Hei! :)
Datele problemei:
- ΔABC dreptunghic in A, A=90°
- AD⊥BC
- AD= 24 cm
- DC= 32 cm
Rezolvare :
Fie Δ ADC dreptunghic in D, D=90° (deoarece AD⊥BC) => din Teorema lui Pitagora
=> [tex]AC^{2} =AD^{2} +DC^{2} \\AC^{2} =24^{2} +32^{2} \\AC^{2} =576+1024\\AC^{2} =1600\\AC=40 cm[/tex]
Fie Δ ABC dreptunghic in A, A=90°
AD- inaltime => din Teorema inaltimii
[tex]AD^{2} =BD *CD\\24^{2} =BD*32 \\576=BD*32 \\BD=576:32=18[/tex]
=> [tex]BC=BD+DC=18+32=50 cm[/tex]
Fie Δ ABC dreptunghic in A, A=90° => din Teorema lui Pitagora
[tex]BC^{2} =AB^{2} +AC^{2} \\50^{2} =AB^{2}+40^{2} \\2500=AB^{2}+1600\\AB^{2}=2500-1600=900\\=> AB= 30 cm[/tex]
