Răspuns:
Explicație pas cu pas:
Ex5. Fie BC=a, AC=b, MP=MN=x, AM=40, BM=30.
ΔABC~ΔAMP ⇒a/x=70/40, ⇒a=(7/4)·x
ΔABC~ΔMBN ⇒b/x=70/30, ⇒b=(7/3)·x
a²+b²=70² ⇒(7/4)²x²+(7/3)²x²=70² ⇒
[tex]7^{2}*(\frac{1}{4^{2}}+\frac{1}{3^{2}}) }x^{2}=7^{2}*10^{2},~~ \frac{3^{2}+4^{2}}{3^{2}*4^{2}} x^{2}=10^{2}~~x^{2}=\frac{10^{2}*3^{2}*4^{2}}{5^{2}}[/tex]
Deci x=(10·3·4)/5=24. Atunci a=(7/4)·24=7·6=42 cm, iar b=(7/3)·24=56cm.
Ex6. Fie AD=4,5cm, BE=6cm, AD∩BE=O. Deoarece O punct de inters a medianelor ⇒AO=3cm, DO=1,5cm, BO=4cm, EO=2cm. Deoarece AD⊥BE ⇒Dun ΔABO, AB²=3²+4²=5², deci AB=5cm.
Din ΔAOE, ⇒AE²=3²+2²=13, deci AE=√13 ⇒AC=2√13cm
Din ΔBDO ⇒BD²=4²+(1,5)²=16+2,25=18,25. Deci BD=√(18,25) ⇒BC=2√(18,25) =√(4·18,25)=√73cm
