JEE Mains · Maths · STD 11 - 9. straight line
Let a point A lie between the parallel lines \( L_{1} \) and \( L_{2} \) such that its distances from \( L_{1} \) and \( L_{2} \) 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 \( L_{1} \) and \( L_{2} \) respectively, is:
- A \( 15\sqrt{6} \)
- B 27
- C \( 21\sqrt{3} \)
- D \( 12\sqrt{2} \)
Answer & Solution
Correct Answer
(C) \( 21\sqrt{3} \)
Step-by-step Solution
Detailed explanation
\( \sin~\theta=\frac{3}{a} \) \( \sin(60^{\circ}+\theta)=\frac{9}{a} \) \( \frac{\sqrt{3}}{2}\cos~\theta+\frac{1}{2}\sin~\theta=\frac{9}{a} \) \( \sqrt{3}\sqrt{1-\frac{9}{a^{2}}}+\frac{3}{a}=\frac{18}{a} \) \( a=\sqrt{84} \) Area of…
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