In `Delta PQR, angleP` is a right angle and PT is perpendicular to QR. If `PT=4 cm QR =12cm`, then the value of (`cot Q+cot R`) is

Answer

To solve the problem, we need to find the value of \( \cot Q + \cot R \) in triangle \( \Delta PQR \) where \( \angle P \) is a right angle, \( PT \) is perpendicular to \( QR \), \( PT = 4 \, \text{cm} \), and \( QR = 12 \, \text{cm} \).

### Step-by-step Solution:

1. **Understanding the Triangle**:
- In triangle \( PQR \), since \( \angle P \) is a right angle, we can denote the lengths of the sides as follows:
- Let \( PQ = a \)
- Let \( PR = b \)
- The hypotenuse \( QR = c = 12 \, \text{cm} \)

2. **Using the Definition of Cotangent**:
- The cotangent of an angle in a right triangle is defined as the ratio of the adjacent side to the opposite side.
- Therefore, we have:
\[ \cot Q = \frac{PQ}{PT} \quad \text{and} \quad \cot R = \frac{PR}{PT} \]

3. **Finding the Lengths of \( PQ \) and \( PR \)**:
- From triangle \( PTQ \):
- \( PT = 4 \, \text{cm} \)
- Let \( QT = x \) (the length of \( QT \)).
- Using the Pythagorean theorem in triangle \( PTQ \):
\[ PQ^2 = PT^2 + QT^2 \implies a^2 = 4^2 + x^2 \implies a^2 = 16 + x^2 \quad (1) \]
- From triangle \( PTR \):
- Let \( RT = y \) (the length of \( RT \)).
- Again using the Pythagorean theorem:
\[ PR^2 = PT^2 + RT^2 \implies b^2 = 4^2 + y^2 \implies b^2 = 16 + y^2 \quad (2) \]

4. **Relating \( QT \) and \( RT \)**:
- Since \( QT + RT = QR = 12 \, \text{cm} \):
\[ x + y = 12 \quad (3) \]

5. **Expressing \( y \) in terms of \( x \)**:
- From equation (3), we can express \( y \) as:
\[ y = 12 - x \quad (4) \]

6. **Substituting \( y \) into Equation (2)**:
- Substitute \( y \) from equation (4) into equation (2):
\[ b^2 = 16 + (12 - x)^2 \]
- Expanding this:
\[ b^2 = 16 + (144 - 24x + x^2) \implies b^2 = 160 - 24x + x^2 \quad (5) \]

7. **Finding \( \cot Q + \cot R \)**:
- Now, we can find \( \cot Q + \cot R \):
\[ \cot Q + \cot R = \frac{PQ}{PT} + \frac{PR}{PT} = \frac{PQ + PR}{PT} = \frac{a + b}{4} \]
- From equations (1) and (5), we can find \( a + b \):
- Since \( a^2 + b^2 = c^2 \) (Pythagorean theorem):
\[ a^2 + b^2 = 12^2 = 144 \]
- We can also express \( a + b \) in terms of \( x \):
\[ a + b = \sqrt{16 + x^2} + \sqrt{160 - 24x + x^2} \]

8. **Final Calculation**:
- However, we can simplify our calculation:
\[ \cot Q + \cot R = \frac{QR}{PT} = \frac{12}{4} = 3 \]

### Final Answer:
Thus, the value of \( \cot Q + \cot R \) is \( 3 \).