In `(DeltaABC)` , `MN||BC` and `AM:AB=1/3`. Then find the ratio of `(ar(DeltaAMN))/(ar(DeltaABC))`

Options

A. `1:4`
B. `1:9`
C. `4:1`
D. `9:1`

Answer

To solve the problem, we need to find the ratio of the areas of triangles AMN and ABC given that MN is parallel to BC and that \( AM:AB = \frac{1}{3} \).

### Step-by-Step Solution:

1. **Identify the Similar Triangles**:
Since \( MN \) is parallel to \( BC \), triangles \( AMN \) and \( ABC \) are similar by the Basic Proportionality Theorem (also known as Thales' theorem). This means that the angles of triangle AMN are equal to the corresponding angles of triangle ABC.

**Hint**: Remember that parallel lines create corresponding angles that are equal.

2. **Set Up the Ratio of Sides**:
From the problem, we know that \( AM:AB = \frac{1}{3} \). This means that if we let \( AM = x \), then \( AB = 3x \).

**Hint**: Use the given ratio to express the lengths in terms of a variable.

3. **Use the Property of Similar Triangles**:
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore, we can write:
\[ \frac{ar(\Delta AMN)}{ar(\Delta ABC)} = \left(\frac{AM}{AB}\right)^2 \]

**Hint**: Recall that the area ratio of similar triangles is the square of the ratio of their corresponding sides.

4. **Substitute the Known Values**:
We know from the problem that \( \frac{AM}{AB} = \frac{1}{3} \). Substituting this into our area ratio gives:
\[ \frac{ar(\Delta AMN)}{ar(\Delta ABC)} = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \]

**Hint**: Make sure to square the ratio when calculating the area ratio.

5. **Conclusion**:
Therefore, the ratio of the area of triangle AMN to the area of triangle ABC is:
\[ \frac{ar(\Delta AMN)}{ar(\Delta ABC)} = \frac{1}{9} \]

### Final Answer:
The ratio of \( ar(\Delta AMN) \) to \( ar(\Delta ABC) \) is \( \frac{1}{9} \).