Odpowiedź :
1.
[tex]a = \sqrt{7} \\H = 4\\\\Pp = \frac{a^{2} \sqrt{3} }{4} = \frac{(\sqrt{7})^{2} \sqrt{3} }{4} = \frac{7 \sqrt{3} }{4} \\Pb = 3*4\sqrt{7} = 12\sqrt{7}\\Pc=2Pp + Pb = 2*\frac{7 \sqrt{3} }{4} + 12\sqrt{7} = \frac{7 \sqrt{3} }{2} + 12\sqrt{7}\\V = Pp * H = \frac{7 \sqrt{3} }{4} *4 = 7\sqrt{3}\\[/tex]
2.
[tex]a = 2dm\\b = 14 dm\\h = 8 dm\\H = \sqrt{5}dm\\ramie= z\\z^{2} = 8^{2} + [(14-2)/2]^{2} \\z^{2}=64+6^{2}\\z^{2} = 64+36\\z^{2} = 100\\z=10 dm\\\\Pp = \frac{1}{2} (a+b)h = \frac{1}{2} (14+2)8 = \frac{1}{2} * 16 * 8 = 64[dm^{2}]\\Pb = 2*10\sqrt{5} + 14\sqrt{5} + 2\sqrt{5} = 20\sqrt{5} + 14\sqrt{5} + 2\sqrt{5} = 36\sqrt{5}[dm^{2}] \\Pc = 2Pp + Pb = 2*64 + 36\sqrt{5} = 128 + 36\sqrt{5} [^{2}dm^{2} ]\\V = Pp * H = 64\sqrt{5} dm^{3}[/tex]