Question

Os pontos A = (0,0), B = (3,7) e C = (5, -1) são vértices de um triângulo. O comprimento da mediana vale:

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo-a-passo:

Vamos calcular o ponto médio do segmento BC.

$\sf M_{BC}\left(\dfrac{x_b + x_c}{2};\dfrac{y_b + y_c}{2}\right)$

$\sf M_{BC}\left(\dfrac{3 + 5}{2};\dfrac{7 - 1}{2}\right)$

$\sf M_{BC}\left(\dfrac{8}{2};\dfrac{6}{2}\right)$

$\sf M_{BC}\left(4;3\right)$

Vamos calcular a distância do vértice A ao ponto médio de BC.

$\sf d_{AM} = \sqrt{(x_m - x_a)^2 + (y_m - y_a)^2}$

$\sf d_{AM} = \sqrt{(4 - 0)^2 + (3 - 0)^2}$

$\sf d_{AM} = \sqrt{4^2 + 3^2}$

$\sf d_{AM} = \sqrt{16 + 9}$

$\sf d_{AM} = \sqrt{25}$

$\boxed{\boxed{\sf d_{AM} = 5}}$

Answer 2

Sejam M, N e P os pontos médios de AB, AC e BC, respectivamente.

• Mediana CM

$\sf \Rightarrow~coordenadas~de~M$

$\sf x_M=\dfrac{x_A+x_B}{2}$

$\sf x_M=\dfrac{0+3}{2}$

$\sf x_M=\dfrac{3}{2}$

$\sf y_M=\dfrac{y_A+y_B}{2}$

$\sf y_M=\dfrac{0+7}{2}$

$\sf y_M=\dfrac{7}{2}$

O comprimento da mediana CM é:

$\sf \overline{CM}=\sqrt{(x_C-x_M)^2+(y_C-y_M)^2}$

$\sf \overline{CM}=\sqrt{\Big(5-\dfrac{3}{2}\Big)^2+(-1-\dfrac{7}{2}\Big)^2}$

$\sf \overline{CM}=\sqrt{\Big(\dfrac{10-3}{2}\Big)^2+(\dfrac{-2-7}{2}\Big)^2}$

$\sf \overline{CM}=\sqrt{\Big(\dfrac{7}{2}\Big)^2+(\dfrac{-9}{2}\Big)^2}$

$\sf \overline{CM}=\sqrt{\dfrac{49}{4}+\dfrac{81}{4}}$

$\sf \overline{CM}=\sqrt{\dfrac{130}{4}}$

$\sf \red{\overline{CM}=\dfrac{\sqrt{130}}{2}}$

• Mediana BN

$\sf \Rightarrow~coordenadas~de~N$

$\sf x_N=\dfrac{x_A+x_C}{2}$

$\sf x_N=\dfrac{0+5}{2}$

$\sf x_N=\dfrac{5}{2}$

$\sf y_N=\dfrac{y_A+y_C}{2}$

$\sf y_N=\dfrac{0-1}{2}$

$\sf y_N=\dfrac{-1}{2}$

O comprimento da mediana BN é:

$\sf \overline{BN}=\sqrt{(x_B-x_N)^2+(y_B-y_N)^2}$

$\sf \overline{BN}=\sqrt{\Big(3-\dfrac{5}{2}\Big)^2+(7+\dfrac{1}{2}\Big)^2}$

$\sf \overline{BN}=\sqrt{\Big(\dfrac{6-5}{2}\Big)^2+(\dfrac{14+1}{2}\Big)^2}$

$\sf \overline{BN}=\sqrt{\Big(\dfrac{1}{2}\Big)^2+(\dfrac{15}{2}\Big)^2}$

$\sf \overline{BN}=\sqrt{\dfrac{1}{4}+\dfrac{225}{4}}$

$\sf \overline{BN}=\sqrt{\dfrac{226}{4}}$

$\sf \red{\overline{BN}=\dfrac{\sqrt{226}}{2}}$

• Mediana AP

$\sf \Rightarrow~coordenadas~de~P$

$\sf x_P=\dfrac{x_B+x_C}{2}$

$\sf x_P=\dfrac{3+5}{2}$

$\sf x_P=\dfrac{8}{2}$

$\sf x_P=4$

$\sf y_P=\dfrac{y_B+y_C}{2}$

$\sf y_P=\dfrac{7-1}{2}$

$\sf y_P=\dfrac{6}{2}$

$\sf y_P=3$

O comprimento da mediana AP é:

$\sf \overline{AP}=\sqrt{(x_A-x_P)^2+(y_A-y_P)^2}$

$\sf \overline{AP}=\sqrt{(0-4)^2+(0-3)^2}$

$\sf \overline{AP}=\sqrt{(-4)^2+(-3)^2}$

$\sf \overline{AP}=\sqrt{16+9}$

$\sf \overline{AP}=\sqrt{25}$

$\sf \red{\overline{AP}=5}$