Oblicz pole powierzchni całkowitej ostrosłupa.
[ w załączniku] Poproszę o każde drobne obliczenie.
a) Pp= 6*6=36
Pb= 4*(a² √3)/4
Pb= 4*(6² √3)/4= 4*(36√3)/4= 9√3
Pc= Pp+Pb
Pc= 36+36√3
b) Pp= (a² √3)/4
Pp= (2² √3)/4= (4√3)/4= √3
Pb= 3*(a² √3)/4
Pb= 3*(2² √3)/4= (3*4√3)/4= 12√3/4=3√3
Pc= √3+3√3
c) a²= 3²+4²
a²= 9+16=25
a=5
Pp= 5*5=25
Pb= 4*a*h
Pb= 4*5*12= 240
Pc= 25+240= 265
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a) Pp= 6*6=36
Pb= 4*(a² √3)/4
Pb= 4*(6² √3)/4= 4*(36√3)/4= 9√3
Pc= Pp+Pb
Pc= 36+36√3
b) Pp= (a² √3)/4
Pp= (2² √3)/4= (4√3)/4= √3
Pb= 3*(a² √3)/4
Pb= 3*(2² √3)/4= (3*4√3)/4= 12√3/4=3√3
Pc= Pp+Pb
Pc= √3+3√3
c) a²= 3²+4²
a²= 9+16=25
a=5
Pp= 5*5=25
Pb= 4*a*h
Pb= 4*5*12= 240
Pc= Pp+Pb
Pc= 25+240= 265