Question
Consider ΔABC , with vertices A(0,3) , B(0,1) , and C(−4,2) , and ΔDEF , with vertices D(0,6) , E(0,2) , and F(8,4) . Which transformations could be used to show that the triangles are similar?(1 point)ResponsesΔABC is dilated by a scale factor of 12 and is reflected across the x-axis.triangle upper A upper B upper C is dilated by a scale factor of Start Fraction 1 over 2 End Fraction and is reflected across the x -axis.ΔABC is dilated by a scale factor of −2 and is reflected across the y-axis.triangle upper A upper B upper C is dilated by a scale factor of negative 2 and is reflected across the y -axis.ΔABC is dilated by a scale factor of 2 and is reflected across the y-axis.triangle upper A upper B upper C is dilated by a scale factor of 2 and is reflected across the y -axis.ΔABC is dilated by a scale factor of 14 and is reflected across the y-axis.triangle upper A upper B upper C is dilated by a scale factor of Start Fraction 1 over 4 End Fraction and is reflected across the y -axis.
Answers
GPT-4o mini
To determine if triangles \( \Delta ABC \) and \( \Delta DEF \) are similar and which transformations could be applied to show this, let's first gather the necessary information about both triangles.
**Vertices of \( \Delta ABC \):**
- A(0, 3)
- B(0, 1)
- C(-4, 2)
**Vertices of \( \Delta DEF \):**
- D(0, 6)
- E(0, 2)
- F(8, 4)
Now, let's calculate the side lengths of both triangles.
### Side Lengths of \( \Delta ABC \):
1. \( AB \):
\[
AB = \sqrt{(0-0)^2 + (3-1)^2} = \sqrt{0 + 4} = 2
\]
2. \( BC \):
\[
BC = \sqrt{(0-(-4))^2 + (1-2)^2} = \sqrt{(4)^2 + (-1)^2} = \sqrt{16+1} = \sqrt{17}
\]
3. \( CA \):
\[
CA = \sqrt{(0-(-4))^2 + (3-2)^2} = \sqrt{(4)^2 + (1)^2} = \sqrt{16+1} = \sqrt{17}
\]
**Side lengths of \( \Delta ABC \)** are:
- \( AB = 2 \)
- \( BC = \sqrt{17} \)
- \( CA = \sqrt{17} \)
### Side Lengths of \( \Delta DEF \):
1. \( DE \):
\[
DE = \sqrt{(0-0)^2 + (6-2)^2} = \sqrt{0 + 16} = 4
\]
2. \( EF \):
\[
EF = \sqrt{(0-8)^2 + (2-4)^2} = \sqrt{(-8)^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68}
\]
3. \( FD \):
\[
FD = \sqrt{(0-8)^2 + (6-4)^2} = \sqrt{(-8)^2 + (2)^2} = \sqrt{64 + 4} = \sqrt{68}
\]
**Side lengths of \( \Delta DEF \)** are:
- \( DE = 4 \)
- \( EF = \sqrt{68} \)
- \( FD = \sqrt{68} \)
### Ratios of Corresponding Side Lengths:
Now, let's see if we get a consistent ratio between the side lengths when comparing triangles.
1. Ratio of corresponding sides \( AB \) and \( DE \):
\[
\text{Ratio} = \frac{AB}{DE} = \frac{2}{4} = \frac{1}{2}
\]
2. Ratio of corresponding sides \( BC \) and \( EF \):
\[
\text{Ratio} = \frac{BC}{EF} = \frac{\sqrt{17}}{\sqrt{68}} = \frac{\sqrt{17}}{2\sqrt{17}} = \frac{1}{2}
\]
3. Ratio of corresponding sides \( CA \) and \( FD \):
\[
\text{Ratio} = \frac{CA}{FD} = \frac{\sqrt{17}}{\sqrt{68}} = \frac{\sqrt{17}}{2\sqrt{17}} = \frac{1}{2}
\]
All corresponding side lengths have the same ratio of \( \frac{1}{2} \), indicating that \( \Delta ABC \) and \( \Delta DEF \) are indeed similar.
