[tex]zad.1\\Do ~~zadania~~jest~~dodany~~zalacznik~~wraz~~z~~oznaczeniami\\Pc ~~~~pole ~~powierzchni~~calkowitej~~ostroslupa\\Pb~~pole~~powierzchni~~bocznej\\Pp~~pole~~postawy\\Pc=Pp+Pb\\Pp=a^{2} ~~\land a=6~cm\\\\Pp=36~~cm^{2} \\\\Aby ~~obliczyc~~Pb~~oblicze~~wysokosc~~sciany~~bocznej,\\aby~~to~~zrobic~~skorzystam~~z~~tw.~~Pitagorasa\\h^{2} +(\frac{a}{2} )^{2} =b^{2} ~~\land~~\frac{a}{2} =3cm~~\land b=5cm\\h^{2} =25cm^{2} -9cm^{2} \\h^{2} =16cm^{2} ~~\land h>0\\h=4cm\\\\[/tex]
[tex]P\Delta = \frac{1}{2} \cdot 4cm\cdot 6cm\\\\P\Delta= 12~~cm^{2} \\\\Pc=Pb+Pp\\Pc=Pp+4\codt P\Delta\\Pc=36cm^{2} +48cm^{2} \\\\Pc=84~cm^{2}[/tex]
[tex]zad.2\\Do ~~zadania~~jest~~dodany~~zalacznik~~wraz~~z~~oznaczeniami\\V=\frac{1}{3} \cdot Pp \cdot H\\H=8cm\\Pp=6\cdoy P\Delta \\P\Delta -~~pole~~trojkata~~rownobocznego\\P\Delta=\frac{a^{2} \sqrt{3} }{4} ~~\land~~a=6cm~~\Rightarrow~~P\Delta=9\sqrt{3}cm^{2} \\\\Pp=6\cdot P\Delta~~\land~~P\Delta=9\sqrt{3}cm^{2}~~\Rightarrow~~Pp=54\sqrt{3} cm^{2}\\\\V=\frac{1}{3} \cdot 54\sqrt{3} \cdot 8~~cm^{3} \\\\V=144\sqrt{3} ~~cm^{3}[/tex]
[tex]Obliczam ~~teraz ~~h~~wysokosc~~sciany~~bocznej ~~trojkata~~,\\korzystam~~z~~tw.~~Pitagorasa\\\\H^{2} +h_{1} ^{2}= h^{2} ~~\land~~H=8cm~~\land~~h_{1} =\frac{a\sqrt{3} }{2} =3\sqrt{3} cm\\\\\\h^{2}=64cm^{2} +27cm^{2} \\\\h^{2}=81cm^{2} ~~\land ~~h>0\\h=9cm\\\\P\Delta=Pjednej~~sciany~~bocznej=\frac{1}{2} \cdot 9cm \cdot 6cm\\\\P\Delta=27~cm^{2} \\\\Pb=6 \cdot P\Delta~~\land~~P\Delta=27~cm^{2}\\\\Pb=162~cm^{2}[/tex]
