Question

Calcular a área do triângulo de vértices A(8,3), B(4,7) e C(2,1).

Answer 1

$d(AB)=\sqrt{(8-4)^2+(7-3)^2}$

$d(AB)=\sqrt{4^2+4^2}$

$d(AB)=\sqrt{16+16}$

$d(AB)=\sqrt{32}$

$d(AC)=\sqrt{(8-2)^2+(3-1)^2}$

$d(AC)=\sqrt{6^2+2^2}$

$d(AC)=\sqrt{36+4}$

$d(AC)=\sqrt{40}$

$d(BC)=\sqrt{(4-2)^2+(7-1)^2}$

$d(BC)=\sqrt{2^2+6^2}$

$d(BC)=\sqrt{4+36}$

$d(BC)=\sqrt{40}$

$p=\frac{\sqrt{32}+\sqrt{40}+\sqrt{40} }{2}$

$p=\frac{\sqrt{4.8}+2\sqrt{40} }{2}$

$p=\frac{2\sqrt{8}+2\sqrt{40} }{2}$

$p=\sqrt{8}+\sqrt{40}$

$area=\sqrt{(\sqrt{8}+\sqrt{40}).(\sqrt{8}+\sqrt{40}-\sqrt{32}).(\sqrt{8}+\sqrt{40}-\sqrt{40}).(\sqrt{8}+\sqrt{40}-\sqrt{40})}$

$area=\sqrt{(\sqrt{8}+\sqrt{40}).(\sqrt{8}+\sqrt{40}-\sqrt{4.8}).\sqrt{8}.\sqrt{8}}$

$area=\sqrt{(\sqrt{8}+\sqrt{40}).(\sqrt{8}+\sqrt{40}-2\sqrt{8}).8}$

$area=\sqrt{(\sqrt{8}+\sqrt{40}).(\sqrt{40}-\sqrt{8}).8}$

$area=\sqrt{(\sqrt{8}.\sqrt{40}-8+40-\sqrt{8}.\sqrt{40}).8}$

$area=\sqrt{32.8}$

$area=\sqrt{256}$

$area=16$