Two parallelograms are similar. A pair of corresponding sides measures 8 cm on the first and 6 cm on the second. Compare the first to the second and give the ratio of their perimeters in simplest terms.

Step-by-step solution

1. 1

   Recognise which ratio you actually need. The question asks for perimeters, not areas, and that distinction decides everything. Perimeter is a sum of lengths, so scaling every length of a figure by a factor $k$ scales the whole sum by the same $k$. Areas would pick up $k^2$ instead.

2. 2

   Write the scale factor from the given side pair. Similar figures have all corresponding sides in one common ratio, so a single pair is enough:

   $k=\frac{\text{side of first}}{\text{side of second}}=\frac{8}{6}$

3. 3

   Justify that the perimeter ratio equals $k$. If the second parallelogram has sides $a$ and $b$, its perimeter is $2(a+b)$. The first has sides $ka$ and $kb$, so its perimeter is $2(ka+kb)=k\cdot 2(a+b)$. Dividing the two perimeters cancels $2(a+b)$ completely:

   $\frac{P_1}{P_2}=\frac{k\cdot 2(a+b)}{2(a+b)}=k=\frac{8}{6}$

   Notice you never needed the second side length.

4. 4

   Reduce to simplest terms. Both $8$ and $6$ are divisible by $2$:

   $\frac{8}{6}=\frac{8\div 2}{6\div 2}=\frac{4}{3}$

5. 5

   State the answer as a ratio in the requested order. "First to second" means the first figure's perimeter goes in front:

   $P_1:P_2=4:3$

   As a sanity check, the first figure has the longer given side, so its perimeter should be the larger of the two, and $4:3$ is indeed greater than $1$.