a, `d/f=40/30=4/3`

`=>` OA=4 ô li

OF=3 ô li

b, Xét `\triangleOAB` $\backsim$ `\triangleOA'B'`. Ta có:

`{OA}/{OA'}={AB}/{A'B'}`          `(1)`

Xét `\triangleOIF'` $\backsim$ `\triangleA'B'F'`. Ta có:

`{OF'}/{A'F'}={OI}/{A'B'}`          

Mà OI=AB

`=>` `{OF'}/{A'F'}={AB}/{A'B'}`          `(2)`

`(1)(2).` Ta có:

`{OA}/{OA'}={OF'}/{A'F'}`

`=>` `OA.A'F'=OA'.OF'`

`=>` `OA(OA'-OF')=OA'.OF'`

`=>` `40(OA'-30)=30OA'`

`=>` `OA'=120cm`

Thay vào `(1)`. Ta có:

`{OA}/{OA'}={AB}/{A'B'}`

`<=>` `40/120={10}/{A'B'}`

`=>` `A'B'=30cm`

Đáp án:

 \(A'B' = 30cm\)

Giải thích các bước giải:

a) Hình vẽ

b) Ta có: 

\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{{40}}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' - OF'}} = \dfrac{{30}}{{OA' - 30}}\\
 \Rightarrow \dfrac{{40}}{{OA'}} = \dfrac{{30}}{{OA' - 30}} \Rightarrow OA' = 120cm\\
 \Rightarrow \dfrac{{AB}}{{A'B'}} = \dfrac{{40}}{{120}} = \dfrac{1}{3} \Rightarrow A'B' = 30cm
\end{array}\)