Đáp án:
* TKHT: \(OA' = 30cm\) ; \(A'B' = 2,5cm\)
* TKPK: \(OA' = 10cm\) ; \(A'B' = \dfrac{5}{6}cm\)
Giải thích các bước giải:
* TH1: TKHT
Ta có:
\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{{30}}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' - OF'}} = \dfrac{{15}}{{OA' - 15}}\\
\Rightarrow \dfrac{{30}}{{OA'}} = \dfrac{{15}}{{OA' - 15}} \Rightarrow OA' = 30cm \Rightarrow A'B' = 2,5cm
\end{array}\)
* TH2: TKPK
Ta có:
\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{{30}}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OF' - OA'}} = \dfrac{{15}}{{15 - OA'}}\\
\Rightarrow \dfrac{{30}}{{OA'}} = \dfrac{{15}}{{15 - OA'}} \Rightarrow OA' = 10cm\\
\Rightarrow A'B' = \dfrac{5}{6}cm
\end{array}\)
