Szczegółowe wyjaśnienie:
Zad.1 Patrz załącznik.
Zad.2
Korzystamy z twierdzenia Pitagorasa.
[tex]x^2=9^2+12^2\\x^2=81+144\\x^2=225\to x=\sqrt{225}\\\boxed{x=15(cm)}\\\\y^2=5^2+12^2\\y^2=25+144\\y^2=169\yo y=\sqrt{169}\\\boxed{y=13(cm)}[/tex]
[tex]e^2=4^2+8^2\\e^2=16+64\\e^2=80\to e=\sqrt{80}\\e=\sqrt{16\cdot5}\\e=\sqrt{16}\cdot\sqrt5\\\boxed{e=4\sqrt5(cm)}\\\\f^2=6^2+8^2\\f^2=36+64\\f^2=100\to f=\sqrt{100}\\\boxed{f=10(cm)}[/tex]
[tex]d^2=10^2+10^2\\d^2=100+100\\d^2=100\cdot2\to d=\sqrt{100\cdot2}\\d=\sqrt{100}\cdot\sqrt2\\\boxed{d=10\sqrt2(cm)}\\\\k^2=10^2+3^2\\k^2=100+9\\k^2=109\to \boxed{k=\sqrt{109}(cm)}[/tex]