### Transformations:
Given that the triangles are similar with a ratio of \( 1:2 \), the best transformation would be to dilate \( \Delta ABC \) by a scale factor of \( 2 \).
The correct transformation is therefore:
**ΔABC is dilated by a scale factor of \( 2 \) and is reflected across the y-axis.**
This option matches with the change needed to achieve the similarity correctly. All other options either propose incorrect scaling factors or incorrect reflections.
**Vertices of \( \Delta ABC \):**
- A(0, 3)
- B(0, 1)
- C(-4, 2)
**Vertices of \( \Delta DEF \):**
- D(0, 6)
- E(0, 2)
- F(8, 4)
Now, let's calculate the side lengths of both triangles.
### Side Lengths of \( \Delta ABC \):
1. \( AB \):
\[
AB = \sqrt{(0-0)^2 + (3-1)^2} = \sqrt{0 + 4} = 2
\]
2. \( BC \):
\[
BC = \sqrt{(0-(-4))^2 + (1-2)^2} = \sqrt{(4)^2 + (-1)^2} = \sqrt{16+1} = \sqrt{17}
\]
3. \( CA \):
\[
CA = \sqrt{(0-(-4))^2 + (3-2)^2} = \sqrt{(4)^2 + (1)^2} = \sqrt{16+1} = \sqrt{17}
\]
**Side lengths of \( \Delta ABC \)** are:
- \( AB = 2 \)
- \( BC = \sqrt{17} \)
- \( CA = \sqrt{17} \)
### Side Lengths of \( \Delta DEF \):
1. \( DE \):
\[
DE = \sqrt{(0-0)^2 + (6-2)^2} = \sqrt{0 + 16} = 4
\]
2. \( EF \):
\[
EF = \sqrt{(0-8)^2 + (2-4)^2} = \sqrt{(-8)^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68}
\]
3. \( FD \):
\[
FD = \sqrt{(0-8)^2 + (6-4)^2} = \sqrt{(-8)^2 + (2)^2} = \sqrt{64 + 4} = \sqrt{68}
\]
**Side lengths of \( \Delta DEF \)** are:
- \( DE = 4 \)
- \( EF = \sqrt{68} \)
- \( FD = \sqrt{68} \)
### Ratios of Corresponding Side Lengths:
Now, let's see if we get a consistent ratio between the side lengths when comparing triangles.
1. Ratio of corresponding sides \( AB \) and \( DE \):
\[
\text{Ratio} = \frac{AB}{DE} = \frac{2}{4} = \frac{1}{2}
\]
2. Ratio of corresponding sides \( BC \) and \( EF \):
\[
\text{Ratio} = \frac{BC}{EF} = \frac{\sqrt{17}}{\sqrt{68}} = \frac{\sqrt{17}}{2\sqrt{17}} = \frac{1}{2}
\]
3. Ratio of corresponding sides \( CA \) and \( FD \):
\[
\text{Ratio} = \frac{CA}{FD} = \frac{\sqrt{17}}{\sqrt{68}} = \frac{\sqrt{17}}{2\sqrt{17}} = \frac{1}{2}
\]
All corresponding side lengths have the same ratio of \( \frac{1}{2} \), indicating that \( \Delta ABC \) and \( \Delta DEF \) are indeed similar.
### Transformations:
Given that the triangles are similar with a ratio of \( 1:2 \), the best transformation would be to dilate \( \Delta ABC \) by a scale factor of \( 2 \).
The correct transformation is therefore:
**ΔABC is dilated by a scale factor of \( 2 \) and is reflected across the y-axis.**
This option matches with the change needed to achieve the similarity correctly. All other options either propose incorrect scaling factors or incorrect reflections.