1. Rysunek w zalacznikach.
Okregi przecinaja sie w dwoch punktach.
2.
[tex]r = \frac{a+b-c}{2}\\r=2\\c = 10\\2 = \frac{a+b-10}{2}\\4=a+b-10\\14=a+b\\14-b=a\\a^2+b^2=c^2\\(14-b)^2+b^2=10^2\\196-28b+b^2+b^2=100\\2b^2-28b+196-100=0\\2b^2-28b+96=0\\2(b^2-14b+48)=0 /:2\\b^2-14b+48=0\\\Delta=(-14)^2-4*1*48\\\Delta=196-192=4\\\sqrt{\Delta}=2\\b_1=\frac{14-2}{2}=6\\b_2=\frac{14+2}{2}=8\\a_1=14-b_1\\a_1=14-6=8\\a_2=14-b_2\\a_2=14-8=6[/tex]
Odp. Trojkat ma przyprostokatne 6 i 8.
3.
a = 6
b = 6
c = 8
[tex]p = \frac{a+b+c}{2}\\r=\sqrt{\frac{(p-a)(p-b)(p-c)}{p}}\\p=\frac{6+6+8}{2}=\frac{20}{2}=10\\r=\sqrt{\frac{(10-6)(10-6)(10-8)}{10}}=\sqrt{\frac{4*4*2}{10}}=\sqrt{\frac{32}{10}}=\frac{4\sqrt{2}}{\sqrt{10}}=\frac{4\sqrt{20}}{10}= \frac{4*2\sqrt{5}}{10}=\frac{4\sqrt{5}}{5}[/tex]
4.
[tex]r = 4\\P_k=\pi*4^2=16\pi\\P_w=12\pi\\\frac{P_w}{P_k}=\frac{12\pi}{16\pi}=\frac{3}{4}\\360*\frac{3}{4}=90*3=270[/tex]
Odp. Kat wyznaczajacy odcinek wynosi 270 stopni.
5.
[tex]r = 3\\c = 2r\\c = 6\\a = 3b\\a^2+b^2=c^2\\(3b)^2+b^2=6^2\\9b^2+b^2=36\\10b^2=36 /:10\\b^2=3,6\\b=\sqrt{\frac{36}{10}} = \frac{6}{\sqrt{10}}=\frac{6\sqrt{10}}{10} = \frac{3\sqrt{10}}{5}\\a = 3*\frac{3\sqrt{10}}{5}=\frac{9\sqrt{10}}{5}\\Ob = a+b+c\\Ob=\frac{9\sqrt{10}}{5}+\frac{3\sqrt{10}}{5}+6=\frac{12\sqrt{10}}{5}+6[/tex]
6.
Pole pierscienia - 48π
Pole wiekszego kola: πR²
Pole mniejszego kola: πr²
Pole pierscienia - 48π = πR²-πr^2 | 48π = π(R^2-r^2)
[tex]R = \frac23h\\r = \frac13h\\h = \frac{a\sqrt3}{2}[/tex]
[tex]R = \frac23*\frac{a\sqrt3}{2}=\frac{a\sqrt3}{3}\\r = \frac13*\frac{a\sqrt3}2=\frac{a\sqrt3}{6}\\48\pi=\pi(R^2-r^2)\\48=R^2-r^2\\48=(\frac{a\sqrt3}3)^2-(\frac{a\sqrt3}{6})^2\\[/tex]
[tex]48=\frac{3a^2}{9}-\frac{3a^2}{36}\\48=\frac{a^2}{3}-\frac{a^2}{12}\\48=\frac{4a^2}{12}-\frac{a^2}{12}\\48=\frac{3a^2}{12} /*12\\576=3a^2 /:3\\192=a^2\\a=\sqrt{192}=\sqrt{8*8*3}=8\sqrt{3}[/tex]
7.
Szesciokat foremny
[tex]P = 12\sqrt3cm^2\\P=6*\frac{a^2\sqrt3}4=\frac{3a^2\sqrt3}2\\12\sqrt{3}=\frac{3a^2\sqrt3}{2} /*2\\24\sqrt{3}=3a^2\sqrt3\\24=3a^2 /:3\\8=a^2\\a=\sqrt{8}=2\sqrt{3}cm[/tex]
Okrag opisany na szesciokacie foremnym:
[tex]R = a\\R = 2\sqrt3cm\\L = 2\pi R\\L=2\pi*2\sqrt3cm\\L=4\sqrt3 \pi cm[/tex]
Okrag wpisany w szesciokat foremny
[tex]r = h\\h = \frac{a\sqrt3}2\\r = \frac{a\sqrt3}{2}\\r=\frac{2\sqrt3cm*\sqrt3}2\\r=\sqrt{3}cm*\sqrt3=3cm\\l = 2\pi r\\l = 2\pi*3cm\\l = 6\pi cm[/tex]
[tex]L-l=4\sqrt3\pi cm-6\pi cm = \pi(4\sqrt3-6)cm[/tex]
