Question

se dau punctele A(1,3) B(3,-3) C(-5,1) .Calculati perimetrul triunghiului ABC

Answer 1

$A(1,3)$

$B(3,-3)$

$C(-5,1)$

$AB=\sqrt{{(x_{B}-x_{A})}^{2}+{( y_{B}-y_{A})}^{2}}$

$AB=\sqrt{{(3-1)}^{2}+{( - 3-3)}^{2}}$

$AB=\sqrt{{2}^{2}+{( - 6)}^{2}}$

$AB=\sqrt{4+36}$

$AB=\sqrt{40}$

$AB=2\sqrt{10}$

$AC=\sqrt{{(x_{C}-x_{A})}^{2}+{(y_{C}-y_{A})}^{2}}$

$AC=\sqrt{{( - 5-1)}^{2}+{(1-3)}^{2}}$

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

$AC=\sqrt{36+4}$

$AC=\sqrt{40}$

$AC=2\sqrt{10}$

$BC=\sqrt{{(x_{C}-x_{B})}^{2}+{(y_{C}-y_{B})}^{2}}$

$BC=\sqrt{{( - 5-3)}^{2}+{(1 + 3)}^{2}}$

$BC=\sqrt{{( - 8)}^{2}+{4}^{2}}$

$BC=\sqrt{64+16}$

$BC=\sqrt{80}$

$BC=4\sqrt{5}$

$AB=AC=>\Delta \: ABC \:isoscel$

$P=AB+AC+BC$

$P=2AB+BC$

$P=2AC+BC$

$P=2 \times 2 \sqrt{10} + 4 \sqrt{5}$

$P=4 \sqrt{10} + 4 \sqrt{5}$

$P = 4( \sqrt{10} + \sqrt{5} )$