Đáp án:
a) \(\dfrac{{20}}{3}cm\) và \(\dfrac{4}{3}cm\)
b) 12cm và 8cm
c) 4cm và 4cm
Giải thích các bước giải:
a) Ta có:
\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{{10}}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' - OF'}} = \dfrac{4}{{OA' - 4}}\\
\Rightarrow \dfrac{{10}}{{OA'}} = \dfrac{4}{{OA' - 4}} \Rightarrow OA' = \dfrac{{20}}{3}cm\\
\Rightarrow A'B' = \dfrac{4}{3}cm
\end{array}\)
b) Ta có:
\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{3}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' + OF'}} = \dfrac{4}{{OA' + 4}}\\
\Rightarrow \dfrac{3}{{OA'}} = \dfrac{4}{{OA' + 4}} \Rightarrow OA' = 12cm\\
\Rightarrow A'B' = 8cm
\end{array}\)
c) Ta có:
\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{2}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' + OF'}} = \dfrac{4}{{OA' + 4}}\\
\Rightarrow \dfrac{2}{{OA'}} = \dfrac{4}{{OA' + 4}} \Rightarrow OA' = 4cm\\
\Rightarrow A'B' = 4cm
\end{array}\)
