The perpendicular from P and QR in the triangle jPQR meets QR at S. Alsl, T is a point on SR sch that QS `=3 cm, ST = 4 cm, and T = 8 cm.` Find the ratio of the areas of the triangles PQS , PST and PTR.`

Answer

To find the ratio of the areas of triangles PQS, PST, and PTR, we can follow these steps:

### Step 1: Understand the Problem
We have triangle PQR with a perpendicular from point P to line QR meeting at point S. Point T is on segment SR. We know the lengths of segments QS, ST, and TR.

### Step 2: Identify Given Values
- QS = 3 cm
- ST = 4 cm
- TR = 8 cm

### Step 3: Calculate Length of SR
To find the length of segment SR, we can add the lengths of QS and ST:
\[ SR = QS + ST = 3 \text{ cm} + 4 \text{ cm} = 7 \text{ cm} \]

### Step 4: Determine the Length of QR
The total length of QR can be calculated by adding the lengths of QS, ST, and TR:
\[ QR = QS + ST + TR = 3 \text{ cm} + 4 \text{ cm} + 8 \text{ cm} = 15 \text{ cm} \]

### Step 5: Use the Area Formula
The area of a triangle is given by the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the height (PS) is common for all triangles.

### Step 6: Write the Area Expressions
- Area of triangle PQS:
\[ \text{Area}_{PQS} = \frac{1}{2} \times QS \times PS = \frac{1}{2} \times 3 \times PS \]

- Area of triangle PST:
\[ \text{Area}_{PST} = \frac{1}{2} \times ST \times PS = \frac{1}{2} \times 4 \times PS \]

- Area of triangle PTR:
\[ \text{Area}_{PTR} = \frac{1}{2} \times TR \times PS = \frac{1}{2} \times 8 \times PS \]

### Step 7: Set Up the Ratios
Now we can set up the ratios of the areas:
\[ \text{Area}_{PQS} : \text{Area}_{PST} : \text{Area}_{PTR} = (3 \times PS) : (4 \times PS) : (8 \times PS) \]

### Step 8: Simplify the Ratios
Since PS is common in all three areas, it cancels out:
\[ 3 : 4 : 8 \]

### Final Answer
The ratio of the areas of triangles PQS, PST, and PTR is:
\[ \text{Ratio} = 3 : 4 : 8 \]