Question

Determinar a área do triângulo A(1; 3),B(4; 7) e C(6; 5)

Answer 1

Resposta:

7

Explicação passo-a-passo:

A= I detA I / 2

$\left[\begin{array}{ccc}XA&YA&1\\XB&YB&1\\XC&YB&1\end{array}\right]$

$\left[\begin{array}{ccc}1&3&1\\4&7&1\\6&5&1\end{array}\right]$

detA= 7+8+20-42-5-12

detA= -14

$A=\frac{14}{2} =7$

Answer 2

Resposta:

Área desse triângulo será dada por:

$\sf A = \dfrac{1}{2} \cdot \mid D \mid$

Onde,

$\sf D =\begin{array}{|ccc|cc|} \sf x_A & \sf y_A & 1 & \sf x_A & \sf y_A \\\sf x_B &\sf y_B &\sf1 & \sf x_B &\sf y_B \\\sf x_C & \sf y_C & 1 & \sf x_C & \sf y_C\\\end{array}$

Resolução:

Aplicando a regra de Sarrus temos:

$D = \begin{array}{|ccc|cc|}1 & 3 & 1 & 1 & 3 \\4 & 7 & 1 & 4 & 7 \\6 & 5 & 1 & 6 & 5 \\\end{array}$

$\sf D = 1*7*1+3*1*6+1*4*5-6*7*1-5*1*1-1*4*3=-14$

$\sf A = \dfrac{1}{2} \cdot \mid D \mid$

$\sf A = \dfrac{1}{2} \cdot \mid - 14 \mid$

$\sf A = \dfrac{1 \cdot 14 }{2}$

$\sf A = \dfrac{14 }{2}$

$\boxed{ \boxed { \boldsymbol{ \sf \displaystyle A = 7 }}} \quad \gets \mathbf{ Resposta }$

Explicação passo-a-passo: