Đáp án:
$d'=24cm$
$h'=1cm$
Giải thích các bước giải:
\(\left\{ \begin{array}{l}OF=OF'=f=12cm\\AO=d=24cm\\AB=h=1cm\\A'O=d'=...cm\\A'B'=h'=...cm\end{array} \right.\)
$∆ABO\sim∆A'B'O(g.g)$
$→\dfrac{AB}{A'B'}=\dfrac{AO}{A'O}(1)$
$∆OIF'\sim∆A'B'F'(g.g)$
$→\dfrac{OI}{A'B'}=\dfrac{OF'}{A'F'}$
Có: $OI=AB;A'F'=A'O-OF'$
$→\dfrac{AB}{A'B'}=\dfrac{OF'}{A'O-OF'}(2)$
Từ (1) và (2) suy ra:
$\dfrac{AO}{A'O}=\dfrac{OF'}{A'O-OF'}$
$→\dfrac{24}{A'O}=\dfrac{12}{A'O-12}$
$→A'O=24cm$
$AO=A'O=24cm→A'B'=AB=1cm$