Răspuns:
Explicație pas cu pas:
Desenul il ai in poza
MB=MC, fiind ΔABC isoscel
[tex]R=\frac{abc}{4A}[/tex], unde R este raza cercului circumscris
a,b,c laturile triunghiului si A aria
semiperimetru: [tex]p=\frac{a+b+c}{2}=\frac{128}{2} =64[/tex]
[tex]A=\sqrt{p(p-a)(p-b)(p-c)} =\sqrt{64\cdot24\cdot24\cdot16} =768cm^{2}[/tex]
[tex]R=\frac{40\cdot40\cdot48}{4\cdot768}=25 cm[/tex]
b. d(M,BC)=MN
AN inaltime in ΔABC⇒[tex]AN\cdot BC=2A[/tex]
AN=32cm⇒ON=AN-AO=32-25=7cm
IN ΔMON aplicam pitagora
MN²=ON²+MO²
MN²=49+576
MN=25 cm
c. d(O,(MBC))=OP
OP⊥MN
MN⊂(MBC)
[tex]A_{MON} =\frac{MO\cdot ON}{2} =\frac{OP\cdot MN}{2}[/tex]
24×7=OP×25
[tex]OP=\frac{168}{25}[/tex]
d. MN⊥BC
MN⊂(MBC)
AN⊥BC
AN⊂(ABC)⇒ ∡((MBC),(ABC))=∡ANM=∡ONM
tg∡ONM=[tex]\frac{MO}{ON} =\frac{24}{7}[/tex]
