To solve the problem where triangles [tex]\( ABC \)[/tex] and [tex]\( A'B'C' \)[/tex] are similar, we can use the properties of similar triangles. Similar triangles have corresponding sides that are proportional. Given that [tex]\( AB = 6 \)[/tex] feet, [tex]\( BC = 8 \)[/tex] feet, and [tex]\( B'C' = 11 \)[/tex] feet, we need to find the length of [tex]\( A'B' \)[/tex].
We start by setting up the proportion from the similar triangles. For similar triangles, the ratio of the corresponding sides of one triangle is equal to the ratio of the corresponding sides of the other triangle. Thus:
[tex]\[
\frac{AB}{A'B'} = \frac{BC}{B'C'}
\][/tex]
Substituting the known values:
[tex]\[
\frac{6}{A'B'} = \frac{8}{11}
\][/tex]
To find [tex]\( A'B' \)[/tex], we cross-multiply to solve for the unknown:
[tex]\[
6 \cdot 11 = 8 \cdot A'B'
\][/tex]
[tex]\[
66 = 8 \cdot A'B'
\][/tex]
Now, divide both sides by 8 to isolate [tex]\( A'B' \)[/tex]:
[tex]\[
A'B' = \frac{66}{8}
\][/tex]
Simplifying the fraction:
[tex]\[
A'B' = 8.25 \text{ feet}
\][/tex]
Thus, the length of [tex]\( A'B' \)[/tex] is:
[tex]\[
C) 8 \frac{1}{4} \text{ feet}
\][/tex]
