High School

Find the square of the side length of equilateral triangle ABC, where:

- Point B is on line L1.
- Point C is on line L2.
- Point A is between the lines, one unit away from L1.

The distance between the two parallel lines L1 and L2 is 5.

Answer :

To solve this problem, let's break it down step-by-step:

1. Understand the problem:
- We have two parallel lines, L1 and L2, separated by a distance of 5 units.
- Points B and C lie on lines L1 and L2, respectively.
- Point A is located between these lines and is 1 unit away from line L1.

2. Find position of point A:
- Since A is 1 unit away from L1, the distance from A to L2 becomes:
- Total distance between L1 and L2 = 5 units.
- Distance from line L1 to point A = 1 unit.
- Thus, distance from point A to L2 = [tex]\(5 - 1 = 4\)[/tex] units.

3. In equilateral triangle ABC:
- Since A, B, and C form the vertices of an equilateral triangle, all sides are equal in length.
- Therefore, the distance from A to B is equal to the distance from A to C.
- We established that distance from A to C (across the space to L2) is 4 units.

4. Calculate the side length:
- Therefore, the side length of the equilateral triangle ABC is 4 units.

5. Find the square of the side length:
- The square of the side length is calculated as:
- Side length = 4 units.
- Side length squared = [tex]\(4^2 = 16\)[/tex].

Thus, the square of the side length of the equilateral triangle ABC is 16.

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