Question

look at this square: 33 in 33 in if the side lengths are tripled, then which of the following statements about its perimeter will be true? the new perimeter will be \frac{1}{2} of the old perimeter. the new perimeter will be 4 times the old perimeter.

Explanation:

Step1: Calculate the old perimeter

The perimeter formula for a square is \(P = 4s\). Given \(s = 33\) in, so \(P_{old}=4\times33 = 132\) in.

Step2: Calculate the new side length and perimeter

The new side length \(s_{new}=3\times33 = 99\) in. Then \(P_{new}=4\times99=396\) in.

Step3: Find the ratio of new to old perimeter

\(\frac{P_{new}}{P_{old}}=\frac{396}{132}=3\). But let's check another way. If original side is \(s\), original perimeter \(P = 4s\). New side \(s'=3s\), new perimeter \(P'=4\times(3s)=3\times(4s)=3P\). Wait, maybe there was a mis - read of options. Let's re - check using the formula \(P = 4s\). If we assume the general case:
Let the original side of the square be \(a\). Original perimeter \(P_1=4a\). New side \(a_2 = 3a\). New perimeter \(P_2=4\times(3a)=12a\). \(\frac{P_2}{P_1}=\frac{12a}{4a}=3\). But if we use the values:
Old perimeter \(P_{old}=4\times33 = 132\). New perimeter \(P_{new}=4\times(3\times33)=4\times99 = 396\). \(\frac{396}{132}=3\). But maybe the user had a typo in options. Wait, if we use the formula \(P = 4s\), if \(s\) is tripled (\(s'=3s\)), \(P'=4\times3s = 3\times(4s)\). So the new perimeter is 3 times the old. But since the options are wrong in the problem (maybe a mis - copy), but if we assume the intended formula application:
Let’s use the property of linear scaling. For a square (a 2D shape with perimeter depending linearly on side length). If the side length is scaled by a factor \(k\) (here \(k = 3\)), the perimeter is scaled by \(k\).

Answer:

None of the given options (The new perimeter will be 3 times the old perimeter)