dado o paralelepipedo abaixo, de dimensões 8 cm, 10cm e 12cm, determine:
A- a diagonal de base;
B- a diaginal do paralelepipedo;
C- a area da base;
D- a area lateral;
E- a area total;
F- o volume ;
Respostas
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124
DIAGONAL DE BASE |====>
d = \/ a^2 + b^2
d = \/ (8)^2 + (10)^2
d = \/ 64 + 100
d = \/ 164cm
d = 12,80cm
DIAGONAL DO PARALELEPIPEDO |===>
d = \/(a)^2 + (b)^2 + (c)^2
d = \/ (8)^2 + (10)^2 + (12)^2
d = \/ 64 + 100 + 144
d = \/ 164 + 144
d = \/308 cm
d = 17,54cm
AREA DA BASE |=====>
Ab = a x b
Ab = 8 x 10
Ab = 80 cm^2
AREA LATERAL |=====>
Al = 2 ( ac + bc )
Al = 2 (8x12 + 10x12)
Al = 2( 96 + 120)
Al = 2x (216)
Al = 432cm
AREA TOTAL |====>
At = 2 x (ab + ac + bc)
At = 2 x ( 8.10 + 8.12 + 10.12)
At = 2x ( 80 + 96 + 120)
At = 2 x ( 176 + 120)
At = 2 x ( 296)
At = 592cm^2
VOLUME |=====>
V = a x b x c
V = 8 x 10 x 12
V = 80 x 12
V = 960cm^3
d = \/ a^2 + b^2
d = \/ (8)^2 + (10)^2
d = \/ 64 + 100
d = \/ 164cm
d = 12,80cm
DIAGONAL DO PARALELEPIPEDO |===>
d = \/(a)^2 + (b)^2 + (c)^2
d = \/ (8)^2 + (10)^2 + (12)^2
d = \/ 64 + 100 + 144
d = \/ 164 + 144
d = \/308 cm
d = 17,54cm
AREA DA BASE |=====>
Ab = a x b
Ab = 8 x 10
Ab = 80 cm^2
AREA LATERAL |=====>
Al = 2 ( ac + bc )
Al = 2 (8x12 + 10x12)
Al = 2( 96 + 120)
Al = 2x (216)
Al = 432cm
AREA TOTAL |====>
At = 2 x (ab + ac + bc)
At = 2 x ( 8.10 + 8.12 + 10.12)
At = 2x ( 80 + 96 + 120)
At = 2 x ( 176 + 120)
At = 2 x ( 296)
At = 592cm^2
VOLUME |=====>
V = a x b x c
V = 8 x 10 x 12
V = 80 x 12
V = 960cm^3
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