[tex]1. \\\text{Jezeli n oznaczymy ilosc bokow wielokata w podstawie ostroslupa, to ostroslup ma}\\\text{n+1 wierzcholkow}\\\text{2n krawedzi}\\\text{n+1 scian}\\n+1=15 /-1\\n=14\\2n=2*14=28\\\text{Ostroslup ma 28 krawedzi}[/tex]
[tex]2. \\Pp=a^2\\4Pb=108cm^2\\Pc=144cm^2\\Pc=Pp+4Pb\\144cm^2=Pp+108cm^2 /-108cm^2\\36cm^2=Pp\\a^2=36cm^2\\a=\sqrt{36cm^2}=6cm\\\\4Pb=108cm^2 /:4\\Pb=27cm^2\\Pb=\frac{ah}2\\27cm^2=\frac{6cm*h}2 /*2\\54cm^2=6cm*h /:6cm\\9cm=h\\(\frac12a)^2+h^2=b^2\\(3cm)^2+(9cm)^2=b^2\\9cm^2+81cm^2=b^2\\90cm^2=b^2\\b=\sqrt{90cm^2}=\sqrt{9*10cm^2}=3\sqrt{10}cm\\\text{Podstawa ma krawedz 6cm a krawedz boczna ostroslupa to }3\sqrt{10}cm[/tex]
[tex]3. \\H=6\\a=2\\b=4\\h=1.5\\Pp=\frac{(a+b)*h}2\\V=\frac13Pp*H\\V=\frac13*\frac{(a+b)*h}2*H\\V=\frac13*\frac{(2+4)*1.5}2*6\\V=\frac{6*1.5}2*2\\V=6*1.5=9[/tex]
[tex]4. \\\text{Czworoscian to ostroslup o podstawie trojkata}\\\text{Jezeli bylby to czworoscian foremny, to wszystkie jego krawedzie bylyby rowne}\\a=4cm\\Pp=\frac{a^2\sqrt3}4\\V=10cm^3\\V=\frac13Pp*H\\\\Pp=\frac{(4cm)^2\sqrt3}4=\frac{16\sqrt3cm^2}4=4\sqrt3cm^2\\10cm^3=\frac13*4\sqrt3cm^2*H /*3\\30cm^3=4\sqrt3cm^2*H /:4\sqrt3cm^2\\\frac{30cm}{4\sqrt3}=H\\H=\frac{30\sqrt3cm}{12}=\frac{5\sqrt3cm}2=\frac{5\sqrt3}2cm\\Odp. \text{Ten ostroslup ma wysokosc rowna }\frac{5\sqrt3}2cm[/tex]
[tex]5. \\Pp_1=(2x)^2=4x^2\\Pp_1=x^2\\\\V_1=\frac13*4x^2*H=\frac{4x^2H}3\\V_2=\frac13*x^2*\frac12H=\frac{x^2H}6\\\frac{V_1}{V_2}=\frac{4x^2H}3:\frac{x^2H}6=\frac{4x^2H}3*\frac6{x^2H}=\frac43*\frac61=\frac41*\frac21=8\\\text{Gorna czesc stanowi } \frac18 \text{ calej objetosci ostroslupa. Zostal wiec on podzielony w stosunku 1:7}[/tex]
