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
- \(\sqrt{6}\)
- \(\sqrt{5}\)
- \(\sqrt{8}\)
- \(\sqrt{7}\)
The Correct Option is D
Solution and Explanation
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.
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