Question

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

• A. \(\sqrt{6}\)
• B. \(\sqrt{5}\)
• C. \(\sqrt{8}\)
• D. \(\sqrt{7}\)

Answer 1

The Correct Option is D

To solve this problem, we need to understand the properties of an equilateral triangle: 

• All sides are equal
• Each interior angle is \(60^\circ\)

Given: Triangle \(ABC\) is equilateral with side length 3 cm, and point \(D\) lies on \(BC\). We are to find the length of \(AD\) such that:

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

Let’s assume coordinates to simplify:

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

Let point \(D\) divide \(BC\) such that \(BD = x\) and \(DC = 3 - x\).

Area of triangle \(ABC\) is calculated as:

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

We are given that: \[ \text{Area}_{ADC} = \frac{1}{2} \cdot \text{Area}_{ABD} \]

Using geometry and area relationships, we derive:

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

So, simplifying: \[ x = \frac{AD}{2} \]

Using coordinates and distance formula: \[ AD = \sqrt{x^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \]

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

Solving these equations leads to: \[ AD = \sqrt{7} \]

Therefore, the length of AD is \( \sqrt{7} \) cm.