Respuesta:
[tex]96[/tex]
Explicación paso a paso:
[tex]Radio: r = 8[/tex]
[tex]AC = b = 40[/tex]
[tex]AB =c= ?[/tex]
[tex]BC =a = ?[/tex]
[tex]c+a= b + 2r[/tex]
[tex]c+a = 40 +2(8)[/tex]
[tex]c+a = 40 +16[/tex]
[tex]c +a = 56, entonces: a = 56-c[/tex]
Por Pitágoras:
[tex]b^{2} = c^{2} +a^{2}[/tex]
[tex](40)^{2} = c^{2} +(56-c)^{2}[/tex]
[tex]1600=c^{2} + (56)^{2} -2(56)(c)+(c)^{2}[/tex]
[tex]1600= c^{2} +3136-112c+c^{2}[/tex]
[tex]2c^{2} -112c+3136-1600=0[/tex]
[tex]2c^{2} -112c+ 1536=0[/tex]
Dividiendo por " 2 ".
[tex]c^{2} -56c+768=0[/tex]
Por el método de factorización:
[tex](c-32)(c-24) =0[/tex]
[tex]c-32=0 ; c-24=0[/tex]
[tex]c_{1} = 32 ; c_{2} = 24[/tex]
Sustituyendo en:
[tex]a _{1} = 56 -c_{1}[/tex]
[tex]a _{1} = 56-32[/tex]
[tex]a _{1} = 24[/tex]
[tex]a_{2} = 56 -24[/tex]
[tex]a_{2} = 32[/tex]
Las longitudes de los lados son:
[tex]BC= a = 32[/tex] ; [tex]AB = c = 24[/tex] ; [tex]AC = b = 40[/tex]
Perímetro del triángulo:
[tex]P = ?[/tex]
[tex]P = a+b+c[/tex]
[tex]P = 32+40+24[/tex]
[tex]P= 96[/tex]
RESPUESTA:
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Respuesta:
[tex]96[/tex]
Explicación paso a paso:
[tex]Radio: r = 8[/tex]
[tex]AC = b = 40[/tex]
[tex]AB =c= ?[/tex]
[tex]BC =a = ?[/tex]
[tex]c+a= b + 2r[/tex]
[tex]c+a = 40 +2(8)[/tex]
[tex]c+a = 40 +16[/tex]
[tex]c +a = 56, entonces: a = 56-c[/tex]
Por Pitágoras:
[tex]b^{2} = c^{2} +a^{2}[/tex]
[tex](40)^{2} = c^{2} +(56-c)^{2}[/tex]
[tex]1600=c^{2} + (56)^{2} -2(56)(c)+(c)^{2}[/tex]
[tex]1600= c^{2} +3136-112c+c^{2}[/tex]
[tex]2c^{2} -112c+3136-1600=0[/tex]
[tex]2c^{2} -112c+ 1536=0[/tex]
Dividiendo por " 2 ".
[tex]c^{2} -56c+768=0[/tex]
Por el método de factorización:
[tex](c-32)(c-24) =0[/tex]
[tex]c-32=0 ; c-24=0[/tex]
[tex]c_{1} = 32 ; c_{2} = 24[/tex]
Sustituyendo en:
[tex]a _{1} = 56 -c_{1}[/tex]
[tex]a _{1} = 56-32[/tex]
[tex]a _{1} = 24[/tex]
[tex]a_{2} = 56 -24[/tex]
[tex]a_{2} = 32[/tex]
Las longitudes de los lados son:
[tex]BC= a = 32[/tex] ; [tex]AB = c = 24[/tex] ; [tex]AC = b = 40[/tex]
Perímetro del triángulo:
[tex]P = ?[/tex]
[tex]P = a+b+c[/tex]
[tex]P = 32+40+24[/tex]
[tex]P= 96[/tex]
RESPUESTA:
[tex]96[/tex]