There is a point P inside an equilateral `DeltaABC` of side a whose distances from vertices A, B and C are 3, 4 and 5, respectively. Rotate the triangle and P through `60^(@)` about C. Let A go to A' and P to P'. Then the area of `DeltaPAP'` (in sq. units) is

Options

A. 8
B. 12
C. 16
D. 6

Explanation

To solve the problem, we need to find the area of triangle \( \Delta PAP' \) after rotating point \( P \) and vertex \( A \) through \( 60^\circ \) about point \( C \).

### Step-by-Step Solution:

1. **Identify the Given Distances**:
- The distances from point \( P \) to vertices \( A \), \( B \), and \( C \) are given as:
- \( AP = 3 \)
- \( BP = 4 \)
- \( CP = 5 \)

2. **Understand the Rotation**:
- When we rotate the triangle \( ABC \) and point \( P \) through \( 60^\circ \) about point \( C \), the following transformations occur:
- Vertex \( A \) moves to \( A' \).
- Point \( P \) moves to \( P' \).

3. **Calculate Distances After Rotation**:
- After rotation, the distances remain the same:
- \( AP' = AP = 3 \)
- \( BP' = BP = 4 \)
- \( CP' = CP = 5 \)

4. **Determine the Angle**:
- The angle \( \angle CPP' \) is \( 60^\circ \) due to the rotation.
- Since \( CP = CP' \) and \( AP = AP' \), we can analyze triangle \( CPP' \).

5. **Use the Law of Cosines**:
- In triangle \( CPP' \), we can find \( PP' \) using the Law of Cosines:
\[ PP' = \sqrt{CP^2 + CP'^2 - 2 \cdot CP \cdot CP' \cdot \cos(60^\circ)} \]
- Substituting the values:
\[ PP' = \sqrt{5^2 + 5^2 - 2 \cdot 5 \cdot 5 \cdot \frac{1}{2}} = \sqrt{25 + 25 - 25} = \sqrt{25} = 5 \]

6. **Calculate the Area of Triangle \( PAP' \)**:
- The area of triangle \( PAP' \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times AP \times AP' \times \sin(\angle PAP') \]
- Since \( \angle PAP' = 90^\circ \) (as derived from the properties of the triangle), we have:
\[ \text{Area} = \frac{1}{2} \times 3 \times 4 \times 1 = \frac{1}{2} \times 12 = 6 \]

7. **Final Answer**:
- The area of triangle \( PAP' \) is \( 6 \) square units.