Tabla Areas Y Volumenes

Páginas: 2 (258 palabras) Publicado: 4 de marzo de 2013
TABLA DE AREAS Y VOLUMENES Cuadrado Triángulo

A = a2
Rectángulo

A=

B⋅h 2

A = B⋅h
Rombo

Romboide

A = B⋅hA=

D⋅d 2

Trapecio

A=

(B + b ) ⋅ h
2

Polígono regular

P⋅a A= 2

P = 2π ⋅ R A = π ⋅ R2
Sector circularCírculo

a es la apotema / P es el perímetro (suma de la longitud de los lados)

Corona circular

A=π ⋅ R −r
2

(

2

)A=

π ⋅ R2 ⋅ n
360

Cubo

A = 6a 2 V = a3

A = 2π ⋅ R(h + R ) V = π ⋅ R2 ⋅ h
Prisma recto

Cilindro

A = 2 ⋅ (ab+ ac + bc ) V = a ⋅b⋅c
A = π ⋅ R (h + g )
2

Ortoedro

AB es área de la base

A = P ⋅ (h + a ) V = AB ⋅ h

Cono

A =p ⋅ g ⋅ (r + R ) + r2 + R 2 V = p ⋅ h ⋅ R + r + Rr 3
2 2

Tronco de cono

[(

g es la generatriz

V =

π ⋅R ⋅h
2

()

)]

3
2

Tetraedro regular

Esfera

A = 3⋅a
V =

2 ⋅ a2 12

A = 4 ⋅π ⋅ R2 4 ⋅π ⋅ R3 V = 3
Huso: cuñaesférica

Pirámide recta

P ⋅ (a + a ′) A= 2 A ⋅h V = B 3

4π ⋅ R 2 ⋅ n A= 360 V ⋅n V= E 360
Casquete esférico

Octaedroregular

A = 2 3⋅a
V =

2

2 ⋅ a3 3

A = 2π ⋅ R ⋅ h π ⋅ h 2 ⋅ (3R − h ) V = 3
Zona esférica

′ (P + P ′) ⋅ a ⋅ AB + AB A=2 ′ ′ A + AB + AB ⋅ AB ⋅ h V = B 3

Tronco de pirámide

(

)

A = 2π ⋅ R ⋅ h π ⋅ h ⋅ (h 2 + 3r 2 + 3r ′ 2 ) V = 6

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