Respuesta :
The dimensions of triangle ΔDEF can be obtained from ΔABC by applying a scale factor of dilation of [tex]\displaystyle \frac{1}{2}[/tex].
- The conclusion which is true about the triangles is D. They are similar by the definition of similarity in terms of dilation.
Reasons:
The given vertices of the triangles are;
ΔABC: A(1, 5), B(3, 9), C(5, 3)
ΔDEF: D(-3, 3), E(-2, 5), and F(-1, 2)
The lengths of the sides of ΔABC are
AB = √((1 - 3)² + (5 - 9)²) = √20 = 2·√5
BC = √((5 - 3)² + (3 - 9)²) = √40 = 2·√10
AC = √((1 - 5)² + (5 - 3)²) = √20 = 2·√10
The lengths of the sides of ΔDEF are;
DE = √((-3 - (-2))² + (3 - 5)²) = √5
EF = FE = √((-1 - (-2))² + (2 - 5)²) = √10
DF = √((-3 - (-1))² + (3 - 2)²) = √5
Therefore;
[tex]\displaystyle The \ dilation \ that \ gives \ \Delta ABC = \frac{\overline {EF}}{\overline {BC}} = \frac{\overline {DE}}{\overline {AB}} = \frac{\overline {DF}}{\overline {AC}} = \frac{\sqrt{10} }{2\cdot \sqrt{10} } = \frac{1}{2}[/tex]
The ratio of the sides of ΔABC to the corresponding sides of ΔDEF are equal to [tex]\displaystyle \frac{1}{2}[/tex].
Which gives;
ΔDEF is a dilation of ΔABC, and ΔDEF and ΔABC are similar by the
definition of similarity in terms of a dilation.
Learn more here:
https://brainly.com/question/14380197
Answer:
Step-by-step explanation:
They are congruent by the definition of congruence in terms of rigid motions.