The problem involves a cube where each edge has a length of 6 cm. We are asked to find the distances between specific points in the cube. Let's analyze each option:

1. A ke F: Points A and F are on different edges of the cube. Using the Pythagorean theorem for the diagonal of a face (AF), the distance can be found.

   $AF = \sqrt{AB^2 + BF^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2} \, \text{cm}$

2. B ke G: This involves finding the space diagonal from B to G. The formula for a space diagonal in a cube is:

   $d = \sqrt{AB^2 + BF^2 + BG^2} = \sqrt{6^2 + 6^2 + 6^2} = \sqrt{108} = 6\sqrt{3} \, \text{cm}$

3. A ke G: This is also a space diagonal from A to G. Since the cube's space diagonal formula is:

   $AG = \sqrt{6^2 + 6^2 + 6^2} = \sqrt{108} = 6\sqrt{3} \, \text{cm}$

4. A ke P: A to P is a face diagonal. The distance can be calculated similarly to the A to F case, but on a different face.

   $AP = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2} \, \text{cm}$

5. O ke B: O is the center of the cube, and B is a vertex. The distance from the center to a vertex is half the space diagonal, so:

   $OB = \frac{6\sqrt{3}}{2} = 3\sqrt{3} \, \text{cm}$

Answer for each option:

• A ke F = $6\sqrt{2}$ cm
• B ke G = $6\sqrt{3}$ cm
• A ke G = $6\sqrt{3}$ cm
• A ke P = $6\sqrt{2}$ cm
• O ke B = $3\sqrt{3}$ cm

Let me know if you'd like further details or have any questions!

Related Questions:

1. What is the distance between point A and point E?
2. How do you calculate the space diagonal of any cube with a side length $s$?
3. What is the formula for the diagonal of a square face?
4. If the side length of the cube is doubled, how would the space diagonal change?
5. How does the Pythagorean theorem apply in 3D geometry?

Tip:

In any cube, the diagonal across the face is always $s\sqrt{2}$, and the space diagonal is $s\sqrt{3}$, where $s$ is the length of the edge.