Câu 10: 

Ta có: 

\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{{30}}{{18}} = \dfrac{{15}}{9} \Rightarrow A'B' = 2,7cm\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' - OF'}} = \dfrac{f}{{18 - f}}\\
 \Rightarrow \dfrac{{15}}{9} = \dfrac{f}{{18 - f}} \Rightarrow f = 11,25cm
\end{array}\)

Câu 11: 

Ta có: 

\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{{OA' + 48}}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF}}{{ - OA' + OF}} = \dfrac{{36}}{{36 - OA'}}\\
 \Rightarrow \dfrac{{OA' + 48}}{{OA'}} = \dfrac{{36}}{{36 - OA'}} \Rightarrow OA' = 24cm\\
 \Rightarrow OA = 72cm
\end{array}\)