To find the possible values of length ([tex]\( l \)[/tex]) and width ([tex]\( w \)[/tex]) that give an area of [tex]\( 68 \, \text{cm}^2 \)[/tex], we need to find pairs of integers [tex]\(( l, w )\)[/tex] such that their product is equal to [tex]\( 68 \)[/tex]. Specifically, we are looking for pairs [tex]\(( l, w )\)[/tex] satisfying the equation:
[tex]\[ l \times w = 68 \][/tex]
Let's identify all pairs of integers that satisfy this equation:
1. Start by taking [tex]\( l = 1 \)[/tex]:
[tex]\[
l = 1 \implies w = \frac{68}{1} = 68
\][/tex]
So, one pair is [tex]\((1, 68)\)[/tex].
2. Next, consider [tex]\( l = 2 \)[/tex]:
[tex]\[
l = 2 \implies w = \frac{68}{2} = 34
\][/tex]
This gives us another pair [tex]\((2, 34)\)[/tex].
3. Now, let's try [tex]\( l = 4 \)[/tex]:
[tex]\[
l = 4 \implies w = \frac{68}{4} = 17
\][/tex]
Thus, another pair is [tex]\((4, 17)\)[/tex].
4. Continue with [tex]\( l = 17 \)[/tex]:
[tex]\[
l = 17 \implies w = \frac{68}{17} = 4
\][/tex]
This results in the pair [tex]\((17, 4)\)[/tex].
5. Next, consider [tex]\( l = 34 \)[/tex]:
[tex]\[
l = 34 \implies w = \frac{68}{34} = 2
\][/tex]
This gives us the pair [tex]\((34, 2)\)[/tex].
6. Finally, let [tex]\( l = 68 \)[/tex]:
[tex]\[
l = 68 \implies w = \frac{68}{68} = 1
\][/tex]
This gives the pair [tex]\((68, 1)\)[/tex].
Putting it all together, the possible integer pairs of [tex]\((l, w)\)[/tex] that give an area of [tex]\( 68 \, \text{cm}^2 \)[/tex] are:
[tex]\[ (1, 68), (2, 34), (4, 17), (17, 4), (34, 2), (68, 1) \][/tex]
Thus, these are the sets of possible values for length and width that satisfy the condition [tex]\( l \times w = 68 \)[/tex].
