wysokosc walca jest 2razy wieksza od srednicy podstawy.pole podstawy wynosi 9pi. oblicz pole calkowite

h = 2 * 2r = 4r Pp = 9π = πr² r = 3 h = 2 * 6 = 12 Pb = 2πr * h = 6π * 12 = 72π P = 2Pp + Pb = 2 * 9π + 72π = 90π

H = 2*śr Pp = 9π Pp = πr² πr² = 9π |:π r² = 9 r = √9 r = 3 średnica = 2r śr = 2*3 śr = 6 H = 2*śr H = 2*6 = 12 Pc = 2πrH+2πr² Pc = 2π*3*12+2π*3² Pc = 72π+2π*9 Pc = 72π+18π Pc = 90π

Pp = πr² 9π =πr² / π 9 = r² r = √9 r = 3 2r=6 6 * 2 = 12 H = 12 Pc = 2πr² + 2πrH Pc = 2π* 3² + 2π*3*12 Pc= 2π *9 + 2π * 36 Pc= 18π + 72π Pc= 90π