ajutati maaaaa ex 13

Răspuns:
a) BC=75cm, AD=36cm, AB=45cm, AC=60cm;
b) se rezolva analog ...
BC=BD+CD=27+48=75, DUPA TEOREMA INALTIMEI AVEM: AD^2=BD*CD=27*48, De unde AD=RADICAL(27*48)=RADICAL(9*3*16*3)= 3*4*3=36, DECI AD=36.
a CUM FOLOSIM TEOREMA CATETEI: AB^2=BD*BC=27*75, DECI AB=RADICAL(27*75)=RADICAL(9*3*25*3)=3*5*3=45, DECI AB=45
ANALOG AFLAM CATETA AC: AC^2=CD*CB=48*75, DE UNDE AC=RADICAL(48*75)=RADICAL(16*3*25*3)=4*5*3=60, DECI AC=60
Explicație pas cu pas:
Răspuns:
a) BC=75cm, AD=36cm, AB=45cm, AC=60cm;
b) BC=50cm; BD=18cm; AD=24cm; AB=30cm;
Explicație pas cu pas:
[tex]\boxed{REZOLVARE\ a)}[/tex]
Teorema inaltimii:
AD²=BD*CD
[tex]AD=\sqrt{BD*CD}[/tex]=[tex]=\sqrt{27*48} =\sqrt{676} =36[/tex]
Calculam BC
BC=BD+CD=27+48=75
TEOREMA CATETEI:
AB²=BD*BC
AB=[tex]\sqrt{27*75} =\sqrt{2025}=45[/tex]
AC²=DC*BC
AC=[tex]\sqrt{48*75} =\sqrt{3600}=60[/tex]
[tex]\boxed{REZOLVARE\ b)}[/tex]
TEOREMA CATETEI:
AC²=DC*BC
1600=32*BC
BC=50
BD=BC-CD=50-32=18
Teorema inaltimii:
AD²=BD*CD
[tex]AD=\sqrt{18*32} =24[/tex]
TEOREMA CATETEI:
AB²=BD*BC
[tex]AB=\sqrt{18*50}=30[/tex]