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Sagot :
To solve this problem, we need to identify the scale factor used to dilate triangle [tex]\( ABC \)[/tex] to form triangle [tex]\( A'B'C' \)[/tex].
Given:
- Original triangle [tex]\( ABC \)[/tex] has vertices at [tex]\( A(-3, -3) \)[/tex], [tex]\( B(3, 3) \)[/tex], and [tex]\( C(0, 3) \)[/tex].
- Dilated triangle [tex]\( A'B'C' \)[/tex] has vertices at [tex]\( A'(-15, -15) \)[/tex], [tex]\( B'(15, 15) \)[/tex], and [tex]\( C'(0, 15) \)[/tex].
The scale factor (SF) can be determined by comparing the distances from the origin of each corresponding pair of vertices of the original and dilated triangles.
1. Vertices [tex]\( A \)[/tex] and [tex]\( A' \)[/tex]:
- Distance from origin to [tex]\( A \)[/tex]: [tex]\[ \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \][/tex]
- Distance from origin to [tex]\( A' \)[/tex]: [tex]\[ \sqrt{(-15)^2 + (-15)^2} = \sqrt{225 + 225} = \sqrt{450} = 15\sqrt{2} \][/tex]
- Scale factor from [tex]\( A \)[/tex] to [tex]\( A' \)[/tex]: [tex]\[ \text{SF}_A = \frac{15\sqrt{2}}{3\sqrt{2}} = 5 \][/tex]
2. Vertices [tex]\( B \)[/tex] and [tex]\( B' \)[/tex]:
- Distance from origin to [tex]\( B \)[/tex]: [tex]\[ \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \][/tex]
- Distance from origin to [tex]\( B' \)[/tex]: [tex]\[ \sqrt{15^2 + 15^2} = \sqrt{225 + 225} = \sqrt{450} = 15\sqrt{2} \][/tex]
- Scale factor from [tex]\( B \)[/tex] to [tex]\( B' \)[/tex]: [tex]\[ \text{SF}_B = \frac{15\sqrt{2}}{3\sqrt{2}} = 5 \][/tex]
3. Vertices [tex]\( C \)[/tex] and [tex]\( C' \)[/tex]:
- Distance from origin to [tex]\( C \)[/tex]: [tex]\[ \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \][/tex]
- Distance from origin to [tex]\( C' \)[/tex]: [tex]\[ \sqrt{0^2 + 15^2} = \sqrt{225} = 15 \][/tex]
- Scale factor from [tex]\( C \)[/tex] to [tex]\( C' \)[/tex]: [tex]\[ \text{SF}_C = \frac{15}{3} = 5 \][/tex]
Based on the distances calculated, the scale factors [tex]\( \text{SF}_A \)[/tex], [tex]\( \text{SF}_B \)[/tex], and [tex]\( \text{SF}_C \)[/tex] are all consistent and equal to 5.
Thus, the scale factor used to dilate the triangle is [tex]\( \boxed{5} \)[/tex].
Given:
- Original triangle [tex]\( ABC \)[/tex] has vertices at [tex]\( A(-3, -3) \)[/tex], [tex]\( B(3, 3) \)[/tex], and [tex]\( C(0, 3) \)[/tex].
- Dilated triangle [tex]\( A'B'C' \)[/tex] has vertices at [tex]\( A'(-15, -15) \)[/tex], [tex]\( B'(15, 15) \)[/tex], and [tex]\( C'(0, 15) \)[/tex].
The scale factor (SF) can be determined by comparing the distances from the origin of each corresponding pair of vertices of the original and dilated triangles.
1. Vertices [tex]\( A \)[/tex] and [tex]\( A' \)[/tex]:
- Distance from origin to [tex]\( A \)[/tex]: [tex]\[ \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \][/tex]
- Distance from origin to [tex]\( A' \)[/tex]: [tex]\[ \sqrt{(-15)^2 + (-15)^2} = \sqrt{225 + 225} = \sqrt{450} = 15\sqrt{2} \][/tex]
- Scale factor from [tex]\( A \)[/tex] to [tex]\( A' \)[/tex]: [tex]\[ \text{SF}_A = \frac{15\sqrt{2}}{3\sqrt{2}} = 5 \][/tex]
2. Vertices [tex]\( B \)[/tex] and [tex]\( B' \)[/tex]:
- Distance from origin to [tex]\( B \)[/tex]: [tex]\[ \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \][/tex]
- Distance from origin to [tex]\( B' \)[/tex]: [tex]\[ \sqrt{15^2 + 15^2} = \sqrt{225 + 225} = \sqrt{450} = 15\sqrt{2} \][/tex]
- Scale factor from [tex]\( B \)[/tex] to [tex]\( B' \)[/tex]: [tex]\[ \text{SF}_B = \frac{15\sqrt{2}}{3\sqrt{2}} = 5 \][/tex]
3. Vertices [tex]\( C \)[/tex] and [tex]\( C' \)[/tex]:
- Distance from origin to [tex]\( C \)[/tex]: [tex]\[ \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \][/tex]
- Distance from origin to [tex]\( C' \)[/tex]: [tex]\[ \sqrt{0^2 + 15^2} = \sqrt{225} = 15 \][/tex]
- Scale factor from [tex]\( C \)[/tex] to [tex]\( C' \)[/tex]: [tex]\[ \text{SF}_C = \frac{15}{3} = 5 \][/tex]
Based on the distances calculated, the scale factors [tex]\( \text{SF}_A \)[/tex], [tex]\( \text{SF}_B \)[/tex], and [tex]\( \text{SF}_C \)[/tex] are all consistent and equal to 5.
Thus, the scale factor used to dilate the triangle is [tex]\( \boxed{5} \)[/tex].
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