Jakie pole powierzchni ma ostrosłup prawidłowy trójkątny o wysokości 5 cm i krawędzi podstawy 9 cm? PILNE!!
Pc= Pp +Pb
h=a√3/2 cm h=9√3/2cm
1/3*9√3/2 = 3√3/25²+(3√3/2)²=h²
h²=25+27/4 h²=25+27/4 h²=127/4 h=√127/√4 h=√127/2cm
Pp=a²√3/4 Pp=9²√3/4 Pp=81√3/4cm²
Pb=1/2*a*h Pb=1/2*9*√127/2 Pp=9√127/4cm²
Pc=81√3/4+3*9√127/4 Pc=(81√3+27√127) /4Pc=27(3√3+√127) /4 cm²
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Pc= Pp +Pb
h=a√3/2 cm
h=9√3/2cm
1/3*9√3/2 = 3√3/2
5²+(3√3/2)²=h²
h²=25+27/4
h²=25+27/4
h²=127/4
h=√127/√4
h=√127/2cm
Pp=a²√3/4
Pp=9²√3/4
Pp=81√3/4cm²
Pb=1/2*a*h
Pb=1/2*9*√127/2
Pp=9√127/4cm²
Pc=81√3/4+3*9√127/4
Pc=(81√3+27√127) /4
Pc=27(3√3+√127) /4 cm²