Let a point A lie between the parallel lines L_1$L_{1}$ and L_2$L_{2}$ such that its distances from L_1$L_{1}$ and L_2$L_{2}$ 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 L_1$L_{1}$ and L_2$L_{2}$ respectively, is:

### Related Formula textArea of Equilateral Triangle = fracsqrt34 a^2$$\text{Area of Equilateral Triangle} = \frac{\sqrt{3}}{4} a^2$$ Where a$a$ is the side length. ### Core Logic Let the side of the equilateral triangle be a$a$. Let theta$\theta$ be the angle between the side AC$AC$ and the parallel line L_2$L_2$. Then, the angle between the side AB$AB$ and the parallel line L_1$L_1$ can be expressed via alternate geometry. Given distances from A$A$ to the lines form right-angled triangles. ### Step 1: Set up geometric projections

From vertex C$C$ to line passing through A$A$ parallel to L_1, L_2$L_1, L_2$, the perpendicular distance is 3$3$. In the right triangle formed, we have: sin theta = frac3a$$\sin \theta = \frac{3}{a}$$ Similarly, point B$B$ lies on L_1$L_1$. The perpendicular distance from A$A$ to L_1$L_1$ is 6$6$. However, combining the overall heights between the parallel lines, the total distance between L_1$L_1$ and L_2$L_2$ is 6 + 3 = 9$6 + 3 = 9$. The projection of side BC$BC$ (which connects L_1$L_1$ and L_2$L_2$) vertically is 9$9$. sin(60^circ + theta) = frac9a$$\sin(60^{\circ} + \theta) = \frac{9}{a}$$ ### Step 2: Solve the trigonometric system Expand sin(60^circ + theta)$\sin(60^{\circ} + \theta)$: fracsqrt32 cos theta + frac12 sin theta = frac9a$$\frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta = \frac{9}{a}$$ Substitute sin theta = frac3a$\sin \theta = \frac{3}{a}$ and cos theta = sqrt1 - sin^2 theta = sqrt1 - frac9a^2$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{9}{a^2}}$: fracsqrt32 sqrt1 - frac9a^2 + frac12 left(frac3aright) = frac9a$$\frac{\sqrt{3}}{2} \sqrt{1 - \frac{9}{a^2}} + \frac{1}{2} \left(\frac{3}{a}\right) = \frac{9}{a}$$ sqrt3 sqrt1 - frac9a^2 + frac3a = frac18a$$\sqrt{3} \sqrt{1 - \frac{9}{a^2}} + \frac{3}{a} = \frac{18}{a}$$ sqrt3 sqrt1 - frac9a^2 = frac15a$$\sqrt{3} \sqrt{1 - \frac{9}{a^2}} = \frac{15}{a}$$ Squaring both sides: 3 left(1 - frac9a^2right) = frac225a^2$$3 \left(1 - \frac{9}{a^2}\right) = \frac{225}{a^2}$$ 3 - frac27a^2 = frac225a^2$$3 - \frac{27}{a^2} = \frac{225}{a^2}$$ 3 = frac252a^2 Rightarrow a^2 = 84$$3 = \frac{252}{a^2} \Rightarrow a^2 = 84$$ ### Step 3: Calculate Area textArea of Delta ABC = fracsqrt34 a^2$$\text{Area of } \Delta ABC = \frac{\sqrt{3}}{4} a^2$$ = fracsqrt34 times 84 = 21sqrt3$$= \frac{\sqrt{3}}{4} \times 84 = 21\sqrt{3}$$ ### Pattern Recognition When a rigid polygon (like an equilateral triangle or square) is wedged between parallel lines, set a base orientation angle theta$\theta$ for one edge and use rotational shifts (e.g., 60^circ + theta$60^{\circ} + \theta$) to project heights. Expanding the sine addition formula instantly yields the side length. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines Class 11 Maths: Trigonometric Functions