A line PQ is drawn parallel to the base BC of `Delta ABC ` which meets sides AB and AC at points P and Q respectively. If `AP = 1/3 PB`, find the value of :
`("Area of " Delta ABC)/("Area of "Delta APQ)`

Answer

To find the ratio of the area of triangle ABC to the area of triangle APQ, we can follow these steps:

### Step 1: Understand the Given Information
We have triangle ABC with a line PQ drawn parallel to the base BC. The line intersects sides AB and AC at points P and Q, respectively. We know that \( AP = \frac{1}{3} PB \).

### Step 2: Express AB in Terms of AP
Since \( AP = \frac{1}{3} PB \), we can express PB in terms of AP:
- Let \( AP = x \).
- Then \( PB = 3x \) (because \( AP \) is one-third of \( PB \)).
- Therefore, the total length of \( AB \) is:
\[ AB = AP + PB = x + 3x = 4x \]

### Step 3: Use the Properties of Similar Triangles
Since line PQ is parallel to BC, triangles APQ and ABC are similar by the AA (Angle-Angle) criterion. The corresponding sides of similar triangles are in proportion.

### Step 4: Set Up the Ratio of the Sides
The ratio of the sides of triangle ABC to triangle APQ can be expressed as:
\[ \frac{AB}{AP} = \frac{4x}{x} = 4 \]

### Step 5: Find the Ratio of the Areas
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore:
\[ \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle APQ} = \left(\frac{AB}{AP}\right)^2 = 4^2 = 16 \]

### Final Answer
Thus, the value of \(\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle APQ}\) is:
\[ \boxed{16} \]

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