Two sides of a triangle are `2sqrt2 and 2sqrt3cm` and the angle opposite to the shorter side of the two is `pi/4`. The largest possible length of the third side is

Options

A. `(sqrt6+sqrt2)cm`
B. `(sqrt6+sqrt2)cm`
C. `(sqrt6-sqrt2)cm`
D. none of these

Answer

To find the largest possible length of the third side of a triangle given two sides and an angle, we can use the Law of Cosines. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Let \( a = 2\sqrt{2} \) cm (the shorter side).
- Let \( b = 2\sqrt{3} \) cm (the longer side).
- The angle opposite the shorter side \( a \) is \( A = \frac{\pi}{4} \).

2. **Apply the Law of Cosines:**
The Law of Cosines states:
\[ c^2 = a^2 + b^2 - 2ab \cos(A) \]
where \( c \) is the length of the third side.

3. **Calculate \( \cos(A) \):**
Since \( A = \frac{\pi}{4} \):
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

4. **Substitute Values into the Law of Cosines:**
\[ c^2 = (2\sqrt{2})^2 + (2\sqrt{3})^2 - 2(2\sqrt{2})(2\sqrt{3})\left(\frac{1}{\sqrt{2}}\right) \]

5. **Calculate Each Term:**
- \( (2\sqrt{2})^2 = 8 \)
- \( (2\sqrt{3})^2 = 12 \)
- \( 2(2\sqrt{2})(2\sqrt{3})\left(\frac{1}{\sqrt{2}}\right) = 4\sqrt{6} \)

6. **Combine the Terms:**
\[ c^2 = 8 + 12 - 4\sqrt{6} \]
\[ c^2 = 20 - 4\sqrt{6} \]

7. **Rearranging the Equation:**
To find \( c \), we need to take the square root:
\[ c = \sqrt{20 - 4\sqrt{6}} \]

8. **Finding the Maximum Length of \( c \):**
We can also express \( c \) in terms of a quadratic equation. The maximum length occurs when we consider the roots of the corresponding quadratic equation derived from the Law of Cosines.

The quadratic equation is:
\[ \sqrt{2}c^2 - 4\sqrt{3}c + 4 = 0 \]

9. **Use the Quadratic Formula:**
The roots are given by:
\[ c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = \sqrt{2}, b = -4\sqrt{3}, c = 4 \).

10. **Calculate the Discriminant:**
\[ b^2 - 4ac = (4\sqrt{3})^2 - 4(\sqrt{2})(4) = 48 - 16\sqrt{2} \]

11. **Final Calculation:**
\[ c = \frac{4\sqrt{3} \pm \sqrt{48 - 16\sqrt{2}}}{2\sqrt{2}} \]
Simplifying gives us two potential values for \( c \):
\[ c = 2\sqrt{3} + \sqrt{6} \quad \text{and} \quad c = 2\sqrt{3} - \sqrt{6} \]

12. **Choose the Maximum Value:**
The largest possible length of the third side is:
\[ c = 2\sqrt{3} + \sqrt{6} \]

### Final Answer:
Thus, the largest possible length of the third side \( c \) is:
\[ \boxed{\sqrt{6} + 2} \]