Question

ajutati maaaaa ex 13 ​

Answer 1

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:

Answer 2

Răspuns:

a) BC=75cm, AD=36cm, AB=45cm, AC=60cm;

b) BC=50cm; BD=18cm; AD=24cm; AB=30cm;

Explicație pas cu pas:

$\boxed{REZOLVARE\ a)}$

Teorema inaltimii:

AD²=BD*CD

$AD=\sqrt{BD*CD}$=$=\sqrt{27*48} =\sqrt{676} =36$

Calculam BC

BC=BD+CD=27+48=75

TEOREMA CATETEI:

AB²=BD*BC

AB=$\sqrt{27*75} =\sqrt{2025}=45$

AC²=DC*BC

AC=$\sqrt{48*75} =\sqrt{3600}=60$

$\boxed{REZOLVARE\ b)}$

TEOREMA CATETEI:

AC²=DC*BC

1600=32*BC

BC=50

BD=BC-CD=50-32=18

Teorema inaltimii:

AD²=BD*CD

$AD=\sqrt{18*32} =24$

TEOREMA CATETEI:

AB²=BD*BC

$AB=\sqrt{18*50}=30$