An ant has to go from one corner to the farthest corner of a canister of size `6xx6xx12`. What's the minimum diatance it has to cover, (All the lengths are in cm)?

Options

A. `6(sqrt2+2)`
B. `6sqrt6`
C. `12sqrt2`
D. `13(sqrt2+1)`

Answer

To find the minimum distance the ant has to cover from one corner of a cuboidal canister to the farthest corner, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Dimensions of the Canister:**
The canister has dimensions of 6 cm x 6 cm x 12 cm.

2. **Label the Corners of the Cuboid:**
Let's label the corners of the cuboid:
- A (0, 0, 0) - one corner
- B (6, 0, 0)
- C (6, 6, 0)
- D (0, 6, 0)
- E (0, 0, 12)
- F (6, 0, 12)
- G (6, 6, 12)
- H (0, 6, 12)

The ant starts at corner A and needs to reach corner F.

3. **Determine the Path:**
The ant can take the following path:
- From A to C (moving along the base)
- From C to F (moving vertically)

4. **Calculate the Distance AC:**
The distance AC can be calculated using the Pythagorean theorem:
\[ AC = \sqrt{(AB)^2 + (BC)^2} \]
Here, AB = 6 cm and BC = 6 cm.
\[ AC = \sqrt{(6)^2 + (6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \text{ cm} \]

5. **Calculate the Distance CF:**
The distance CF is simply the height of the canister:
\[ CF = 12 \text{ cm} \]

6. **Total Distance:**
The total distance the ant needs to cover from A to F is:
\[ \text{Total Distance} = AC + CF = 6\sqrt{2} + 12 \text{ cm} \]

7. **Final Expression:**
We can factor out the common terms:
\[ \text{Total Distance} = 6\sqrt{2} + 12 = 6(\sqrt{2} + 2) \text{ cm} \]

### Conclusion:
The minimum distance the ant has to cover is \( 6(\sqrt{2} + 2) \) cm.