[tex]\huge{\boxed{P_c = 436cm^2}}[/tex]
[tex]\huge{\boxed{P_c = 2P_p + P_b}}[/tex]
Pole podstawy:
[tex]a= 6cm\\b=12cm\\h = 4cm\\\\P_p = \dfrac{(a+b) \cdot h}{2} = \dfrac{(6cm+12cm)\cdot 4cm}{2} = \dfrac{18cm \cdot \not 4 ^2cm}{\not 2_1} = \large{\boxed{36cm^2}}[/tex]
Pole powierzchni bocznej:
[tex]a=6cm\\b = 12cm\\c=5cm\\d=5cm\\H=13cm\\\\P_b = a \cdot H + b \cdot H + c \cdot H + d \cdot H = 6cm \cdot 13cm + 12cm \cdot 13cm + 5cm \cdot 13cm +\\+ 5cm \cdot 13cm = 78cm^2 +156cm^2 + 65cm^2 + 65cm^2 = \large{\boxed{364cm^2}}[/tex]
Pole całkowite:
[tex]P_p = 36cm^2\\P_b = 364cm^2\\\\P_c = 2P_p+P_b = 2 \cdot 36cm^2 + 364cm^2 = 72cm^2 + 364cm^2 = \huge{\boxed{436cm^2}}[/tex]
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Odpowiedź:
[tex]\huge{\boxed{P_c = 436cm^2}}[/tex]
Szczegółowe wyjaśnienie:
Wzór na pole powierzchni całkowitej graniastosłupa:
[tex]\huge{\boxed{P_c = 2P_p + P_b}}[/tex]
Liczymy pole powierzchni całkowitej graniastosłupa:
Pole podstawy:
[tex]a= 6cm\\b=12cm\\h = 4cm\\\\P_p = \dfrac{(a+b) \cdot h}{2} = \dfrac{(6cm+12cm)\cdot 4cm}{2} = \dfrac{18cm \cdot \not 4 ^2cm}{\not 2_1} = \large{\boxed{36cm^2}}[/tex]
Pole powierzchni bocznej:
[tex]a=6cm\\b = 12cm\\c=5cm\\d=5cm\\H=13cm\\\\P_b = a \cdot H + b \cdot H + c \cdot H + d \cdot H = 6cm \cdot 13cm + 12cm \cdot 13cm + 5cm \cdot 13cm +\\+ 5cm \cdot 13cm = 78cm^2 +156cm^2 + 65cm^2 + 65cm^2 = \large{\boxed{364cm^2}}[/tex]
Pole całkowite:
[tex]P_p = 36cm^2\\P_b = 364cm^2\\\\P_c = 2P_p+P_b = 2 \cdot 36cm^2 + 364cm^2 = 72cm^2 + 364cm^2 = \huge{\boxed{436cm^2}}[/tex]