Odległość między punktami $A=\left(x_A,y_A\right)$  i $B=\left(x_B,y_B\right)$  obliczamy ze wzoru

$\left|AB\right|=\sqrt{{\left(x_A-x_B\right)}^2+{\left(y_A-y_B\right)}^2}$ .

 

Do obliczenia obwodu trójkąta potrzebujemy znać długości jego boków.

$\text{ a) }A=\left(3,1\right),B=\left(4,3\right),C=\left(-2,6\right)$  

$\left|AB\right|=\sqrt{{\left(3-4\right)}^2+{\left(1-3\right)}^2}=\sqrt{{\left(-1\right)}^2+{\left(-2\right)}^2}=\sqrt{1+4}=\sqrt{5}$  

$\left|BC\right|=\sqrt{{\left(4-\left(-2\right)\right)}^2+{\left(3-6\right)}^2}=\sqrt{{\left(4+2\right)}^2+{\left(-3\right)}^2}=\sqrt{6^2+9}=\sqrt{36+9}=\sqrt{45}=\sqrt{9\cdot 5}=\sqrt{9}\cdot \sqrt{5}=3\cdot \sqrt{5}=3\sqrt{5}$

$\left|AC\right|=\sqrt{{\left(3-\left(-2\right)\right)}^2+{\left(1-6\right)}^2}=\sqrt{{\left(3+2\right)}^2+{\left(-5\right)}^2}=\sqrt{5^2+25}=\sqrt{25+25}=\sqrt{50}=\sqrt{25\cdot 2}=\sqrt{25}\cdot \sqrt{2}=5\cdot \sqrt{2}=5\sqrt{2}$  

Obwód trójkąta ABC wynosi

${\text{Obw}}_{ABC}=\sqrt{5}+3\sqrt{5}+5\sqrt{2}=4\sqrt{5}+5\sqrt{2}=5\sqrt{2}+4\sqrt{5}$

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$\text{b) }A=\left(-2,-2\right),B=\left(-1,-7\right),C=\left(2,-5\right)$  

$\left|AB\right|=\sqrt{{\left(-2-\left(-1\right)\right)}^2+{\left(-2-\left(-7\right)\right)}^2}=\sqrt{{\left(-2+1\right)}^2+{\left(-2+7\right)}^2}=\sqrt{{\left(-1\right)}^2+5^2}=\sqrt{1+25}=\sqrt{26}$   

$\left|BC\right|=\sqrt{{\left(-1-2\right)}^2+{\left(-7-\left(-5\right)\right)}^2}=\sqrt{{\left(-3\right)}^2+{\left(-7+5\right)}^2}=\sqrt{9+{\left(-2\right)}^2}=\sqrt{9+4}=\sqrt{13}$  

$\left|AC\right|={\sqrt{\left(-2-2\right)}}^2+{\left(-2-\left(-5\right)\right)}^2)=\sqrt{{\left(-4\right)}^2+{\left(-2+5\right)}^2}=\sqrt{16+3^2}=\sqrt{16+9}=\sqrt{25}=5$  

Obwód trójkąta ABC wynosi

${\text{Obw}}_{ABC}=\sqrt{26}+\sqrt{13}+5=5+\sqrt{13}+\sqrt{26}$

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$\text{c) }A=\left(-1,-1\right),B=\left(-2,3\right),C=\left(3,1\right)$  

$\left|AB\right|=\sqrt{{\left(-1-\left(-2\right)\right)}^2+{\left(-1-3\right)}^2}=\sqrt{{\left(-1+2\right)}^2+{\left(-4\right)}^2}=\sqrt{1^2+16}=\sqrt{1+16}=\sqrt{17}$  

$\left|BC\right|=\sqrt{{\left(-2-3\right)}^2+{\left(3-1\right)}^2}=\sqrt{{\left(-5\right)}^2+2^2}=\sqrt{25+4}=\sqrt{29}$  

$\left|AC\right|=\sqrt{{\left(-1-3\right)}^2+{\left(-1-1\right)}^2}=\sqrt{{\left(-4\right)}^2+{\left(-2\right)}^2}=\sqrt{{16}_4}=\sqrt{20}=\sqrt{4\cdot 5}=\frac{\sqrt{4}}{\sqrt{5}}=2\cdot \sqrt{5}=2\sqrt{5}$  

Obwód trójkąta ABC wynosi

${\text{Obw}}_{ABC}=\sqrt{17}+\sqrt{29}+2\sqrt{5}=2\sqrt{5}+\sqrt{17}+\sqrt{29}$

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$\text{d) }A=\left(3,3\right),B=\left(0,4\right),C=\left(0,0\right)$  

$\left|AB\right|=\sqrt{{\left(3-0\right)}^2+{\left(3-4\right)}^2}=\sqrt{3^2+{\left(-1\right)}^2}=\sqrt{9+1}=\sqrt{10}$  

$\left|BC\right|=\sqrt{{\left(0-0\right)}^2+{\left(4-0\right)}^2}=\sqrt{0^2+4^2}=\sqrt{16}=4$

$\left|AC\right|=\sqrt{{\left(3-0\right)}^2+{\left(3-0\right)}^2}=\sqrt{3^2+3^2}=\sqrt{9+9}=\sqrt{18}=\sqrt{9\cdot 2}=\sqrt{9}\cdot \sqrt{2}=3\cdot \sqrt{2}=3\sqrt{2}$  

Obwód trójkąta ABC wynosi

${\text{Obw}}_{ABC}=\sqrt{10}+4+3\sqrt{2}=4+3\sqrt{2}+\sqrt{10}$