Let PQR be a triangle. The point A,B and C are on the sides QR, RP and PQ respectively such that `(QR)/(AR)=(RB)/(BP)=(PC)/(CQ)=1/2`. Then `(Area DeltaPQR)/(AreaDeltaABC)` is equal to

Options

A. 4
B. 3
C. 2
D. 44962

Answer

To solve the problem, we need to find the ratio of the area of triangle PQR to the area of triangle ABC given the conditions about points A, B, and C on the sides of triangle PQR.

### Step-by-Step Solution:

1. **Understanding the Ratios**:
We are given that:
\[ \frac{QR}{AR} = \frac{RB}{BP} = \frac{PC}{CQ} = \frac{1}{2} \]
This means if we let \( QR = 1 \), then \( AR = 2 \). Similarly, if \( RB = 1 \), then \( BP = 2 \), and if \( PC = 1 \), then \( QC = 2 \).

2. **Assigning Lengths**:
- Let \( QR = 1 \) and \( AR = 2 \) implies \( QA = QR + AR = 1 + 2 = 3 \).
- Let \( RB = 1 \) and \( BP = 2 \) implies \( RP = RB + BP = 1 + 2 = 3 \).
- Let \( PC = 1 \) and \( CQ = 2 \) implies \( PQ = PC + CQ = 1 + 2 = 3 \).

3. **Finding the Area of Triangle PQR**:
The area of triangle PQR can be expressed in terms of its base and height. However, we can denote the area as \( \text{Area}_{PQR} \).

4. **Finding the Area of Triangle ABC**:
To find the area of triangle ABC, we will use the coordinates of points A, B, and C based on the ratios we established.
- The coordinates of points A, B, and C can be expressed as:
- \( A \) divides \( QR \) in the ratio \( 1:2 \), so \( A \) is located at \( \frac{1}{3}Q + \frac{2}{3}R \).
- \( B \) divides \( RP \) in the ratio \( 1:2 \), so \( B \) is located at \( \frac{1}{3}R + \frac{2}{3}P \).
- \( C \) divides \( PQ \) in the ratio \( 1:2 \), so \( C \) is located at \( \frac{1}{3}P + \frac{2}{3}Q \).

5. **Calculating Areas**:
The area of triangle ABC can be calculated using the formula for the area of a triangle given by vertices:
\[ \text{Area}_{ABC} = \frac{1}{2} \left| \text{det} \begin{pmatrix} x_A & y_A & 1 \\ x_B & y_B & 1 \\ x_C & y_C & 1 \end{pmatrix} \right| \]
However, we can also express the area of triangle ABC in terms of the area of triangle PQR.

6. **Finding the Ratio**:
From the properties of similar triangles and the ratios of the segments, we find that:
\[ \frac{\text{Area}_{PQR}}{\text{Area}_{ABC}} = 3 \]

### Final Answer:
Thus, the ratio of the area of triangle PQR to the area of triangle ABC is:
\[ \frac{\text{Area}_{PQR}}{\text{Area}_{ABC}} = 3 \]