To find the possible lengths of the longest side of a rectangular room with a perimeter of 70 m, we can use the formula for the perimeter of a rectangle:
\[ P = 2L + 2W \]
where \(L\) is the length and \(W\) is the width.
Given that the perimeter \(P = 70\) m, we can simplify this to:
\[ 2L + 2W = 70 \]
Dividing the entire equation by 2 gives us:
\[ L + W = 35 \]
This means that the sum of the length and width of the room must equal 35 m.
To find the maximum possible length, we can set \(W\) to be as small as possible. The smallest value for \(W\) would be just above 0 (but we can't have a width of 0 since it is a room).
If we consider the following values for the longest side (L):
- If \(L = 35\), then \(W = 0\), which is not possible.
- If \(L = 34\), then \(W = 1\).
- If \(L = 33\), then \(W = 2\).
- Continuing this way, if we keep reducing \(W\), we can have:
- \(L = 30\), \(W = 5\)
- \(L = 25\), \(W = 10\)
- \(L = 20\), \(W = 15\)
- \(L = 15\), \(W = 20\)
- etc.
However, if we are given possible lengths (11 m, 12 m, 24 m, and 40 m), we can check each to see if it satisfies the condition and is reasonable given our parameters:
- 11 m: \(W = 35 - 11 = 24\) (valid)
- 12 m: \(W = 35 - 12 = 23\) (valid)
- 24 m: \(W = 35 - 24 = 11\) (valid)
- 40 m: \(W = 35 - 40 = -5\) (invalid, as width cannot be negative)
From the options given, the maximum valid length (that still keeps \(W\) positive) would be 24 m.
Therefore, the length of the longest side of the room could be 24 m.