PQR is a triangle, whose are is `180 cm^(2)` . S is a point on side QR, such that PS is the angle bisector of ` angle QPR` . If PQ : PR = 2 : 3 , then what is the area (in `cm^(2)`) triangle PSR ?

Options

A. 90
B. 108
C. 144
D. 72

Answer

To find the area of triangle PSR, we can use the information given in the problem along with the Angle Bisector Theorem. Here’s how we can solve it step by step:

### Step 1: Understand the Problem
We have triangle PQR with an area of 180 cm². PS is the angle bisector of angle QPR, and the ratio of the lengths of sides PQ and PR is given as 2:3.

### Step 2: Apply the Angle Bisector Theorem
According to the Angle Bisector Theorem, the ratio of the areas of triangles formed by an angle bisector is equal to the ratio of the other two sides. Therefore, we can express the areas of triangles PQS and PSR in terms of the ratio of PQ and PR.

Let:
- Area of triangle PQS = A1
- Area of triangle PSR = A2

According to the Angle Bisector Theorem:
\[ \frac{A1}{A2} = \frac{PQ}{PR} = \frac{2}{3} \]

### Step 3: Express Areas in Terms of A2
From the ratio, we can express A1 in terms of A2:
\[ A1 = \frac{2}{3} A2 \]

### Step 4: Relate Areas to Total Area
The total area of triangle PQR is the sum of the areas of triangles PQS and PSR:
\[ A1 + A2 = 180 \text{ cm}^2 \]

Substituting A1 from the previous step:
\[ \frac{2}{3} A2 + A2 = 180 \]

### Step 5: Combine Like Terms
To combine the terms, we can express A2 in a common fraction:
\[ \frac{2}{3} A2 + \frac{3}{3} A2 = 180 \]
\[ \frac{5}{3} A2 = 180 \]

### Step 6: Solve for A2
Now, we can solve for A2:
\[ A2 = 180 \times \frac{3}{5} \]
\[ A2 = \frac{540}{5} = 108 \text{ cm}^2 \]

### Conclusion
The area of triangle PSR is 108 cm².