The length, breadth and height of a room are 1050 cm, 750 cm, and 425 cm respectively. Find the length of the longest tape which can measure the three dimenstions of the room exactly.

Answer

To find the length of the longest tape that can measure the dimensions of the room exactly, we need to calculate the highest common factor (HCF) of the three dimensions: 1050 cm, 750 cm, and 425 cm. We will use the prime factorization method to find the HCF.

### Step-by-Step Solution:

1. **Find the Prime Factorization of 1050 cm:**
- Divide 1050 by 2:
\[ 1050 \div 2 = 525 \]
- Divide 525 by 3:
\[ 525 \div 3 = 175 \]
- Divide 175 by 5:
\[ 175 \div 5 = 35 \]
- Divide 35 by 5:
\[ 35 \div 5 = 7 \]
- Finally, divide 7 by 7:
\[ 7 \div 7 = 1 \]
- Thus, the prime factorization of 1050 is:
\[ 1050 = 2^1 \times 3^1 \times 5^2 \times 7^1 \]

2. **Find the Prime Factorization of 750 cm:**
- Divide 750 by 2:
\[ 750 \div 2 = 375 \]
- Divide 375 by 3:
\[ 375 \div 3 = 125 \]
- Divide 125 by 5:
\[ 125 \div 5 = 25 \]
- Divide 25 by 5:
\[ 25 \div 5 = 5 \]
- Finally, divide 5 by 5:
\[ 5 \div 5 = 1 \]
- Thus, the prime factorization of 750 is:
\[ 750 = 2^1 \times 3^1 \times 5^3 \]

3. **Find the Prime Factorization of 425 cm:**
- Divide 425 by 5:
\[ 425 \div 5 = 85 \]
- Divide 85 by 5:
\[ 85 \div 5 = 17 \]
- Finally, divide 17 by 17:
\[ 17 \div 17 = 1 \]
- Thus, the prime factorization of 425 is:
\[ 425 = 5^2 \times 17^1 \]

4. **Identify the Common Prime Factors:**
- The prime factorizations are:
- 1050: \(2^1 \times 3^1 \times 5^2 \times 7^1\)
- 750: \(2^1 \times 3^1 \times 5^3\)
- 425: \(5^2 \times 17^1\)
- The common prime factor is \(5\).

5. **Determine the HCF:**
- The minimum power of the common prime factor \(5\) is \(2\).
- Therefore, the HCF is:
\[ HCF = 5^2 = 25 \]

### Conclusion:
The length of the longest tape that can measure the three dimensions of the room exactly is **25 cm**.