Question

Nos triângulos abaixo, DE//BC. Nessas condições, determine:

I) a medida de x:

II) o perímetro de cada triângulo:

Answer 1

Resposta:

ei cara você falou em Pi não não falou coberta de. açúcar

Answer 2

Como os segmentos DE e BC são paralelos (DE//BC), podemos afirmar que, nas duas figuras, os triângulos ABC e ADE são semelhantes.

Com isso, aplicando o Teorema de Talles, poderemos achar relações entre as medidas dos dois triângulos.

a)

Aplicando o teorema de Talles, teremos as seguintes relações:

$\boxed{\sf \dfrac{AD}{AB}~=~\dfrac{AE}{AC}}$

Substituindo as medidas dos segmentos:

$\sf \dfrac{x-1}{(x-1)+3}~=~\dfrac{x+4}{(x+4)+x}\\\\\\(x+2)\cdot (x+4)~=~(x-1)\cdot (2x+4)\\\\\\x^2~+~6x~+~8=~2x^2~+~2x~-~4\\\\\\\boxed{\sf x^2~-~4x~-~12=~0}$

Aplicando Bhaskara à equação de 2° grau:

$\sf \Delta~=~(-4)^2-4\cdot 1\cdot (-12)\\\\\Delta ~=~16~+~48\\\\\boxed{\sf \Delta ~=~64}$

$\sf x'~=~\dfrac{4+\sqrt{64}}{2\cdot 1}~=~\dfrac{4+8}{2}~=~\dfrac{12}{2}~~\Rightarrow~~\boxed{\sf x'~=~6~cm}\\\\\\\sf x''~=~\dfrac{4-\sqrt{64}}{2\cdot 1}~=~\dfrac{4-8}{2}~=~\dfrac{-4}{2}~~\Rightarrow~~\boxed{\sf x''~=\,-2~cm}$

Como as medidas de comprimento não podem ser negativas, descartaremos a raiz x''=-2, logo a resposta para o item (I) será x=6cm.

Vamos aplicar novamente o Teorema de Talles para determinar agora a medida do segmento DE no triângulo menor:

$\sf \dfrac{AD}{AB}~=~\dfrac{DE}{BC}\\\\\\\dfrac{x-1}{(x-1)+3}~=~\dfrac{DE}{8}\\\\\\\dfrac{6-1}{(6-1)+3}~=~\dfrac{DE}{8}\\\\\\\dfrac{5}{8}~=~\dfrac{DE}{8}\\\\\\8\cdot DE~=~5\cdot 8\\\\\\DE~=~\dfrac{40}{8}\\\\\\\boxed{\sf DE~=~5~cm}$

Falta calcularmos os perímetros dos triângulos:

$\sf Perimetro~Triangulo~ADE~=~AD~+~DE~+~AE\\\\\\\sf Perimetro~Triangulo~ADE~=~(6-1)~+~DE~+~(6+4)\\\\\\Perimetro~Triangulo~ADE~=~(6-1)~+~5~+~(6+4)\\\\\\Perimetro~Triangulo~ADE~=~5~+~5~+~10\\\\\\\boxed{\sf Perimetro~Triangulo~ADE~=~20~cm}$

$\sf Perimetro~Triangulo~ABC~=~AB~+~BC~+~AC\\\\\\\sf Perimetro~Triangulo~ABC~=~(x-1)~+~3~+~BC~+~x~+~(x+4)\\\\\\\sf Perimetro~Triangulo~ABC~=~(6-1)~+~3~+~8~+~6~+~(6+4)\\\\\\\sf Perimetro~Triangulo~ABC~=~5~+~3~+~8~+~6~+~10\\\\\\\boxed{\sf Perimetro~Triangulo~ABC~=~32~cm}$

b)

Aplicando o teorema de Talles, teremos as seguintes relações:

$\boxed{\sf \dfrac{AD}{AB}~=~\dfrac{AE}{AC}}$

Substituindo as medidas dos segmentos:

$\sf \dfrac{x+4}{(x+4)+(x+1)}~=~\dfrac{6}{6+(x-1)}\\\\\\(x+4)\cdot (x+5)~=~6\cdot (2x+5)\\\\\\x^2~+~9x~+~20=~12x~+~30\\\\\\\boxed{\sf x^2~-~3x~-~10=~0}$

Aplicando Bhaskara à equação de 2° grau:

$\sf \Delta~=~(-3)^2-4\cdot 1\cdot (-10)\\\\\Delta ~=~9~+~40\\\\\boxed{\sf \Delta ~=~49}$

$\sf x'~=~\dfrac{3+\sqrt{49}}{2\cdot 1}~=~\dfrac{3+7}{2}~=~\dfrac{10}{2}~~\Rightarrow~~\boxed{\sf x'~=~5~cm}\\\\\\\sf x''~=~\dfrac{3-\sqrt{49}}{2\cdot 1}~=~\dfrac{3-7}{2}~=~\dfrac{-4}{2}~~\Rightarrow~~\boxed{\sf x''~=\,-2~cm}$

Como as medidas de comprimento não podem ser negativas, descartaremos a raiz x''=-2, logo a resposta para o item (I) será x=5cm.

Vamos aplicar novamente o Teorema de Talles para determinar agora a medida do segmento DE no triângulo menor:

$\sf \dfrac{AD}{AB}~=~\dfrac{DE}{BC}\\\\\\\dfrac{x+4}{(x+4)+(x+1)}~=~\dfrac{DE}{11}\\\\\\\dfrac{5+4}{(5+4)+(5+1)}~=~\dfrac{DE}{11}\\\\\\\dfrac{9}{15}~=~\dfrac{DE}{11}\\\\\\15\cdot DE~=~9\cdot 11\\\\\\DE~=~\dfrac{99}{15}\\\\\\\boxed{\sf DE~=~6,6~cm}$

Falta calcularmos os perímetros dos triângulos:

$\sf Perimetro~Triangulo~ADE~=~AD~+~DE~+~AE\\\\\\\sf Perimetro~Triangulo~ADE~=~(5+4)~+~6,6~+~6\\\\\\Perimetro~Triangulo~ADE~=~9~+~6,6~+~6\\\\\\\boxed{\sf Perimetro~Triangulo~ADE~=~21,6~cm}$

$\sf Perimetro~Triangulo~ABC~=~AB~+~BC~+~AC\\\\\\\sf Perimetro~Triangulo~ABC~=~(x+4)~+~(x+1)~+~11~+~(x-1)~+~6\\\\\\\sf Perimetro~Triangulo~ABC~=~(5+4)~+~(5+1)~+~11~+~(5-1)~+~6\\\\\\\sf Perimetro~Triangulo~ABC~=~9~+~6~+~11~+~4~+~6\\\\\\\boxed{\sf Perimetro~Triangulo~ABC~=~36~cm}$

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