(15) Benzerlik Problemleri-Çözümlü

${\displaystyle \left|AL\right|=k}$ olsun
${\displaystyle K}$  ve  ${\displaystyle L}$  açıları yöndeş


${\displaystyle BKE\cong BLD}$

${\displaystyle \frac{\left|LD\right|}{\left|KE\right|}=\frac{\left|LB\right|}{\left|KB\right|}}$

${\displaystyle \Rightarrow \frac{\left|LD\right|}{2}=\frac{3k}{k}}$

${\displaystyle \Rightarrow \left|LD\right|=6}$

${\displaystyle C}$  ve  ${\displaystyle D}$  açıları yöndeş

${\displaystyle ALD\cong ABC}$

${\displaystyle \frac{\left|BC\right|}{\left|LD\right|}=\frac{\left|AB\right|}{\left|AL\right|}}$

${\displaystyle \Rightarrow \frac{x}{6}=\frac{4k}{k}}$

${\displaystyle \Rightarrow x=24}$

${\displaystyle \left|CF\right|=2k}$  diyelim

${\displaystyle B}$  ve  ${\displaystyle D}$  açıları yöndeş

${\displaystyle ABC\cong FDC}$

${\displaystyle \frac{\left|AC\right|}{\left|CF\right|}=\frac{\left|AB\right|}{\left|DF\right|}}$

${\displaystyle \Rightarrow \frac{\left|AC\right|}{2k}=\frac{7}{2}}$

${\displaystyle \Rightarrow \left|AC\right|=\mathrm{7k}}$

${\displaystyle \Rightarrow \left|EF\right|=\left|AC\right|-\left|CF\right|-\left|AE\right|}$

${\displaystyle =7k-4k}$

${\displaystyle =3k}$

${\displaystyle E}$  ve  ${\displaystyle F}$  açıları yöndeş

${\displaystyle ADF\cong AGE}$

${\displaystyle \frac{\left|GE\right|}{\left|DF\right|}=\frac{\left|AE\right|}{\left|AF\right|}}$

${\displaystyle \Rightarrow \frac{x}{2}=\frac{2k}{5k}}$

${\displaystyle x=\frac{4}{5}}$

${\displaystyle \Rightarrow x=0,8}$

${\displaystyle m\left(E\right)=m\left(C\right)}$

${\displaystyle \Rightarrow ABC\cong DBE}$

${\displaystyle A\left(ABC\right)=A\left(BED\right)+A\left(AEDC\right)}$

${\displaystyle =2A\left(BED\right)}$


Alanların oranı benzerlik oranının karesidir.


${\displaystyle \frac{\left|BC\right|}{\left|EB\right|}=\sqrt{\frac{A\left(ABC\right)}{A\left(DBE\right)}}}$

${\displaystyle \Rightarrow \frac{\left|BC\right|}{4}=\sqrt{2}}$

${\displaystyle \Rightarrow \left|BC\right|=4\sqrt{2}}$