Question

Considere o triângulo de vértices A(6,8),B(2,3) e C(4,5). O valor da medida da mediana AM do triângulo ABC é:

Answer 1

Veja imagem para melhor entendimento.


A (6 ; 8)  ;  B (2 ; 3)  ;  C (4 ; 5)

BC = a
AB = b
AC = c

1°, calcular distância entre BC, AB e AC, pra saber qual é a  medida de a, b e c, respectivamente:


$Fo\´rmula~\to~~~D_{pq}= \sqrt{(x_q-x_p)^2+(y_q-y_p)^2}$




$D_{bc}= \sqrt{(4-2)^2+(5-3)^2} \to \\ D_{bc}= \sqrt{2^2+2^2} \to \\D_{bc}= \sqrt{4+4} \to \\ D_{bc}= \sqrt{8} \to \\D_{bc}= \sqrt{2^2.2} \to \\ \boxed{D_{bc}= 2\sqrt{2}}$



$D_{ab}= \sqrt{(2-6)^2+(3-8)^2} \to \\ D_{ab}= \sqrt{(-4)^2+(-5)^2} \to \\D_{ab}= \sqrt{16+25} \to \\ \boxed{D_{ab}= \sqrt{41}}$



$D_{ac}= \sqrt{(4-6)^2+(5-8)^2} \to \\ D_{ac}= \sqrt{(-2)^2+(-3)^2} \to \\D_{ac}= \sqrt{4+9} \to \\ \boxed{D_{ac}= \sqrt{13}}$





$Ent\~ao~~ temos:\\\\a= \sqrt{41}\\b=2 \sqrt{2} \\c= \sqrt{13}\\\\\\\\ \boxed{m= \sqrt{ \dfrac{2b^2+2c^2-a^2}{4} }}~~~~ m= \sqrt{ \dfrac{2(2 \sqrt{2} )^2+2(\sqrt{13})^2-( \sqrt{41} )^2}{4} } \to\\\\\\ m= \sqrt{ \dfrac{2(4 .2 )+2.13-41}{4} } \to ~~m= \sqrt{ \dfrac{2.8+26-41}{4} } \to \\\\\\ m= \sqrt{ \dfrac{16+26-41}{4} } \to ~~ m= \sqrt{ \dfrac{42-41}{4} } \to ~~ m= \sqrt{ \dfrac{1}{4} } \to ~~ \large\boxed{\boxed{m= \frac{1}{2} }}$




$R.: ~~O~~ valor~~da~~mediana~~e\´~~ \dfrac{1}{2} ~um$

$[ obs:~um=unidade~~de~~medida ]$