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Let a point A lie between the parallel lines L 1 and L ... | null (Mathematics) | DivineJEE

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Mathematics

JEE Mains 2026 21st January Shift 1

Mathematics

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 :

A

15 \sqrt{6}

B

27

C

21 \sqrt{3}

D

12 \sqrt{2}

Explanation

Question Image

\sin \theta=\frac{3}{\mathrm{a}}\\

\sin \left(60^{\circ}+\theta\right)=\frac{9}{\mathrm{a}}\\

\frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta=\frac{9}{\mathrm{a}}\\

\sqrt{3} \sqrt{1-\frac{9}{\mathrm{a}^{2}}}+\frac{3}{\mathrm{a}}=\frac{18}{\mathrm{a}}\\

\mathrm{a}=\sqrt{84}\\

Area of

\triangle \mathrm{ABC}=\frac{\sqrt{3}}{4} \mathrm{a}^{2} \\=\frac{\sqrt{3}}{4} \times 84=21 \sqrt{3}

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