Question
In rectangle ABCD, AB = 10 cm and BC = 14 cm. If rectangle EFGH is the image after a dilation, which could be its side lengths?
EF = 6 cm and FG = 10 cm
EF = 7 cm and FG = 3 cm
EF = 15 cm and FG = 20 cm
EF = 25 cm and FG = 35 cm
EF = 6 cm and FG = 10 cm
EF = 7 cm and FG = 3 cm
EF = 15 cm and FG = 20 cm
EF = 25 cm and FG = 35 cm
Answers
GPT-4o mini
To determine the possible side lengths of rectangle EFGH after dilation, we first need to find the scale factor of the dilation based on the original dimensions of rectangle ABCD. The original dimensions are:
- AB = 10 cm (width)
- BC = 14 cm (height)
Now, the area of rectangle ABCD can be calculated as:
\[
\text{Area}_{ABCD} = AB \times BC = 10 \, \text{cm} \times 14 \, \text{cm} = 140 \, \text{cm}^2.
\]
The dimensions of rectangle EFGH must maintain the same ratio of sides as rectangle ABCD since both are rectangles. This means the ratio \( \frac{EF}{FG} \) must equal \( \frac{10}{14} = \frac{5}{7} \).
Now let’s check the options given:
1. **EF = 6 cm and FG = 10 cm**
\[
\frac{EF}{FG} = \frac{6}{10} = \frac{3}{5}, \text{ which does not equal } \frac{5}{7}.
\]
2. **EF = 7 cm and FG = 3 cm**
\[
\frac{EF}{FG} = \frac{7}{3}, \text{ which does not equal } \frac{5}{7}.
\]
3. **EF = 15 cm and FG = 20 cm**
\[
\frac{EF}{FG} = \frac{15}{20} = \frac{3}{4}, \text{ which does not equal } \frac{5}{7}.
\]
4. **EF = 25 cm and FG = 35 cm**
\[
\frac{EF}{FG} = \frac{25}{35} = \frac{5}{7}, \text{ which equals } \frac{5}{7}.
\]
From the above analysis, the only option that maintains the ratio of the sides consistent with the original rectangle ABCD is:
**EF = 25 cm and FG = 35 cm.**
- AB = 10 cm (width)
- BC = 14 cm (height)
Now, the area of rectangle ABCD can be calculated as:
\[
\text{Area}_{ABCD} = AB \times BC = 10 \, \text{cm} \times 14 \, \text{cm} = 140 \, \text{cm}^2.
\]
The dimensions of rectangle EFGH must maintain the same ratio of sides as rectangle ABCD since both are rectangles. This means the ratio \( \frac{EF}{FG} \) must equal \( \frac{10}{14} = \frac{5}{7} \).
Now let’s check the options given:
1. **EF = 6 cm and FG = 10 cm**
\[
\frac{EF}{FG} = \frac{6}{10} = \frac{3}{5}, \text{ which does not equal } \frac{5}{7}.
\]
2. **EF = 7 cm and FG = 3 cm**
\[
\frac{EF}{FG} = \frac{7}{3}, \text{ which does not equal } \frac{5}{7}.
\]
3. **EF = 15 cm and FG = 20 cm**
\[
\frac{EF}{FG} = \frac{15}{20} = \frac{3}{4}, \text{ which does not equal } \frac{5}{7}.
\]
4. **EF = 25 cm and FG = 35 cm**
\[
\frac{EF}{FG} = \frac{25}{35} = \frac{5}{7}, \text{ which equals } \frac{5}{7}.
\]
From the above analysis, the only option that maintains the ratio of the sides consistent with the original rectangle ABCD is:
**EF = 25 cm and FG = 35 cm.**