Question

If \( \triangle ABC \) is an equilateral triangle such that \( AD \perp BC \), then \( AD^2 \) is equal to:

• A. \( 3 \, DC^2 \)
• B. \( 2 \, DC^2 \)
• C. \( \frac{3}{2} \, DC^2 \)
• D. \( 4 \, DC^2 \)

Hint:

In a 30-60-90 triangle, the ratio of the sides is \( 1 : \sqrt{3} : 2 \). The altitude can be found using this ratio.

Answer 1

The Correct Option is C

Let \( \triangle ABC \) be an equilateral triangle, and let \( AD \) be the perpendicular from \( A \) to \( BC \). In an equilateral triangle, the altitude \( AD \) divides the triangle into two congruent 30-60-90 right triangles.
Step 1: Relationship between the sides.
In a 30-60-90 triangle, the sides are in the ratio: \[ 1 : \sqrt{3} : 2 \]
where:
- The side opposite the 30° angle is half the hypotenuse (this is \( DC \)), - The side opposite the 60° angle is the altitude \( AD \). Thus, \( AD = \frac{\sqrt{3}}{2} \times BC \).
Step 2: Use Pythagoras' Theorem.
We can relate the side \( BC \) with \( DC \) (half of \( BC \)) and use the properties of the equilateral triangle. \[ AD^2 = \frac{3}{4} BC^2 = \frac{3}{4} \times 4 DC^2 = 3 DC^2 \]
Step 3: Conclusion.
Therefore, \( AD^2 = \frac{3}{2} DC^2 \).