30 jan 2018, 18:36
31 jan 2018, 18:28
31 jan 2018, 23:40
01 fev 2018, 00:13
jorgeluis Escreveu:MariaDuarte1,
se,
\(\overline{OA} =\overline{OB} =\overline{OC} = \overline{OG} = {r} {(raio.da.circunferencia)}
\overline{CP} = {3}
\overline{GP} = {4}
{\bf \overline{FP} = {1}
\overline{PE} = {x}
\overline{OP} = {y}
\overline{OE}^2 = {y}^2-{x}^2
\overline{OF}^2 = {y}^2-1 }\)
então, podemos extrair 2 triângulos retangulos, cujas hipotenusas é o raio da circunferência. Assim,
em,
\(\Delta\overline{OCE}:
(\overline{OC})^2=(\overline{CE})^2+(\overline{OE})^2
(\overline{OC})^2=(\overline{CP}+\overline{PE})^2+(\overline{OE})^2
(r)^2=(3+x)^2+{\bf y^2-x^2}
(r)^2={\bf 6x+9+y^2}\)
e
\(\Delta\overline{OGF}:
(\overline{OG})^2=(\overline{GF})^2+(\overline{OF})^2
(\overline{OG})^2=(\overline{GP}+\overline{PF})^2+(\overline{OF})^2
(r)^2=(4+1)^2+{\bf y^2-1}
(r)^2={\bf 24+y^2}\)
logo,
\({\bf 6x+9+y^2=24+y^2}
6x=15
x=\frac{5}{2}\)