In `DeltaABC` line PQ is drawn parallel to side BC where P and Q are respectively lie on side AB and AC. If AB = 3AP, what is the ratio of area of `DeltaAPQ` to area of `DeltaABC` ?

Options

A. a)`1:3`
B. b)`1:5`
C. c)`1:7`
D. d)`1:9`

Answer

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

### Step 1: Understand the Given Information
We are given that line \( PQ \) is drawn parallel to side \( BC \) of triangle \( ABC \), and points \( P \) and \( Q \) lie on sides \( AB \) and \( AC \) respectively. We also know that \( AB = 3AP \).

### Step 2: Establish the Ratios of the Sides
Since \( PQ \) is parallel to \( BC \), triangles \( APQ \) and \( ABC \) are similar by the Basic Proportionality Theorem (or Thales's theorem).

From the given information, we can express the lengths:
- Let \( AP = x \).
- Then \( AB = AP + PB = x + 3x = 4x \) (since \( AB = 3AP \)).

Thus, the ratio of the lengths \( AP \) to \( AB \) is:
\[ \frac{AP}{AB} = \frac{x}{4x} = \frac{1}{4} \]

### Step 3: Use the Property of Similar Triangles
The areas of similar triangles are proportional to the square of the ratio of their corresponding sides. Therefore, the ratio of the areas of triangles \( APQ \) and \( ABC \) can be expressed as:
\[ \frac{Area(\Delta APQ)}{Area(\Delta ABC)} = \left(\frac{AP}{AB}\right)^2 \]

### Step 4: Substitute the Ratio
Substituting the ratio we found:
\[ \frac{Area(\Delta APQ)}{Area(\Delta ABC)} = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]

### Step 5: Conclusion
Thus, the ratio of the area of triangle \( \Delta APQ \) to the area of triangle \( \Delta ABC \) is:
\[ \frac{Area(\Delta APQ)}{Area(\Delta ABC)} = \frac{1}{16} \]

### Final Answer
The ratio of the area of \( \Delta APQ \) to the area of \( \Delta ABC \) is \( 1:16 \).

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