[tex]BC=\frac{4AB}{3}[/tex]
AB=AC=24 cm
⇒BC=4×8=32 cm
a)
Cand cunoastem toate cele trei laturi ale triunghiului putem folosi formula lui Heron
[tex]A=\sqrt{p(p-a)(p-b)(p-c)}[/tex]
p=semiperimetrul, iar a,b si c sunt laturile triunghiului
[tex]p=\frac{24+24+32}{2}= 40\ cm\\\\A=\sqrt{40\cdot 16\cdot 16\cdot 8}= 128\sqrt{5}\ cm^2[/tex]
b)
Fie BE⊥AC
d(B,AC)=BE
Folosim aria in doua moduri:
[tex]A=\frac{BE\cdot AC}{2}\\\\ 128\sqrt{5} =\frac{BE\cdot 24}{2} \\\\BE=\frac{32\sqrt{5} }{3} \ cm[/tex]
c)
Raza cercului inscris:
[tex]r=\frac{A}{p} =\frac{128\sqrt{5} }{40}=\frac{16\sqrt{5} }{5}[/tex]
Fie AD⊥BC, AD inaltime si bisectoare
I∈AD
DC=32:2=16 cm
[tex]A=\frac{AD\cdot BC}{2}\\\\ 128\sqrt{5}=AD\cdot 16\\\\ AD=8\sqrt{5} \ cm[/tex]
ΔAIN~ΔADC
⇒ [tex]\frac{AI}{AD} =\frac{IN}{DC}[/tex]
ID=raza=r
[tex]AI=AD-ID=8\sqrt{5} -\frac{16\sqrt{5} }{5} =\frac{24\sqrt{5} }{5}[/tex]
[tex]\frac{\frac{24\sqrt{5} }{5} }{8\sqrt{5} } =\frac{IN}{16}\\\\ IN=\frac{\frac{24\sqrt{5} }{5} \cdot 16}{8\sqrt{5} }=\frac{48}{5} \\\\ MN=2IN=\frac{96}{5} \ cm[/tex]
Un exercitiu de geometrie gasesti aici: https://brainly.ro/tema/4720633
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