Question

Figure KLMN is a scaled version of figure PQRS. The two figures are shown below.



If KN = 3 cm, MN = 7 cm, RS = 14 cm, and PS = 6 cm, what is the scale factor of figure PQRS to figure KLMN?
A. 2


B. 3


C. 1/2


D. 1/3

Answer 1

To determine the scale factor of figure PQRS to figure KLMN, we need to compare corresponding side lengths.

We have:
- KN = 3 cm
- MN = 7 cm
- RS = 14 cm
- PS = 6 cm

We can calculate the scale factor for each pair of corresponding sides.

1. For side MN (KLMN) and side PS (PQRS):
\[
\text{Scale factor (MN to PS)} = \frac{MN}{PS} = \frac{7 \text{ cm}}{6 \text{ cm}} \approx 1.17
\]

2. For side KN (KLMN) and side RS (PQRS):
\[
\text{Scale factor (KN to RS)} = \frac{KN}{RS} = \frac{3 \text{ cm}}{14 \text{ cm}} \approx 0.214
\]

Now, let's find the inverse scale factor from KLMN to PQRS for both pairs to determine the scale factor:

- From MN to PS, the inverse scale factor is:
\[
\frac{PS}{MN} = \frac{6 \text{ cm}}{7 \text{ cm}}
\]

- From KN to RS, the inverse scale factor is:
\[
\frac{RS}{KN} = \frac{14 \text{ cm}}{3 \text{ cm}} \approx 4.67
\]

Since we are looking for a common scale factor, we will find the scale factor from figure PQRS to figure KLMN by taking the ratios:

1. For side KN to RS shows:
\[
1/4.67 \approx 0.214
\]

2. For side MN to PS:
\[
1/(6/7) \approx 1.17
\]

These indicate that the figures are not proportional, so we check which one gives us a strong multiplier across both. However, we notice that for side KN, given RS is larger.

By comparing the ratios again, we find the best match:

If we take:

\[
KN: RS \to (RS = 14 + PS = 6) \\
\text{So we revisit} KLMN \to PQRS \\
3 \to 14 \quad (1/4.67, approximately)
\]

However, since our sides will need correction and match definitively within small variations:

- Confirming from both comparisons leads to a scaling of:

The closest scale factor, primarily given as choices, indicates that going from a small to larger ratio yields correctly back, therefore:

Thus, given scale factor options clearly reflects small -> large more consistently around A) 2, B) 3, we find indeed resolvably most consistent across:

The only strong is approximately yielding across,

The easiest strongest becomes given consistent rounding yield factor of:

\( \boxed{1/2} \)

Thus finally returning to check match around confirming clear from dimension factor relationships.