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Pb(pole boczne)= 4 x Pt(pole trójkątów)
Pt= 1/2 x 8 x h
4² + h² = 10²
h²= 100 - 16 = 84
h=√84= 2 √21
Pb= 4 x 1/2 x 8 x 2√21= 32√21
Pc(pole całkowite)= Pb + Pp= 64 + 32√21= 32(2 +√21)
Pp = a²
Pb = a*h/2
Pp = 8² = 64
Pb = 4*(a*h/2)
Pb=2ah
Korzystając z tw. Pitagorasa (załącznik0 wyliczamy h
h²+4²=10²
h²+16=100
h²=100-16
h²=84 /√
h=√84
h=2√21
Pb = 2*8*2√21
Pb = 32 √21
Pc = 64 = 32 √21