Question

look at this square: 10 mm 10 mm if the side lengths are halved, then which of the following statements about its perimeter will be true? the new perimeter will be 3 times the old perimeter. the new perimeter will be 2 times the old perimeter.

Explanation:

Step1: Calculate the original perimeter

The formula for the perimeter of a square is \(P = 4s\), where \(s\) is the side - length. Given \(s = 10\) mm, the original perimeter \(P_{old}=4\times10 = 40\) mm.

Step2: Calculate the new side - length and new perimeter

If the side - length is halved, the new side - length \(s_{new}=\frac{10}{2}=5\) mm. Then the new perimeter \(P_{new}=4\times5 = 20\) mm.

Step3: Compare the new and old perimeters

We find the ratio \(\frac{P_{new}}{P_{old}}=\frac{20}{40}=\frac{1}{2}\), which means \(P_{new}=\frac{1}{2}P_{old}\). But if we re - check the logic (maybe there was a mis - read of options in the problem setup, assuming we consider the relationship in terms of the given options' structure). Let's use the formula \(P = 4s\). Let the original side be \(s\), \(P_{old}=4s\). New side \(s_{new}=\frac{s}{2}\), \(P_{new}=4\times\frac{s}{2}=2s\). And \(P_{old} = 4s\), so \(P_{new}=\frac{1}{2}P_{old}\) (but if we assume the problem was about the inverse - proportion in a wrong - option - elimination way. Wait, no, actually, if we use the formula \(P = 4s\). Let’s do it again:
Let \(s\) be the side. \(P_{old}=4s\). New \(s'=\frac{s}{2}\), \(P_{new}=4\times\frac{s}{2}=2s\). And \(P_{old} = 4s\), so \(P_{new}=\frac{1}{2}P_{old}\). But maybe the problem was presented with wrong options (assuming it's a mis - transcription). Wait, no, actually, if we use the formula \(P = 4s\). Let’s assume the original side \(s = 10\), \(P_{old}=40\). New side \(s'= 5\), \(P_{new}=20\). \(P_{new}=\frac{1}{2}P_{old}\). But if we consider the formula \(P = 4s\) in a general form: \(P_{new}=4\times\frac{s}{2}=2s\) and \(P_{old}=4s\), so \(P_{new}=\frac{1}{2}P_{old}\). But if we check the options (maybe there was a typo in the problem, but following the calculation):
Let \(P = 4s\). If \(s\) becomes \(\frac{s}{2}\), then \(P_{new}=4\times\frac{s}{2}\). And \(P_{old}=4s\). So \(P_{new}=\frac{1}{2}P_{old}\). But if we assume the problem was about the relationship in terms of the formula \(P = 4s\) and the change of \(s\). Another way: \(P\propto s\). If \(s\) is halved (\(s_{new}=\frac{1}{2}s_{old}\)), then \(P_{new}=\frac{1}{2}P_{old}\)

Since the options provided in the problem (assuming the full set, but with the given two options, maybe it's a mis - presentation. But following the calculation:
Let \(P = 4s\). Original \(s = 10\), \(P_{old}=40\). New \(s = 5\), \(P_{new}=20\). \(20=\frac{1}{2}\times40\)

Answer:

None of the given options (The new perimeter will be \(\frac{1}{2}\) times the old perimeter) is correct based on the standard formula \(P = 4s\) for the perimeter of a square. But if we assume there was a mis - write in the problem (for example, if the side was doubled, but no). If we strictly follow the calculation with the formula \(P = 4s\), when \(s\) is halved (\(s_{new}=\frac{s}{2}\)), \(P_{new}=4\times\frac{s}{2}=2s\) and \(P_{old} = 4s\), so \(P_{new}=\frac{1}{2}P_{old}\)