Let a point A lie between the parallel lines L1 and L2 such that its distances from L1 and L2 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 L1 and L2, respectively, is :
- 21 3
- 12 2
- 15 6
- 27
Answer
(1) 21 3
Solution
[figure]
Let side of the triangle = a and let AB make angle with the lines. Then AC makes angle - 60 (measured the same way).
Distance of B from A perpendicular to the lines: a = 6 ...(i)
Distance of C: a (60 - ) = 3 ...(ii)
From (ii): a ( 32 - 12 ) = 3
32 a - 3 = 3 (using (i))
a = 12 3 = 4 3 ...(iii)
Squaring and adding (i) and (iii): a2 = 36 + 48 = 84
Area = 34a2 = 34 84 = 21 3 sq. units
Answer: (1)