Question

6) Um triângulo tem vértices nos pontos A (5,8), B(2,2) e C(8,2). Determine a medida da
mediana relativa ao vértice C.​

Answer 1

Explicação passo-a-passo:

Seja $\sf M$ o ponto médio do lado $\sf \overline{AB}$

Temos que:

• $\sf x_M=\dfrac{x_A+x_B}{2}$

$\sf x_M=\dfrac{5+2}{2}$

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

• $\sf y_M=\dfrac{y_A+y_B}{2}$

$\sf y_M=\dfrac{8+2}{2}$

$\sf y_M=\dfrac{10}{2}$

$\sf y_M=5$

Assim, $\sf M\left(\dfrac{7}{2},5\right)$

A mediana relativa ao vértice C mede:

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

$\sf \overline{CM}=\sqrt{\left(8-\dfrac{7}{2}\right)^2+(2-5)^2}$

$\sf \overline{CM}=\sqrt{\left(\dfrac{16-7}{2}\right)^2+(-3)^2}$

$\sf \overline{CM}=\sqrt{\left(\dfrac{9}{2}\right)^2+9}$

$\sf \overline{CM}=\sqrt{\dfrac{81}{4}+9}$

$\sf \overline{CM}=\sqrt{\dfrac{81+36}{4}}$

$\sf \overline{CM}=\sqrt{\dfrac{117}{4}}$

$\sf \red{\overline{CM}=\dfrac{3\sqrt{13}}{2}}$