Question

Vă rog !!! Dau coroana + 50 de puncte !​

Answer 1

ΔABC-isoscel, BC-baza ⇒AB≡AC

raza cercului circumscris triunghiului este:

$r=\frac{abc}{4A}$

unde a=BC. b=AC, c=AB si A=aria

a)

$BC=8cm$

$AC=4\sqrt{2} cm=AB$

$A=\sqrt{p(p-a)(p-b)(p-c)}$

$p=\frac{P}{2} =\frac{AB+BC+AC}{2} =\frac{8+4\sqrt{2}+4\sqrt{2} }{2} =4+4\sqrt{2}$

$A=\sqrt{4+4\sqrt{2} (4+4\sqrt{2}-8)(4+4\sqrt{2}-4\sqrt{2} )(4+4\sqrt{2} -4\sqrt{2}) }$

$A=\sqrt{4+4\sqrt{2}(-4+4\sqrt{2} )*4*4 }$

$A=\sqrt{4+4^2*4\sqrt{2} (-4+4\sqrt{2} )}$

$A=\sqrt{4+64\sqrt{2}(-4+4\sqrt{2}) }$

$A=2\sqrt{1+16\sqrt{2}(-4+4\sqrt{2} ) }$

$A=2\sqrt{1-64\sqrt{2}+128 }$

$A=2\sqrt{129-64\sqrt{2} }$

$r=\frac{abc}{4A} =\frac{8*4\sqrt{2}*4\sqrt{2} }{4(2\sqrt{129-64\sqrt{2} } )}$ de aici calculezi

b)

$BC=12cm$

$AB=10cm=AC$

$A=\sqrt{p(p-a)(p-b)(p-c)}$

$p=\frac{P}{2} =\frac{AB+BC+AC}{2} =\frac{12+10+10}{2} =\frac{32}{2} =16$

$A=\sqrt{16(16-12)(16-10)(16-10)}$

$A=\sqrt{16*4*6*6}$

$A=4*2*6=48cm^2$

$r=\frac{abc}{4A} =\frac{12*10*10}{4*48} =\frac{1200}{192} =6,25cm$

c)

$BC=24cm$

$AC=15cm=AB$

$A=\sqrt{p(p-a)(p-b)(p-c)}$

$p=\frac{P}{2} =\frac{AB+BC+AC}{2} =\frac{24+15+15}{2} =\frac{54}{2} =27$

$A=\sqrt{27(27-24)(27-15)(27-15)}$

$A=\sqrt{27*3*12*12}$

$A=\sqrt{81*12*12}$

$A=9*12=108cm^2$

$r=\frac{abc}{4A} =\frac{24*15*15}{4*108} =\frac{5400}{432} =12,5cm$

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