Răspuns :
a)AD= inaltimea penpendicular peBC BD=6 DC=24
AD²=BDXBC=6X24=144
AD=√144=12 cm
ΔABD⇒AB²=AD²+DB²=144+36=180 AB=√180=6√5 CATETA
ΔADC⇒AC²=AD²+CD²=12²+24²=144+576=720 AC=√720=12√5 CATETA
b) AD=√12X16=√192=8√3
AB²=(8√3)²+12²=192+144=336 AB=√336=16√21 cateta
AC²=(8√3)²+16²=192+256=448 AC=√448=8√7 CATETA
AD²=BDXBC=6X24=144
AD=√144=12 cm
ΔABD⇒AB²=AD²+DB²=144+36=180 AB=√180=6√5 CATETA
ΔADC⇒AC²=AD²+CD²=12²+24²=144+576=720 AC=√720=12√5 CATETA
b) AD=√12X16=√192=8√3
AB²=(8√3)²+12²=192+144=336 AB=√336=16√21 cateta
AC²=(8√3)²+16²=192+256=448 AC=√448=8√7 CATETA
VEZI FIGURA
ABC este triungi dreptunghic cu unghiul A=90 si AD inaltime
aplici teorema catetei
AB²=BD·BC sau AC²=CD·CB
a)AB²=6·30=180⇒AB=6√5;
AC²=24·30⇒AC=12√5
IDENTIC REZOLVI CELELALTE PUNCTE
ABC este triungi dreptunghic cu unghiul A=90 si AD inaltime
aplici teorema catetei
AB²=BD·BC sau AC²=CD·CB
a)AB²=6·30=180⇒AB=6√5;
AC²=24·30⇒AC=12√5
IDENTIC REZOLVI CELELALTE PUNCTE
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