a, `d/f=10/18=5/9`

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'`

`=>` `10(OA'+18)=18OA'`

`=>` `OA'=22,5cm`

Thay vào `(1)`. Ta có:

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

`<=>` `10/[22,5]={2}/{A'B'}`

`=>` `A'B'=4,5cm`

Đáp án:

b) \(OA' = 22,5cm\)

c) \(A'B' = 4,5cm\)

Giải thích các bước giải:

a) Hình vẽ

b) Ta có: 

\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{{10}}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' + OF'}} = \dfrac{{18}}{{OA' + 18}}\\
 \Rightarrow \dfrac{{10}}{{OA'}} = \dfrac{{18}}{{OA' + 18}} \Rightarrow OA' = 22,5cm
\end{array}\)

c) Ta có: 

\(\dfrac{{AB}}{{A'B'}} = \dfrac{{10}}{{22,5}} = \dfrac{4}{9} \Rightarrow A'B' = 4,5cm\)