The altitudes from the angular points A,B, and C on the opposite sides BC, CA and AB of `DeltaABC` are 210, 195 and 182 respectively. Then the value of a is ____

Explanation

To solve the problem, we need to find the value of side \( a \) of triangle \( \Delta ABC \) given the altitudes from the vertices to the opposite sides. The altitudes are given as follows:

- Altitude from point A to side BC: \( h_a = 210 \)
- Altitude from point B to side CA: \( h_b = 195 \)
- Altitude from point C to side AB: \( h_c = 182 \)

### Step-by-step Solution:

1. **Area of the Triangle**:
The area \( A \) of triangle \( ABC \) can be expressed using the formula involving the side and its corresponding altitude:
\[ A = \frac{1}{2} \times a \times h_a = \frac{1}{2} \times b \times h_b = \frac{1}{2} \times c \times h_c \]

2. **Expressing Area in terms of \( a \)**:
Using the altitude from A:
\[ A = \frac{1}{2} \times a \times 210 \]
Thus,
\[ A = 105a \]

3. **Expressing Area in terms of \( b \)**:
Using the altitude from B:
\[ A = \frac{1}{2} \times b \times 195 \]
Thus,
\[ A = 97.5b \]

4. **Expressing Area in terms of \( c \)**:
Using the altitude from C:
\[ A = \frac{1}{2} \times c \times 182 \]
Thus,
\[ A = 91c \]

5. **Setting the Equations Equal**:
From the area expressions, we can set them equal to each other:
\[ 105a = 97.5b \quad \text{(1)} \]
\[ 105a = 91c \quad \text{(2)} \]

6. **Finding Ratios**:
From equation (1):
\[ b = \frac{105a}{97.5} = \frac{14a}{13} \]
From equation (2):
\[ c = \frac{105a}{91} = \frac{15a}{13} \]

7. **Calculating the Semi-perimeter**:
The semi-perimeter \( s \) is given by:
\[ s = \frac{a + b + c}{2} = \frac{a + \frac{14a}{13} + \frac{15a}{13}}{2} \]
Simplifying:
\[ s = \frac{a + \frac{29a}{13}}{2} = \frac{\frac{42a}{13}}{2} = \frac{21a}{13} \]

8. **Using Heron's Formula for Area**:
The area using Heron's formula is:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
We need to calculate \( s - a \), \( s - b \), and \( s - c \):
\[ s - a = \frac{21a}{13} - a = \frac{8a}{13} \]
\[ s - b = \frac{21a}{13} - \frac{14a}{13} = \frac{7a}{13} \]
\[ s - c = \frac{21a}{13} - \frac{15a}{13} = \frac{6a}{13} \]

9. **Substituting into Heron's Formula**:
\[ A = \sqrt{\frac{21a}{13} \cdot \frac{8a}{13} \cdot \frac{7a}{13} \cdot \frac{6a}{13}} \]
Simplifying:
\[ A = \sqrt{\frac{21 \cdot 8 \cdot 7 \cdot 6 \cdot a^4}{13^4}} = \frac{84a^2}{169} \]

10. **Equating the Two Area Expressions**:
Setting the two expressions for area equal:
\[ 105a = \frac{84a^2}{169} \]
Rearranging gives:
\[ 84a^2 - 105 \cdot 169a = 0 \]
Factoring out \( a \):
\[ a(84a - 105 \cdot 169) = 0 \]
Thus, solving for \( a \):
\[ 84a = 105 \cdot 169 \implies a = \frac{105 \cdot 169}{84} \]

11. **Calculating the Final Value**:
Simplifying:
\[ a = \frac{15 \cdot 169}{12} = \frac{2535}{12} = 211.25 \]

### Final Answer:
The value of \( a \) is \( \frac{845}{4} \).