Question

determinar a área e o perímetro do triângulo, dados os pontos: A(7,5), B (3,2) e C (7,2)

Answer 1

Vamos começar pelo perímetro. Os lados do triangulo serão os segmentos AB, BC e AC.

O comprimento de cada um desses segmentos pode ser calculado pela equação da distancia entre pontos, acompanhe:

$\overline{AB}~=~Distancia_{\,A,B}\\\\\\\overline{AB}~=~\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}\\\\\\\overline{AB}~=~\sqrt{(7-3)^2+(5-2)^2}\\\\\\\overline{AB}~=~\sqrt{4^2+3^2}\\\\\\\overline{AB}~=~\sqrt{25}\\\\\\\boxed{\overline{AB}~=~5~unidades~de~comprimento}$

$\overline{BC}~=~Distancia_{\,B,C}\\\\\\\overline{BC}~=~\sqrt{(x_B-x_C)^2+(y_B-y_C)^2}\\\\\\\overline{BC}~=~\sqrt{(3-7)^2+(2-2)^2}\\\\\\\overline{BC}~=~\sqrt{(-4)^2+0^2}\\\\\\\overline{BC}~=~\sqrt{16}\\\\\\\boxed{\overline{BC}~=~4~unidades~de~comprimento}$

$\overline{AC}~=~Distancia_{\,A,C}\\\\\\\overline{AC}~=~\sqrt{(x_A-x_C)^2+(y_A-y_C)^2}\\\\\\\overline{AC}~=~\sqrt{(7-7)^2+(5-2)^2}\\\\\\\overline{AC}~=~\sqrt{0^2+3^2}\\\\\\\overline{AC}~=~\sqrt{9}\\\\\\\boxed{\overline{AC}~=~3~unidades~de~comprimento}$

Calculando o perímetro:

$Perimetro = \overline{AB}+\overline{BC}+\overline{AC}\\\\\\Perimetro = 5+4+3\\\\\\\boxed{Perimetro~=~12~unidades~de~comprimento}$

Para calcular a área, podemos utilizar o método do determinante como é mostrado abaixo:

$Area~=~\frac{1}{2}\,.\,\left|\left|\begin{array}{ccc}x_A&y_A&1\\x_B&y_B&1\\x_C&y_C&1\end{array}\right|\right|\\\\\\Area~=~\frac{1}{2}\,.\,\left|\left|\begin{array}{ccc}7&5&1\\3&2&1\\7&2&1\end{array}\right|\right|\\\\\\Area~=~\frac{1}{2}\,.\,\left|~(7.2.1+3.2.1+7.5.1)~-~(1.2.7+1.2.7+1.5.3)~\right|\\\\\\Area~=~\frac{1}{2}\,.\,\left|~(14+6+35)~-~(14+14+15)~\right|\\\\\\Area~=~\frac{1}{2}\,.\,\left|~14+6+35-14-14-15~\right|$

$Area~=~\frac{1}{2}\,.\,\left|~12~\right|\\\\\\\boxed{Area~=~6~unidades~de~area}$