Răspuns:
ΔABC isoscel
AB=AC , AB=15cm, BC=24cm
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[tex]h_{A} ,h_{B} ,h_{C} =?[/tex]
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In ΔABC isoscel construim [tex]h_{A}[/tex]⊥ BC, [tex]h_{A}[/tex]∩BC={D} ⇒ ΔADC si ΔADB dreptunghice , D mij lat BC ⇒BD=CD=12cm
In ΔADB dreptunghic, m∡ADB=90°, conform teoremei lui Pitagora ⇒
⇒AB²=AD²+BD² ⇒ AD²=AB²-BD²
AD²=15²-12²=225-144=81
AD=√81= 9cm
[tex]A_{ABC} =\frac{B*h}{2}= \frac{BC*AD}{2}=\frac{24*9}{2}=12*9= 108cm^{2}[/tex]
[tex]A_{ABC}=\frac{AC*h_{B} }{2} =\frac{AB*h_{C} }{2}[/tex]
[tex]108=\frac{15*h_{B} }{2} \\216=15*h_{B}\\ h_{B}=\frac{216}{15}=14,4cm[/tex]
Cum ΔABC este isoscel ⇒[tex]h_{B} =h_{C}=14,4 cm[/tex]
