Question

01. Dada as coordenadas dos vértices de um triângulo ABC, A (5,1), B (5,8) e C
(8,3). Qual distância do baricentro deste triângulo até o seu vértice A?

a) 2
b) 4
e) √10
d) √17
e) 20
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Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{M_{BC} = \{\dfrac{x_B + x_C}{2};\dfrac{y_B + y_C}{2}\}}$

$\mathsf{M_{BC} = \{\dfrac{5 + 8}{2};\dfrac{8 + 3}{2}\}}$

$\mathsf{M_{BC} = \{\dfrac{13}{2};\dfrac{11}{2}\}}$

$\mathsf{d_{AM} = \sqrt{(x_M - x_A)^2 + (y_M - y_A)^2}}$

$\mathsf{d_{AM} = \sqrt{(\dfrac{13}{2} - 5)^2 + (\dfrac{11}{2} - 1)^2}}$

$\mathsf{d_{AM} = \sqrt{(\dfrac{13 - 10}{2})^2 + (\dfrac{11 - 2}{2})^2}}$

$\mathsf{d_{AM} = \sqrt{(\dfrac{3}{2})^2 + (\dfrac{9}{2})^2}}$

$\mathsf{d_{AM} = \sqrt{\dfrac{9}{4} + \dfrac{81}{4}}}$

$\mathsf{d_{AM} = \sqrt{\dfrac{90}{4}}}$

$\mathsf{d_{AM} = \sqrt{\dfrac{3^2.10}{2^2}}}$

$\mathsf{d_{AM} = \dfrac{3\sqrt{10}}{2}}$

$\mathsf{x = \dfrac{2}{3} \times \dfrac{3\sqrt{10}}{2}}$

$\boxed{\boxed{\mathsf{x = \sqrt{10}}}}\leftarrow\textsf{letra C}$

Answer 2

Resposta:

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