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:

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![image.png](https://res.cloudinary.com/aptidude/image/upload/v1743536845/aptidude_questions/hkcysoolyismjjzqjkgr.png)

1. **Key Information and Setup:**  
   - The triangle is equilateral with side length $$3$$.
   - The area of triangle $$ADC$$ is half the area of triangle $$ABD$$.  
   - Drop a perpendicular from $$A$$ to $$BC$$, meeting at point $$E$$. This divides $$BC$$ into two equal parts: $$BE = EC = 1.5$$.

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2. **Area Ratio and Segment Division:**  
   Since the area of $$	riangle ADC = \frac{1}{2} 	imes 	ext{Area of } 	riangle ABD$$, the base $$DC$$ must be half of $$BD$$.  
   - Let $$BD = 2x$$ and $$DC = x$$.  
   - Since $$BD + DC = BC = 3$$:  
     $$
     BD = 2 \, 	ext{cm}, \, DC = 1 \, 	ext{cm}.
     $$

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3. **Calculate Altitude ($$AE$$):**  
   The altitude ($$AE$$) of an equilateral triangle is given by:  
   $$
   AE = \frac{\sqrt{3}}{2} 	imes 	ext{side length}.
   $$  
   Substituting side length $$3$$:  
   $$
   AE = \frac{\sqrt{3}}{2} 	imes 3 = \frac{3\sqrt{3}}{2}.
   $$

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4. **Find Length of $$ED$$:**  
   From the division, $$EC = 1.5$$, and since $$DC = 1$$:  
   $$
   ED = EC - DC = 1.5 - 1 = 0.5.
   $$

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5. **Use Pythagoras Theorem in Triangle $$AED$$:**  
   Triangle $$AED$$ is a right-angled triangle with:  
   - Hypotenuse: $$AD$$,  
   - Perpendicular: $$AE = \frac{3\sqrt{3}}{2}$$,  
   - Base: $$ED = 0.5$$.

Using Pythagoras theorem:  
   $$
   AD^2 = AE^2 + ED^2.
   $$

Substitute values:  
   $$
   AD^2 = \left(\frac{3\sqrt{3}}{2}ight)^2 + (0.5)^2.
   $$

Simplify:  
   $$
   AD^2 = \frac{27}{4} + \frac{1}{4} = \frac{28}{4} = 7.
   $$

Thus:  
$$
AD = \sqrt{7}.
$$

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### Final Answer:  
The length of $$AD$$ is **$$\sqrt{7}$$**, and the correct answer is **D)**.

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