Oblicz długość przekątnej ścian bocznych graniastosłupa prostego o wysokości 6 Cm którego podstawa jest czworokąt o bokach 3 cm 5 Cm 10 cm i 12 Cm

Ś1
[tex] {5}^{2} + {3}^{2} = {d}^{2} \\ 25 + 9 = {d}^{2} \\ 34 = {d}^{2} \\ \boxed{ d = \sqrt{34}cm }[/tex]
Ś2
[tex] {5}^{2} + {5}^{2} = {d}^{2} \\ 25 + 25 = {d}^{2} \\ 50 = {d}^{2} \\ d = \sqrt{50} \\ d = \sqrt{25 \times 2} \\ \boxed{d = 5 \sqrt{2} cm}[/tex]
Ś3
[tex] {5}^{2} + {10}^{2} = {d}^{2} \\ 25 + 100 = {d}^{2} \\ 125 = {d}^{2} \\ d = \sqrt{125} \\ d = \sqrt{25 \times 5} \\ \boxed{ d = 5 \sqrt{5} cm}[/tex]
Ś4
[tex] {5}^{2} + {12}^{2} = {d}^{2} \\ 25 + 144 = {d}^{2} \\ d = \sqrt{169} \\ \boxed{ d = 13cm}[/tex]