Oznaczmy długość przeciwprostokątnej przez $c$. W każdym podpunkcie skorzystamy z twierdzenia Pitagorasa.

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a)

${30}^2+{40}^2=c^2$

$900+1600=c^2$

$2500=c^2|\sqrt{},c>0$

$50=c$

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b)

${\left(\sqrt{5}\right)}^2+2^2=c^2$

$5+4=c^2$

$9=c^2|\sqrt{},c>0$

$3=c$  

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c)

${\left(2\sqrt{2}-1\right)}^2+{\left(2\sqrt{2}+1\right)}^2=c^2$

$\left(4\cdot 2-4\sqrt{2}+1\right)+\left(4\cdot 2+4\sqrt{2}+1\right)=c^2$

$9-4\sqrt{2}+9+4\sqrt{2}=c^2$

$18=c^2|\sqrt{},c>0$

$\sqrt{18}=c$

$3\sqrt{2}=c$  

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d)

${\left(\sqrt{5}-\sqrt{3}\right)}^2+{\left(\sqrt{5}+\sqrt{3}\right)}^2=c^2$

$\left(5-2\sqrt{5}\cdot \sqrt{3}+3\right)+\left(5+2\sqrt{5}\cdot \sqrt{3}+3\right)=c^2$

$8-2\sqrt{15}+8+2\sqrt{15}=c^2$

$16=c^2|\sqrt{},c>0$

$4=c$

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e)

${\left(2\sqrt{2}\right)}^2+{\left(2\sqrt{2}\right)}^2=c^2$

$4\cdot 2+4\cdot 2=c^2$

$16=c^2|\sqrt{},c>0$

$4=c$

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f)

$a=\frac{1}{b}=2$  

$a=2,\frac{1}{b}=2$   

$a=2,b=\frac{1}{2}$

$2^2+{\left(\frac{1}{2}\right)}^2=c^2$

$4+\frac{1}{4}=c^2$

$4\frac{1}{4}=c^2$  

$\frac{17}{4}=c^2|\sqrt{},c>0$

$\frac{\sqrt{17}}{2}=c$