Mate Desarrollo
-
Upload
jesus-cruz-castro -
Category
Documents
-
view
128 -
download
4
Transcript of Mate Desarrollo
![Page 1: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/1.jpg)
1
![Page 2: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/2.jpg)
2
![Page 3: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/3.jpg)
3
![Page 4: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/4.jpg)
4
![Page 5: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/5.jpg)
5
![Page 6: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/6.jpg)
6
![Page 7: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/7.jpg)
7
![Page 8: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/8.jpg)
8
![Page 9: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/9.jpg)
9
![Page 10: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/10.jpg)
10
![Page 11: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/11.jpg)
11
![Page 12: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/12.jpg)
12
![Page 13: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/13.jpg)
13
![Page 14: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/14.jpg)
14
![Page 15: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/15.jpg)
15
![Page 16: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/16.jpg)
16
![Page 17: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/17.jpg)
17
![Page 18: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/18.jpg)
18
![Page 19: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/19.jpg)
CUBO
FORMULAS
V=a3
A=6a2
d=√3a
EJEMPLOSDETERMINAR A Y d EN CADA CASO
1. V=270 cm3
a=√270=6.46 cmA=6a2=6(6.46)2=250.38 cm2
d=√3a=√3(6.46)=4.40 cm2. V=333 cm3
a=√333=6.93cmA=6a2=6(6.93)2=288.25 cm2
d=√3a=√3(6.93)=4.56 cm3. V=500 cm3
a=√500=7.93cmA=6a2=6(7.93)2=377.97 cm2
d=√3a=√3(7.93)=4.87 cm
19
![Page 20: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/20.jpg)
PRISMA RECTANGULAR
FORMULAS
V=a∗b∗c
A=2 (ab+ac+bc )
d=√a2+b2+c2
EJEMPLOSV= 30,000 cm3
A= ? cm2
a= 50 cmb= 30 cmc= 20 cm
A=2 ⌈ (50∗30)+(50∗20 )+(30∗20)⌉A=2 (1500+1000+600 )
A=2 (3100 )A=6200 cm2
CILINDRO RECTO
20
![Page 21: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/21.jpg)
FORMULAS
V= π4d2h
A1=2πrh
A2=2πr (r+h) EJEMPLOS
V= 370 cm3
r= 8 cmh= ? cmd= 16 cm
DESPEJAMOS LA FORMULA DE V Y OBTENEMOS:h= V
π4d2
h= 370π4162
h=¿ 1.84 cm
PRISMA FORMULAS
V=A1h
21
![Page 22: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/22.jpg)
EJEMPLOS
V= 178 cm3
A1= ? cm2
h= ? cm
h= VA1
A1=VhPOR FALTA DE DATOS SUPONEMOS DOS NUMEROS
QUE AL MULTIPLICARSE NOS DEN 178 (44.5 Y 4)Y LOS SUSTITUIMOS EN LA FORMULA PARA COMPROBAR.
