Question:medium

The length of each side of an equilateral triangle ABC is 3 cm. Let D be a point on BC such that the area of triangle ADC is half the area of triangle ABD. Then the length of AD,in cm,is

Updated On: Jan 15, 2026
  • \(\sqrt{6}\)
  • \(\sqrt{5}\)
  • \(\sqrt{8}\)
  • \(\sqrt{7}\)
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The Correct Option is D

Solution and Explanation

To resolve this, we must first acknowledge the characteristics of an equilateral triangle:

  • All sides exhibit equal length.
  • Each internal angle measures \(60^\circ\).

Provided information: Triangle \(ABC\) is equilateral with a side length of 3 cm. Point \(D\) is situated on \(BC\). The objective is to determine the length of \(AD\) under the condition that:

The area of triangle \(ADC\) is precisely half the area of triangle \(ABD\).

For analytical ease, we assign coordinates:

  • \(A = (0, \sqrt{3})\)
  • \(B = (-1.5, 0)\)
  • \(C = (1.5, 0)\)

Let point \(D\) segment \(BC\) such that the length of \(BD\) is \(x\) and the length of \(DC\) is \(3 - x\).

The area of triangle \(ABC\) is computed as:

\[ \text{Area}_{ABC} = \frac{\sqrt{3}}{4} \cdot (3)^2 = \frac{9\sqrt{3}}{4} \]

The given condition is: \[ \text{Area}_{ADC} = \frac{1}{2} \cdot \text{Area}_{ABD} \]

Applying geometric principles and area relationships, we deduce:

\[ \frac{1}{2} \cdot x \cdot h = \frac{1}{4} \cdot AD \cdot 2h \]

Simplifying this equation yields: \[ x = \frac{AD}{2} \]

Utilizing coordinates and the distance formula, we find: \[ AD = \sqrt{x^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \]

Furthermore, derived from triangle geometry: \[ AD = \sqrt{3^2 - x^2} \]

The simultaneous solution of these equations results in: \[ AD = \sqrt{7} \]

Consequently, the length of AD is established as \( \sqrt{7} \) cm.

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