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GEOMETRIA ANALITICA


Enviado por   •  29 de Agosto de 2014  •  1.225 Palabras (5 Páginas)  •  369 Visitas

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Tabla resumen

Número Sección Perímetro (m) Área (m2)

1 Enrocamiento 93.48 200 u2

2 Corazón 100 902.5 u2

3 Filtro 89.6 40 u2

4 Enrocamiento 99.08 180 u2

Enrocamiento

AB=10

BC= B (xb, yb) y C (xc, yc)

Es decir

Xb= 10 xc= 16

Yb= 0 yc= 40

Sustituir en la formula

(BC)2 = (BC)2 X + (BC)2 Y

(BC)2= + (yc-yb)2

(BC)2 =

(BC)2= = √1636

BC= 40.4

(CA)=?

C=(16,40)

A= (0,0)

C= (Xc,Yc) y A (Xa,Ya)

Es decir

Xa = 0 ; Xc = 16

Ya = 0 ; Yc = 40

CA = (CA)x2 + (CA) y2

CA =

CA =

CA =

Perímetro:

AB + BC + CA= 10 + 40.4 + 43.08 =93.48 m

Corazón

DB

B (10,0)

D (13,0)

Dx = 13

Bx = 10

Dy = 0

By = 0

PB = (Dx - Bx) – (Dy - By)

DB = (13 - 10) – (0 - 0)

DB = (3 - 0) = 3

DB =3

ED

E (16, - 5)

D (13 ,0)

Ex = 16

Dx = 13

Ey = -5

Dy = 0

ED = (Ex –Dx ) – ( Ey - Dy)

ED = ( 16 - 13) – ( -5 – (- 5)

ED = (3) – (-5 )

ED = 8

FE

F (20, -5)

E (16, -5)

Fx = 20

Ex = 16

Fy = -5

Ey = -5

FE = (Fx - Ex) – (Fy – Ey)

FE = (20 - 16) – (- 5 – (-5)

FE = 4 – (0)

FE = 4

GF

G = (24,0)

F = (20, - 5)

Gx = 24

Fx = 20

Gy = 0

Fy = -5 GF = (6x – Fx ) – ( Gy – Fy)

GF = (24 -20) – (0-(-5)

GF = 4 – (5) = -1

HJ = H (XH, YH) ; H( 26, 0)

Y J (XJ, YJ) ; (19,40)

XH= 26 XJ= 19

YH= 0 YJ= 40

SUSTITUIR EN FORMULA

(HJ)2= (HJ)2X + (HJ)2 Y

(HJ)2=

(HJ)2=

(HJ)2= =

= 40.60 m

JC

J(19.40)

C=(16,40

J( XJ, YJ) Y C ( XC, YC)

ES DECIR

XJ= 19 XC= 16

YJ= 40 YC= 40 SUSTITUIR EN FORMULA

(JC)2=

(JC)2 =

(JC)2 =

(JC)2 =

= 3

Perímetro Corazón

DB + ED + FE + GF + HG + JH + JC + BC

3 + 8 + 4 + (-1) + 2 + 40.60 + 3 + 40.4 = 100 m

Distancia Filtro

IH =

H= (26, 0)

I = (27, 0)

Hx = 26

Ix = 27

Hy = 0

Iy = 0

HI = ( Ix - Hx) – ( Iy - Hy)

HI = (27 – 26 ) – (0 – 0 )

HI = ( 1 – 0 ) = 1

KI =

K= ( 20, 40)

I = ( 27, 0 )

Kx = 20

Ky = 40

Ix = 27

Iy = 0

KI =(Ix - Kx) – ( Iy – Ky )

KI = ( 27 – 20 ) – (0 – 40 )

KI = 7 – (-40 ) = 47

KJ=

K = ( 20, 40 )

J = ( 19, 40 )

Kx = 20

Ky = 40

Jx = 19

Jy = 40 KJ = ( Kx – Jx ) – ( Jy - Ky)

KJ = ( 20 – 19 ) – ( 40 - 40)

KJ = ( 1) – ( O ) = 1

Perímetro Filtro

HI + IK + KJ + JH = 1+ 47 + 1 + 40.60 = 89.6 m

Enrocamiento

IA =

I ( 27, 0)

A ( 36, 0)

Ax = 36

Ay = 0

Ix = 27

Iy = 0

IA = ( Ax – Ix ) – ( Ay - Iy)

...

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