Question

Under a certain transformation T, AA'BC' is the image of AABC. The perimeter of AA'B'C' is twice the perimeter of ABC. What kind of transformation is T?

Answer 1

The dimensions of the image are twice the dimensions of the preimage,

which indicates that the transformation is a dilation.

• The transformation T is; a dilation transformation with a scale factor of 2, D₂

Reasons:

The transformation applied to ΔABC = T

The image of ΔABC following the transformation, T = ΔA'B'C'

The perimeter of ΔA'B'C' = Twice the perimeter of ΔABC

Therefore, we have;

A'B' + B'C' + A'C' = 2 × (AB + BC + AC)

Which gives

A'B' + B'C' + A'C' = 2·AB + 2·BC + 2·AC

By similar triangles, we have the ratio of corresponding sides as follows;

$\displaystyle \frac{AB}{A'B'} = \frac{BC}{B'C'} = \mathbf{ \frac{AC}{A'C'}}$

Which gives;

$\displaystyle \frac{AB}{A'B'} = \frac{AB}{2 \cdot AB} = \frac{1}{2}$

A'B' = 2·AB

Therefore;

• Triangle ΔA'B'C' is twice the dimension of triangle ΔABC, and T is a dilation transformation with a scale factor of 2, D₂

Learn more about dilation transformation here:

brainly.com/question/2458912