Câu 1:
Ta có:
\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{18}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF}}{{ - OA' + OF}} = \dfrac{{18}}{{18 - OA'}}\\
\Rightarrow \dfrac{18}{{OA'}} = \dfrac{{18}}{{18 - OA'}} \Rightarrow OA' = 9cm
\end{array}\)
Câu 2:
Ta có:
\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{f}{{d'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF}}{{ - OA' + OF}} = \dfrac{f}{{f - d'}}\\
\Rightarrow \dfrac{f}{{d'}} = \dfrac{f}{{f - d'}} \Rightarrow d' = \dfrac{f}{2}\\
\Rightarrow A'B' = \dfrac{{AB}}{2} = 1,5cm
\end{array}\)
Câu 3:
Ta có:
\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{{20}}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' + OF'}} = \dfrac{{40}}{{OA' + 40}}\\
\Rightarrow \dfrac{{20}}{{OA'}} = \dfrac{{40}}{{OA' + 40}} \Rightarrow OA' = 40cm
\end{array}\)