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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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:
Solution
$\sin \theta = \frac{3}{a}$
$\sin (60^\circ + \theta) = \frac{9}{a}$
$\frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta = \frac{9}{a}$
$\sqrt{3}\sqrt{1 - \frac{9}{a^2}} + \frac{3}{a} = \frac{18}{a}$
$a = \sqrt{84}$
Area of $\triangle ABC = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \cdot 84 = 21\sqrt{3}$
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