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 APQ)/("Area of trapezium"PBCQ)`

Answer

To solve the problem, we need to find the ratio of the area of triangle APQ to the area of trapezium PBCQ. Given that AP = 1/3 PB, we can follow these steps:

### Step 1: Understand the relationship between AP and PB
Given that \( AP = \frac{1}{3} PB \), we can express PB in terms of AP:
\[ PB = 3 \cdot AP \]

### Step 2: Express the lengths in terms of a single variable
Let \( AP = x \). Then:
\[ PB = 3x \]
The total length of AB can be expressed as:
\[ AB = AP + PB = x + 3x = 4x \]

### Step 3: Use the properties of similar triangles
Since line PQ is parallel to the base BC of triangle ABC, triangles APQ and ABC are similar by the AA (Angle-Angle) similarity criterion. Therefore, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

### Step 4: Find the ratio of the sides
The ratio of the sides of triangle APQ to triangle ABC is:
\[ \frac{AP}{AB} = \frac{x}{4x} = \frac{1}{4} \]

### Step 5: Find the ratio of the areas
Using the property of similar triangles, the ratio of the areas is:
\[ \frac{\text{Area of } \triangle APQ}{\text{Area of } \triangle ABC} = \left(\frac{AP}{AB}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]

### Step 6: Express the area of trapezium PBCQ
The area of trapezium PBCQ can be found by subtracting the area of triangle APQ from the area of triangle ABC:
\[ \text{Area of trapezium PBCQ} = \text{Area of } \triangle ABC - \text{Area of } \triangle APQ \]

Let the area of triangle ABC be \( A \). Then:
\[ \text{Area of } \triangle APQ = \frac{1}{16} A \]
Thus, the area of trapezium PBCQ is:
\[ \text{Area of trapezium PBCQ} = A - \frac{1}{16} A = \frac{16}{16} A - \frac{1}{16} A = \frac{15}{16} A \]

### Step 7: Find the ratio of the areas
Now we can find the required ratio:
\[ \frac{\text{Area of } \triangle APQ}{\text{Area of trapezium PBCQ}} = \frac{\frac{1}{16} A}{\frac{15}{16} A} = \frac{1}{15} \]

### Final Answer
The value of \( \frac{\text{Area of } \triangle APQ}{\text{Area of trapezium PBCQ}} \) is:
\[ \frac{1}{15} \]