ΔABC-isoscel, BC-baza ⇒AB≡AC
raza cercului circumscris triunghiului este:
[tex]r=\frac{abc}{4A}[/tex]
unde a=BC. b=AC, c=AB si A=aria
a)
[tex]BC=8cm[/tex]
[tex]AC=4\sqrt{2} cm=AB[/tex]
[tex]A=\sqrt{p(p-a)(p-b)(p-c)}[/tex]
[tex]p=\frac{P}{2} =\frac{AB+BC+AC}{2} =\frac{8+4\sqrt{2}+4\sqrt{2} }{2} =4+4\sqrt{2}[/tex]
[tex]A=\sqrt{4+4\sqrt{2} (4+4\sqrt{2}-8)(4+4\sqrt{2}-4\sqrt{2} )(4+4\sqrt{2} -4\sqrt{2}) }[/tex]
[tex]A=\sqrt{4+4\sqrt{2}(-4+4\sqrt{2} )*4*4 }[/tex]
[tex]A=\sqrt{4+4^2*4\sqrt{2} (-4+4\sqrt{2} )}[/tex]
[tex]A=\sqrt{4+64\sqrt{2}(-4+4\sqrt{2}) }[/tex]
[tex]A=2\sqrt{1+16\sqrt{2}(-4+4\sqrt{2} ) }[/tex]
[tex]A=2\sqrt{1-64\sqrt{2}+128 }[/tex]
[tex]A=2\sqrt{129-64\sqrt{2} }[/tex]
[tex]r=\frac{abc}{4A} =\frac{8*4\sqrt{2}*4\sqrt{2} }{4(2\sqrt{129-64\sqrt{2} } )}[/tex] de aici calculezi
b)
[tex]BC=12cm[/tex]
[tex]AB=10cm=AC[/tex]
[tex]A=\sqrt{p(p-a)(p-b)(p-c)}[/tex]
[tex]p=\frac{P}{2} =\frac{AB+BC+AC}{2} =\frac{12+10+10}{2} =\frac{32}{2} =16[/tex]
[tex]A=\sqrt{16(16-12)(16-10)(16-10)}[/tex]
[tex]A=\sqrt{16*4*6*6}[/tex]
[tex]A=4*2*6=48cm^2[/tex]
[tex]r=\frac{abc}{4A} =\frac{12*10*10}{4*48} =\frac{1200}{192} =6,25cm[/tex]
c)
[tex]BC=24cm[/tex]
[tex]AC=15cm=AB[/tex]
[tex]A=\sqrt{p(p-a)(p-b)(p-c)}[/tex]
[tex]p=\frac{P}{2} =\frac{AB+BC+AC}{2} =\frac{24+15+15}{2} =\frac{54}{2} =27[/tex]
[tex]A=\sqrt{27(27-24)(27-15)(27-15)}[/tex]
[tex]A=\sqrt{27*3*12*12}[/tex]
[tex]A=\sqrt{81*12*12}[/tex]
[tex]A=9*12=108cm^2[/tex]
[tex]r=\frac{abc}{4A} =\frac{24*15*15}{4*108} =\frac{5400}{432} =12,5cm[/tex]
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