Ensino Fundamental ⇒ Triângulo Tópico resolvido

[botelho]

Em um triângulo ABC reto em B. A medida do < ACB=15°, exteriormente ao dito triângulo traça um triângulo equilátero ADC. Calcular a distância do baricentro do dito triângulo equilátero a BC, se:AB+BC=36.
a)8
b)9
c)10
d)11
e)12

Resposta

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e

[petras]

botelho,


[tex3]AB=AC sen15\\
BC=AC cos15⇒AB+BC=AC(sen15+cos15)=36\implies AC=\frac{36}{sen15+cos15}(I)\\
AG=GC = \frac{2DH}{3} \implies GH=\frac{DH}{3} \\
\triangle GHA :AG^2=AH^2+GH^2⇒(\frac{2DH}{3})^2=(\frac{AC}{2})^2+(\frac{DH}{3})2\\
DH=\frac{AC\sqrt3}{2} \implies De(I):DH=\frac{18\sqrt3}{sen15+cos15}\\
GE=GC sen45=(\frac{2DH}{3}).sen45=\frac{6\sqrt6}{sen15+cos15} = \frac{6\sqrt6}{\frac{\sqrt6-\sqrt2}{4}+\frac{\sqrt6+\sqrt2}{4}}=\\
\frac{24\sqrt6}{2\sqrt6}=\boxed{12}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]AB=AC sen15\\
BC=AC cos15⇒AB+BC=AC(sen15+cos15)=36\implies AC=\frac{36}{sen15+cos15}(I)\\
AG=GC = \frac{2DH}{3} \implies GH=\frac{DH}{3} \\
\triangle GHA :AG^2=AH^2+GH^2⇒(\frac{2DH}{3})^2=(\frac{AC}{2})^2+(\frac{DH}{3})2\\
DH=\frac{AC\sqrt3}{2} \implies De(I):DH=\frac{18\sqrt3}{sen15+cos15}\\
GE=GC sen45=(\frac{2DH}{3}).sen45=\frac{6\sqrt6}{sen15+cos15} = \frac{6\sqrt6}{\frac{\sqrt6-\sqrt2}{4}+\frac{\sqrt6+\sqrt2}{4}}=\\
\frac{24\sqrt6}{2\sqrt6}=\boxed{12}[/tex3]

(Solução:Delmar-adaptada)