ACTIVIDAD 4 GEOMETRIA ANALITICA
Enviado por kiarasoret • 25 de Abril de 2014 • 699 Palabras (3 Páginas) • 271 Visitas
ACTIVIDAD 4 GEOMETRIA ANALITICA
A(0,0) ; B(10,0) ; C(13,0) ; D(16,-4) ; E(20,-4) ; F(24,0) ; G(26,0) ; H(27,0) ; I(37,0) ; J(20,30) ; K(19,30) ; L(16,30)
PUNTO SECCION PERIMETRO(m) AREA (m2) Volumen (m3) Costo (USD)
A,B,L (1)
Enrocamiento AB+BL+LA
10+30.59+34
SUBTOTAL=74.59 150 150(50)
SUBTOTAL=
7500 500(750)
SUBTOTAL=
3,750,000
B,C,D,E,F,G,K,L (2)
CORAZON
IMPERMEBALE
MATERIAL FINO
BC+CD+DE+EF+FG+GK+KL+LB
3+5+4+5.66+2+30.80+3+30.59
SUBTOTAL= 84.05 315 315(50)
SUBTOTAL=
15,750 200(15,750)
SUBTOTAL=
3,150,000
G,H,J,K (3)
FILTRO DE TRANSICION DE GRAVA Y ARENA GH+HJ+JK+KG
1+30.80+1+30.80
SUBTOTAL=63.60 30 30(50)
SUBTOTAL=
1,500 300(1500)
SUBTOTAL=
450,000
H,I,J (4)
ENROCAMIENTO HI+IJ+JH
10+34.48+30.80
SUBTOTAL= 75.28 150 150(50)
SUBTOTAL=
7,500 400(7500)
SUBTOTAL = 3,000,000
TOTAL PERIMETRO=
29752 TOTAL
AREA=
645 TOTAL VOLUMEN
32,250 COSTO TOTAL= 10,350,000 USD
LINEA Nº X2 X1 Y2 Y1
1 16 0 30 0 =34---------LA
2 16 10 30 0 =30.59-------LB
3 16 13 -4 0 =5---------CD
4 24 20 0 -4 =5.66--------EF
5 19 26 30 0 =30.80-------GK
6 20 27 30 0 =30.80--------HJ
7 20 37 30 0 34.48-------IJ
AB = √(X2) –(X1)²+ (Y2)-Y1)²
1.- AB= √ (16-0)² +(30-0)² =√ (16)² +(30)² =√(256)+(900) = √1156= 34
2.-AB= √(16-10)²+(30)-0)²=√(16)+(30)²=√36+900=√936=30.59
3.- AB= √(16-13)²+(-4-0)²=√(13)²+(-4)²=√9+16=√25=5
4.- AB =√AB(24-20)²+(0(-4)²=√ 16+16)=√32=5.66
5.- AB=√(19-26)²+(30-0)²=√(-7)²+(30)²=√49+900=√949=30.80
6.-AB=√(20-27)²+(30-0)²=√(-7)²+(30)²=√49+900=√949=30.80
7.- AB=√(20-37)²+(30-0)²=√(17)²+(30)²=√289+900=√1189=34.48
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