a)
Obliczmy wysokość h ściany bocznej
[tex] {3}^{2} + {h}^{2} = {6}^{2} \\ 9 + {h}^{2} = 36 \\ {h}^{2} = 27 \\ h = \sqrt{27} \\ h = \sqrt{9 \times 3} \\ \underline{h = 3 \sqrt{3} }[/tex]
Obliczmy wysokość H ostrosłupa
[tex] {3}^{2} + {H}^{2} = (3 \sqrt{3} {)}^{2} \\ 9 + {H}^{2} = 9 \times 3 \\ {H}^{2} = 27 - 9 \\ H = \sqrt{18} \\ H = \sqrt{9 \times 2} \\ \underline{\underline{ H = 3 \sqrt{2} }}[/tex]
Teraz obliczmy pole zacieniowanej figury
[tex] \large{ \frac{a \times h}{2} }[/tex]
[tex] \frac{6 \times 3 \sqrt{2} }{2} = \frac{18 \sqrt{2} }{2} = \boxed{ \boxed{9 \sqrt{2} }}[/tex]
b)
Podstawa zacieniowanej figury to 2a czyli u nas jest to
[tex]2a = 2 \times 4 = \underline{\underline 8}[/tex]
Obliczmy wysokość H ostrosłupa
[tex] {4}^{2} + {H}^{2} = {6}^{2} \\ 16 + {H}^{2} = 36 \\ {H}^{2} = 20 \\ H = \sqrt{20} \\ H = \sqrt{4 \times 5} \\ \underline{ H = 2 \sqrt{5} }[/tex]
Teraz obliczmy pole zacieniowanej figury
[tex] \large{ \frac{a \times h}{2} }[/tex]
[tex] \frac{8 \times 2 \sqrt{5} }{2} = \frac{16 \sqrt{5} }{2} = \boxed{ \boxed{ 8 \sqrt{5} }}[/tex]
