Statement I : The area of the triangle formed by the points A(100, 102), B(102, 105), C(104, 107) is same as the area formed by A'(0, 0), B' (2, 3), C'(4, 5).
Statement II : The area of the triangle is constant wih respect to translation.

Options

A. Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I.
B. Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I.
C. Statement I is true, Statement II is false.
D. Statement I is false, Statement II is true.

Answer

To solve the problem, we need to determine if the area of triangle ABC formed by points A(100, 102), B(102, 105), and C(104, 107) is the same as the area of triangle A'B'C' formed by points A'(0, 0), B'(2, 3), and C'(4, 5). We will also verify the validity of the statements regarding the translation of triangles.

### Step-by-Step Solution:

1. **Calculate the Area of Triangle ABC:**
The area of a triangle given by vertices (x1, y1), (x2, y2), and (x3, y3) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For triangle ABC:
- A(100, 102) → (x1, y1) = (100, 102)
- B(102, 105) → (x2, y2) = (102, 105)
- C(104, 107) → (x3, y3) = (104, 107)

Plugging in the values:
\[ \text{Area}_{ABC} = \frac{1}{2} \left| 100(105 - 107) + 102(107 - 102) + 104(102 - 105) \right| \]
\[ = \frac{1}{2} \left| 100(-2) + 102(5) + 104(-3) \right| \]
\[ = \frac{1}{2} \left| -200 + 510 - 312 \right| \]
\[ = \frac{1}{2} \left| -2 \right| = 1 \]

2. **Calculate the Area of Triangle A'B'C':**
For triangle A'B'C':
- A'(0, 0) → (x1, y1) = (0, 0)
- B'(2, 3) → (x2, y2) = (2, 3)
- C'(4, 5) → (x3, y3) = (4, 5)

Using the same area formula:
\[ \text{Area}_{A'B'C'} = \frac{1}{2} \left| 0(3 - 5) + 2(5 - 0) + 4(0 - 3) \right| \]
\[ = \frac{1}{2} \left| 0 + 10 - 12 \right| \]
\[ = \frac{1}{2} \left| -2 \right| = 1 \]

3. **Compare the Areas:**
We found that:
\[ \text{Area}_{ABC} = 1 \quad \text{and} \quad \text{Area}_{A'B'C'} = 1 \]
Thus, the areas of both triangles are equal.

4. **Verify Statement II:**
Statement II claims that the area of the triangle is constant with respect to translation. This is true because translating a triangle by adding or subtracting a constant from the coordinates of its vertices does not change its area.

For example, translating triangle A'B'C' by adding (100, 102) to each vertex gives us triangle ABC:
- A'(0, 0) + (100, 102) = A(100, 102)
- B'(2, 3) + (100, 102) = B(102, 105)
- C'(4, 5) + (100, 102) = C(104, 107)

### Conclusion:
Both statements are correct:
- Statement I is true because the areas of triangles ABC and A'B'C' are equal.
- Statement II is true because the area remains constant under translation.

### Final Answer:
Both statements are correct, and Statement II correctly explains Statement I.