Write True or False and justify your answer:
The cost of levelling the ground in the form of a triangle having the sides `51` `m`, `37` `m` and `20` `m` at the rate of `Rs` `3` per `m^(2)` is `Rs` `918`.

Explanation

To determine whether the statement is true or false, we need to calculate the area of the triangular ground with sides 51 m, 37 m, and 20 m using Heron's formula, and then find the cost of leveling the ground at the rate of Rs 3 per m².

### Step-by-step Solution:

1. **Identify the sides of the triangle:**
- Let \( a = 51 \, m \)
- Let \( b = 37 \, m \)
- Let \( c = 20 \, m \)

2. **Calculate the semi-perimeter (s):**
\[ s = \frac{a + b + c}{2} = \frac{51 + 37 + 20}{2} = \frac{108}{2} = 54 \, m \]

3. **Apply Heron's formula to find the area (A):**
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
Substituting the values:
\[ A = \sqrt{54(54 - 51)(54 - 37)(54 - 20)} \]
\[ = \sqrt{54 \times 3 \times 17 \times 34} \]

4. **Simplify the expression under the square root:**
- First, calculate \( 54 \times 3 = 162 \)
- Then, calculate \( 17 \times 34 = 578 \)
- Now, calculate \( 162 \times 578 \)

To simplify:
\[ 162 = 9 \times 18 \quad \text{and} \quad 578 = 17 \times 34 \]
Thus,
\[ A = \sqrt{54 \times 3 \times 17 \times 34} = \sqrt{(9 \times 6) \times 3 \times (17 \times 34)} \]
This can be further simplified to:
\[ A = \sqrt{(9 \times 17) \times (6 \times 34)} = \sqrt{153 \times 204} \]

5. **Calculate the area:**
After performing the calculations, we find:
\[ A \approx 306 \, m^2 \]

6. **Calculate the cost of leveling the ground:**
\[ \text{Cost} = \text{Area} \times \text{Cost per m}^2 = 306 \, m^2 \times 3 \, \text{Rs/m}^2 \]
\[ = 918 \, Rs \]

7. **Conclusion:**
Since the calculated cost of leveling the ground is Rs 918, which matches the statement given in the question, we conclude that the statement is **True**.