Desenez triunghiul ABC, dreptunghic în A.
Trasez înălțimea AD, cu D pe BC.
[tex]\it ABC -dr\ \ m(\hat{A})=90^o \ \stackrel{T.Pitagora}{\Longrightarrow} BC^2=AB^2+AC^2 \Longrightarrow
\\\;\\
BC^2=12^2+16^2 =144+256=400\Longrightarrow BC=\sqrt{400} =20cm[/tex]
Determin înălțimea AD cu formula de calcul:
[tex]\it AD = \dfrac{AB\cdot AC}{BC} =\dfrac{12\cdot16}{20}=\dfrac{6\cdot16}{10} =\dfrac{96}{10}=9,6cm[/tex]
[tex]\it DAB-dr\ \ m(\hat{D})=90^o \stackrel{T.Pitagora}{\Longrightarrow} BD^2=AB^2-AD^2 \Longrightarrow
\\\;\\
\Longrightarrow BD^2 = 12^2-9,6^2 =(12-9,6)(12+0,6)=2,4\cdot21,6=
\\\;\\
=\dfrac{24}{10}\cdot\dfrac{216}{10} = \dfrac{6\cdot216}{25}= \dfrac{6\cdot6\cdot36}{25}=\dfrac{36\cdot36}{25} \Longrightarrow BD=\sqrt{\dfrac{36\cdot36}{25}} =
\\\;\\ \\\;\\
= \dfrac{^{2)}36}{\ 5} =\dfrac{72}{10}=7,2 cm[/tex]
DC = BC - BD = 20 - 7,5 = 12,5 cm.
