Let `Q = (3,sqrt(5)),R =(7,3sqrt(5))`. A point P in the XY-plane varies in such a way that perimeter of `DeltaPQR` is 16. Then the maximum area of `DeltaPQR` is

Options

A. 6
B. 12
C. 18
D. 9

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript while ensuring clarity in each step.

### Step 1: Identify Points Q and R
We have two points:
- \( Q = (3, \sqrt{5}) \)
- \( R = (7, 3\sqrt{5}) \)

### Step 2: Calculate the Length of QR
Using the distance formula, we calculate the distance \( QR \):
\[ QR = \sqrt{(7 - 3)^2 + (3\sqrt{5} - \sqrt{5})^2} \]
Calculating the differences:
\[ = \sqrt{(4)^2 + (2\sqrt{5})^2} = \sqrt{16 + 4 \cdot 5} = \sqrt{16 + 20} = \sqrt{36} = 6 \]

### Step 3: Relate QR to the Ellipse
Since \( Q \) and \( R \) are the foci of an ellipse, the distance between the foci \( 2c \) is equal to \( QR \):
\[ 2c = 6 \implies c = 3 \]

### Step 4: Establish the Perimeter Condition
The perimeter condition states that:
\[ PQ + PR = 16 \]
We also know:
\[ PQ + PR = 2a \quad \text{(for an ellipse)} \]
Thus:
\[ 2a = 16 - QR = 16 - 6 = 10 \implies a = 5 \]

### Step 5: Calculate b Using the Relationship Between a, b, and c
We use the relationship:
\[ c^2 = a^2 - b^2 \]
Substituting the known values:
\[ 3^2 = 5^2 - b^2 \implies 9 = 25 - b^2 \implies b^2 = 16 \implies b = 4 \]

### Step 6: Find the Maximum Area of Triangle PQR
The maximum area of triangle \( PQR \) occurs when point \( P \) lies on the minor axis of the ellipse. The height from point \( P \) to line \( QR \) is \( b = 4 \), and the base \( QR \) is \( 6 \).

The area \( A \) of triangle \( PQR \) is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 4 = 12 \]

### Final Answer
The maximum area of triangle \( PQR \) is:
\[ \boxed{12} \]