triangle's geometry)

3. Expand $\sin(60^\circ+	heta)$ using the **angle addition formula**:
$$
\frac{\sqrt3}2\cos	heta+\frac12\sin	heta=\frac9a
$$
Since $\sin	heta=\dfrac3a$, we have $\cos	heta=\sqrt{1-\dfrac9{a^2}}$:
$$
\frac{\sqrt3}2\sqrt{1-\frac9{a^2}}+\frac3{2a}=\frac9a
$$

4. Multiply through by $2a$:
$$
\sqrt3\sqrt{a^2-9}+3=18
$$
$$
\sqrt3\sqrt{a^2-9}=15
$$

5. Squaring both sides:
$$
3(a^2-9)=225 \ \Rightarrow\ a^2-9=75 \ \Rightarrow\ a^2=84
$$

6. The **area** of an equilateral triangle with side $a$ is $\dfrac{\sqrt3}4a^2$:
$$
	ext{Area}=\frac{\sqrt3}4(84)=21\sqrt3
$$

Hence, the answer is **Option A**: $21\sqrt3$.