To determine three possible sets of dimensions for a cuboid with a volume of [tex]\(36 \, \text{cm}^3\)[/tex], we need to find combinations of length, width, and height such that their product equals the volume. Here are three sets of possible dimensions:
### i)
- Length: [tex]\(6 \, \text{cm}\)[/tex]
- Width: [tex]\(6 \, \text{cm}\)[/tex]
- Height: [tex]\(\frac{36}{6 \times 6} = 1 \, \text{cm}\)[/tex]
So, one possible set of dimensions is:
- Length [tex]\( = 6 \, \text{cm} \)[/tex]
- Width [tex]\( = 6 \, \text{cm} \)[/tex]
- Height [tex]\( = 1 \, \text{cm} \)[/tex]
### ii)
- Length: [tex]\(9 \, \text{cm}\)[/tex]
- Width: [tex]\(2 \, \text{cm}\)[/tex]
- Height: [tex]\(\frac{36}{9 \times 2} = 2 \, \text{cm}\)[/tex]
So, another possible set of dimensions is:
- Length [tex]\( = 9 \, \text{cm} \)[/tex]
- Width [tex]\( = 2 \, \text{cm} \)[/tex]
- Height [tex]\( = 2 \, \text{cm} \)[/tex]
### iii)
- Length: [tex]\(3 \, \text{cm}\)[/tex]
- Width: [tex]\(3 \, \text{cm}\)[/tex]
- Height: [tex]\(\frac{36}{3 \times 3} = 4 \, \text{cm}\)[/tex]
So, another possible set of dimensions is:
- Length [tex]\( = 3 \, \text{cm} \)[/tex]
- Width [tex]\( = 3 \, \text{cm} \)[/tex]
- Height [tex]\( = 4 \, \text{cm} \)[/tex]
Therefore, three possible sets of dimensions for the cuboid are:
1. [tex]\((6\, \text{cm}, 6\, \text{cm}, 1\, \text{cm})\)[/tex]
2. [tex]\((9\, \text{cm}, 2\, \text{cm}, 2\, \text{cm})\)[/tex]
3. [tex]\((3\, \text{cm}, 3\, \text{cm}, 4\, \text{cm})\)[/tex]
