Правоъгълен триъгълник

[Гост]

Дадено:AB=88dm, BD=33dm, CF=15dm.Търси се:AC

[Евва]

Нека CD=y и AC=x
AC=x=?

[tex]S_{ABD }[/tex]=[tex]S_{ABC }[/tex]+[tex]S_{CBD }[/tex]

[tex]\frac{AD.BD}{2}[/tex]=[tex]\frac{AB.CF}{2}[/tex]+[tex]\frac{CD.BD}{2}[/tex]

[tex]\frac{(x+y)33}{2}[/tex]=[tex]\frac{88.15}{2}[/tex]+[tex]\frac{y.33}{2}[/tex]

[tex]\frac{33x+33y}{2}[/tex]=[tex]\frac{1 320+33y}{2}[/tex]

33x+33y=1 320+33y
x=1 320:33
x=40 dm.

[Евва]

(2 начин)

[tex]S_{ABC }[/tex]=[tex]\frac{AC.BD}{2}[/tex] и [tex]S_{ABC }[/tex]=[tex]\frac{AB.CF}{2}[/tex]
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[tex]\frac{AC.BD}{2}[/tex]=[tex]\frac{AB.CF}{2}[/tex]

AC.33=88.15

AC=40 dm.