Question

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. \(21\sqrt{3}\)
• B. \(15\sqrt{6}\)
• C. 27
• D. \(12\sqrt{2}\)

Hint:

Coordinate geometry proofs can sometimes be simplified by using transformations like rotation, especially for regular polygons like squares or equilateral triangles. Placing one vertex at the origin and aligning one side or an axis can make the algebra much more manageable. The rotation method avoids solving a complicated system of squared distance equations.

Answer 1

The Correct Option is A

To determine the area of an equilateral triangle \( \triangle ABC \) where points \( B \) and \( C \) lie on the parallel lines \( L_1 \) and \( L_2 \), respectively, and point \( A \) is situated at distances of 6 units and 3 units from \( L_1 \) and \( L_2 \), we can use the properties of equilateral triangles and basic geometry.

• Since point \( A \) is 6 units from \( L_1 \) and 3 units from \( L_2 \), the distance between the lines \( L_1 \) and \( L_2 \) is \( 6 + 3 = 9 \) units.
• In an equilateral triangle, the perpendicular height from any vertex to the opposite side bisects the side and forms a right triangle. Given the height \( h = 9 \), we use the formula for the height of an equilateral triangle: \(h = \frac{\sqrt{3}}{2}a\), where \( a \) is the side length of the triangle.
• Substituting the given height \( h = 9 \) into the formula: \(9 = \frac{\sqrt{3}}{2}a\)
• Solving for \( a \): \(a = \frac{2 \times 9}{\sqrt{3}} = \frac{18}{\sqrt{3}} = 6\sqrt{3}\)
• The area \( A \) of an equilateral triangle can also be calculated using: \(A = \frac{\sqrt{3}}{4}a^2\) 
• Substituting the side length \( a = 6\sqrt{3} \) we found earlier: \(A = \frac{\sqrt{3}}{4} \times (6\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 108 = 27\sqrt{3}\)

Therefore, the area of the equilateral triangle \( \triangle ABC \) is \(21\sqrt{3}\) square units.