Đáp án:

b) \(OA' = 8cm\)

c) \(\dfrac{8}{3}cm\)

Giải thích các bước giải:

a) Hình vẽ

b) Ta có: 

\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{{24}}{{OA'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF}}{{ - OA' + OF}} = \dfrac{{12}}{{12 - OA'}}\\
 \Rightarrow \dfrac{{24}}{{OA'}} = \dfrac{{12}}{{12 - OA'}} \Rightarrow OA' = 8cm
\end{array}\)

c) Ta có: 

\(\dfrac{{AB}}{{A'B'}} = \dfrac{{24}}{8} = 3 \Rightarrow A'B' = \dfrac{8}{3}cm\)

a, `d/f=24/12=2/1`

`=>` OA=2 ô li

OF=1 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(OF'-OA')=OA'.OF'`

`=>` `24(12-OA')=12OA'`

`=>` `OA'=8cm`

Thay vào `(1)`. Ta có:

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

`<=>` `24/8={8}/{A'B'}`

`=>` `A'B'=8/3~~2,667cm`