Answer:
D. [tex]\sf A'C' = 1 ,\quad m \angle A' = 37^\circ [/tex]
Step-by-step explanation:
To find the length of [tex]\sf A'C' [/tex] after the dilation and the measure of [tex]\sf \angle A'' [/tex], let's first understand the concept of dilation.
When a triangle is dilated by a factor [tex]\sf k [/tex], all its side lengths are multiplied by [tex]\sf k [/tex] and all its angles remain the same.
Given:
- [tex]\sf m \angle A = 37^\circ [/tex]
- [tex]\sf AC = 5 [/tex]
- The triangle is dilated by a factor of [tex]\sf \dfrac{1}{5} [/tex] to produce [tex]\sf \triangle A'B'C' [/tex]
Length of [tex]\sf A'C' [/tex]:
Since the triangle is dilated by a factor of [tex]\sf \dfrac{1}{5} [/tex], the length of [tex]\sf A'C' [/tex] will be [tex]\sf \dfrac{1}{5} [/tex] of the length of [tex]\sf AC [/tex]:
[tex]\sf A'C' = \dfrac{1}{5} \times AC \\\\ = \dfrac{1}{5} \times 5 \\\\ = 1 [/tex]
So, the length of [tex]\sf A'C' [/tex] after dilation is 1.
Measure of [tex]\sf m\angle A' [/tex]:
The measure of [tex]\sf \angle A'' [/tex] will remain the same as [tex]\sf m \angle A [/tex] because angles do not change during dilation.
Therefore, [tex]\sf m \angle A' = 37^\circ [/tex].
So, the answer is:
D. [tex]\sf A'C' = 1 ,\quad m \angle A' = 37^\circ [/tex]
