Question

consider that △abc is an equilateral triangle, and (overline{ad}) is a perpendicular bisector of △abc. if (ab = 2x), complete the statements below. (square^2 + (ad)^2 = (2x)^2) (square^2 = square x^2 - x^2) ((ad)^2 = square x^2) (ad = square sqrt{3}) (options for dropdowns: x, 2x, 30, 60)

Explanation:

Step1: Analyze the equilateral triangle

In an equilateral triangle \( \triangle ABC \), \( AD \) is the perpendicular bisector, so \( BD = DC \) and \( \triangle ABD \) is a right triangle with \( AB = 2x \), \( BD = x \) (since \( BC = AB = 2x \) and \( D \) bisects \( BC \)).

Step2: Apply the Pythagorean theorem

For right triangle \( \triangle ABD \), by Pythagoras: \( (BD)^2 + (AD)^2 = (AB)^2 \). Substituting \( BD = x \), \( AB = 2x \), we get \( x^2 + (AD)^2 = (2x)^2 \).

Step3: Solve for \( (AD)^2 \)

\( (AD)^2=(2x)^2 - x^2 = 4x^2 - x^2 = 3x^2 \).

Step4: Solve for \( AD \)

Taking square root: \( AD = \sqrt{3x^2}=x\sqrt{3} \).

Answer:

1. First blank (in \( \square^2 + (AD)^2=(2x)^2 \)): \( x \)
2. Second blank (in \( \square x^2 - x^2 \)): \( 4 \) (since \( (2x)^2 = 4x^2 \))
3. Third blank (in \( (AD)^2=\square x^2 \)): \( 3 \)
4. Fourth blank (in \( AD = \square\sqrt{3} \)): \( x \)