Witam, interesują mnie rozwiązania zadań od 8 do 14 jeśli mogę prosić o pomoc, to bardzo ważne, pozdrawiam. http://iv.pl/images/61jcln8gllobyd0dagn5.jpg

zad 8 2*\pi *6* \frac{120}{360} =2* \pi * r 12* \frac{1}{3}* \pi=2* \pi * r 4* \pi=2* \pi * r 2* \pi * r=4* \pi r=\frac{4* \pi}{2* \pi} r=2

zad 9 l=3cm r=3*\frac{\alpha}{360} P_c=\pi r (l+r) P_c=\pi *3*\frac{\alpha}{360} *(3+3*\frac{\alpha}{360}) a)\alpha=180 P_c=\pi *3*\frac{180}{360} *(3+3*\frac{180}{360}) P_c=\pi *3*\frac{1}{2} *(3+3*\frac{1}{2}) P_c=\pi *3*\frac{1}{2} *(3+3*\frac{1}{2}) P_c=\pi *\frac{3}{2} *(3+1\frac{1}{2}) P_c=\pi *\frac{3}{2} *4\frac{1}{2} P_c=\pi *\frac{3}{2} *\frac{9}{2} P_c=\frac{27}{4}\pi P_c=6\frac{3}{4}\pi b)\alpha=120 P_c=\pi *3*\frac{120}{360} *(3+3*\frac{120}{360}) P_c=\pi *3*\frac{1}{3} *(3+3*\frac{1}{3}) P_c=\pi *1 *(3+1) P_c=\pi *4 P_c=4\pi c)\alpha=60 P_c=\pi *3*\frac{60}{360} *(3+3*\frac{60}{360}) P_c=\pi *3*\frac{1}{6} *(3+3*\frac{1}{6}) P_c=\pi *\frac{1}{2} *(3+\frac{1}{2}) P_c=\pi *\frac{1}{2} *3\frac{1}{2} P_c=\pi *\frac{1}{2} *\frac{7}{2} P_c=\pi *\frac{7}{4} P_c=\pi *1\frac{3}{4} P_c=1\frac{3}{4}\pi

zad 10 P_c=\pi r (l+r) a) P_c=\pi 9 (41+9) P_c=\pi 9*50 P_c=450\pi b) P_c=\pi 9 (15+9) P_c=\pi 9*24 P_c=\pi 216 P_c=216\pi c) h^2+r^2=l^2 20^2+15^2=l^2 400+225=l^2 l^2=400+225 l^2=625 l=\sqrt{625} l=25 P_c=\pi 15 (25+15) P_c=\pi 15*40 P_c=\pi 600 P_c=600\pi

zad 11 l=10 cm 2r=12 r=6 P_c=\pi r (l+r) P_c=\pi 6 (10+6) P_c=\pi 6*16 P_c=\pi 96 P_c=96\pi

zad 12 36\pi=\pi r l \pi r l=36\pi r=\frac{36\pi}{\pi l} r=\frac{36}{l} a) l=6 P_c=\pi r (l+r) P_c=\pi \frac{36}{6} (6+\frac{36}{6}) P_c=\pi 6*(6+6) P_c=\pi 6*12 P_c=\pi 72 P_c=72\pi b) l=9 P_c=\pi r (l+r) P_c=\pi \frac{36}{9} (9+\frac{36}{9}) P_c=\pi 4 (9+4) P_c=\pi 4*13 P_c=\pi 52 P_c=52\pi c) l=12 P_c=\pi r (l+r) P_c=\pi \frac{36}{12} (12+\frac{36}{12}) P_c=\pi 3 (12+3) P_c=\pi 3*15 P_c=\pi 45 P_c=45\pi

zad 13 396\pi=\pi *9* l \pi *9* l=396\pi l=\frac{396\pi}{9\pi} l=\frac{396}{9} l=\frac{396}{9} l=44 h^2+r^2=l^2 h^2+9^2=44^2 h^2+81=1936 h^2=1936-81 h^2=1855 h=\sqrt{1855} h\approx43

zad 14 Obwód =2*\pi*r =55 , \pi=\frac{22}{7} podstawiam do tego równania i obliczam r. 2*(\frac{22}{7})*r = 55 (\frac{44}{7})*r = 55 r = 55*(\frac{7}{44})------skracam przez 11 r =\frac{35}{4} Wysokość stożka,promień i tworząca są bokami trójkąta prostokątnego.Tworzącą obliczę z tw.Pitagorasa. l^2 = r^2 + h^2 l^2 =(\frac{35}{4})^2 + 48^2 =2380,825 l=\sqrt{2380,825}\approx48,8 Pole boczne stożka /pi*r*l = (\frac{22}{7})*(\frac{35}{4})*48,8=[22*35*48,8]/[7*4]= =1342 cm^2)

too7 jesteś wielki !! Wielkie dzięki, pozdrawiam ;] 6!