In `Delta AST,` where `TP bot SA,` area equals 204 sq. cm. Given that `TP = 17 cm and ` area of `Delta TSP:` area of `Delta TAP = 3:5,` find SP and PA.

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript.

### Step 1: Understand the given information
We have a triangle AST with a perpendicular TP drawn to the base SA. The area of triangle AST is given as 204 sq. cm, and the height TP is 17 cm. The area ratio of triangles TSP and TAP is given as 3:5.

### Step 2: Calculate the length of base AS
The area of a triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
For triangle AST:
\[ 204 = \frac{1}{2} \times AS \times TP \]
Substituting the known values:
\[ 204 = \frac{1}{2} \times AS \times 17 \]
Multiplying both sides by 2:
\[ 408 = AS \times 17 \]
Now, divide both sides by 17 to find AS:
\[ AS = \frac{408}{17} = 24 \text{ cm} \]

### Step 3: Express AS in terms of AP and PS
Let:
\[ AS = AP + PS \quad \text{(Equation 1)} \]
Since we found that \(AS = 24\) cm, we can write:
\[ AP + PS = 24 \text{ cm} \]

### Step 4: Set up the ratio of areas
The ratio of the areas of triangles TSP and TAP is given as:
\[ \frac{\text{Area of } TSP}{\text{Area of } TAP} = \frac{3}{5} \]
Using the area formula for both triangles:
\[ \frac{\frac{1}{2} \times TP \times PS}{\frac{1}{2} \times TP \times PA} = \frac{3}{5} \]
The \(\frac{1}{2} \times TP\) cancels out:
\[ \frac{PS}{PA} = \frac{3}{5} \quad \text{(Equation 2)} \]

### Step 5: Express PS in terms of PA
From Equation 2, we can express PS in terms of PA:
\[ PS = \frac{3}{5} PA \]

### Step 6: Substitute PS in Equation 1
Substituting \(PS\) in Equation 1:
\[ AP + \frac{3}{5} PA = 24 \]
Let \(AP = PA - PS\):
\[ PA - \frac{3}{5} PA + \frac{3}{5} PA = 24 \]
This simplifies to:
\[ PA + PS = 24 \]
Now substituting \(PS = \frac{3}{5} PA\):
\[ PA + \frac{3}{5} PA = 24 \]
Combining the terms:
\[ \frac{8}{5} PA = 24 \]

### Step 7: Solve for PA
Multiply both sides by \(\frac{5}{8}\):
\[ PA = 24 \times \frac{5}{8} = 15 \text{ cm} \]

### Step 8: Find SP using Equation 1
Now, substitute \(PA\) back into Equation 1 to find \(SP\):
\[ AP + PS = 24 \]
We know \(PS = \frac{3}{5} \times 15 = 9 \text{ cm}\):
\[ AP + 9 = 24 \]
Thus:
\[ AP = 24 - 9 = 15 \text{ cm} \]

### Final Results
- Length of side \(PA = 15 \text{ cm}\)
- Length of side \(SP = 9 \text{ cm}\)