V=¿44.5)(4) = 178
h=4 cm A1=44.5cm2
CILINDRO HUECO FORMULAS
V= π4h (D2−d2)
EJEMPLOS
22
![Page 23: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/23.jpg)
V= 290 cm3
d= 3 cmD= 5 cmh= ? cm
290=π4h (52−32)
290π4
(52−32)=h
h=23.07cm
PIRAMIDE TRUNCA FORMULAS
V=h3
¿
EJEMPLOSV= 200 cm3
POR FALTA DE DATOS SUPONEMOS LOS VALORES DE A1 Y A2
23
![Page 24: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/24.jpg)
(50 Y 25) DESPEJAMOS h Y TENIENDO EL VALOR DE h, SUSTITUIMOS EN LA FORMULA PARA COMPROBAR
h= 3∗200(50+25+√50∗25)h=5.43698 cm
V=5.436983
(50+25+√50∗25)
V=200 cm3
PRISMA FORMULAS
V=A1∗h
EJEMPLOSV= 178 cm3
POR FALTA DE DATOS SUPONEMOS LOS VALORES DE h (5) DESPEJAMOS A1 Y TENIENDO EL VALOR DE A1, SUSTITUIMOS EN LA FORMULA PARA COMPROBAR
A1=1785
A1=35.6cm
V=35.6∗5V=178 cm3
24
![Page 25: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/25.jpg)
ESFERA FORMULAS
V= 43πr3=1
6πd3
≅ 4.189 r3
A=4 πr2=πd2
EJEMPLOS
V= 570 cm3
V=16πd3
3√ 57016 π=d
d=10.287cm
V=16π (10.287)3=570
PIRAMIDE25
![Page 26: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/26.jpg)
FORMULAS
V= Ah3
EJEMPLOSV= 120 cm3
SUPONEMOS EL VALOR DE h (6), DESPEJAMOS A Y SUSTITUIMOS EN V PARA COMPROBAR
A=3Vh
A=3 (120)6
A=60cm2
V=60∗63
V=120 cm3
CONO TRUNCADO FORMULAS
V= π12h (D2+Dd+d2)
26
![Page 27: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/27.jpg)
A=π2g (D+d )=2πph
g= [ (D−d ) /2 ]2+h
g=AREA DELCONO sinCONTAR LOSCORTES
SEGMENTO ESFERICO TRUNCADO FORMULAS
V= π6h (3a2+3b2+h2)
A=2πrh (ZONA ESFERICA )
27
![Page 28: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/28.jpg)
A=π (2rh+a2+b2)
EJEMPLOSa= 15 cmb= 25 cmr= 25 cmh= 44 cm
V= π6h (3a2+3b2+h2) A=2πr h (ZONA ESFERICA ) A=π (2r h+a2+b2)
V= π644(3 (15)2+3 (25)2+442) A=2π (25∗44) A=π (2(25∗44)+152+252)
V=103313 cm3 A=6911cm2 A=9581.86
SECCION ESFERICA FORMULAS
V=23π r2h
A=π2r (4h+d )
28
![Page 29: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/29.jpg)
EJEMPLOSr= 14 cmd= 28 cmh= 6 cm
V=23π ¿142∗6 A=π
214 (4 (6)+28 )
V=2463.01cm3 A=1143.54cm2
r= 13 cmd= 26 cmh= 14 cm
V=23π ¿132∗14 A=π
213 (4(14 )+26 )
V=4955 .34 cm3 A=1674 .47cm2
ESFERA CON PERFORACION CILINDRICA FORMULAS
V= π6h3
A=2π r h (R+r )
29
![Page 30: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/30.jpg)
ESFERA CON PERFORACION CONICA FORMULAS
V=23π r2h
A=2πr (h+√r2− h24 )
EJEMPLOSr= 44 cmh= 22 cm
V=23π∗442∗22 A=2π∗44 (22+√442−2224 )
30
![Page 31: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/31.jpg)
V=44602.2 cm3 A=6082.12 cm2
r= 13 cmh= 14 cm
V=23π∗18.52∗37 A=2π∗18.5(37+√18.52−3724 )
V=26521.8 cm3 A=4300 .84 cm2
TOROIDE(ANILLO DE SECCION CIRCULAR)
FORMULAS
V= π2
4Dd2
A=π 2Dd
EJEMPLOSd= 4.5 cmD= 12 cm V= π
2
4(12∗4.52) A=π 2¿)
V=599.57 cm3 A=532.95 cm2
d= 5 cmD= 14 cm
V= π2
4(14∗52) A=π 2¿)
31
![Page 32: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/32.jpg)
V=863.59 cm3 A=690.87cm2
BARRIL FORMULAS
V= π12h (2D2+d2)
EJEMPLOSD= 33 cmd= 20 cmh= 50 cm
V=( π12
∗50)(2¿332+202)
V=33745.9cm3
D= 24 cmd= 14 cmh= 33 cm
V=( π12
∗33)(2¿242+142)
V=11645.9 cm3
32
![Page 33: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/33.jpg)
PRISMA RECTANGULAR
FORMULAS33
![Page 34: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/34.jpg)
V=a∗b∗c
A=2 (ab+ac+bc )
d=√a2+b2+c2
PROBLEMACALCULAR d UTILIZANDO LOS VALORES SIGUIENTES
a = 15 cmb = 4 cmc = 6 cm
d=√152+42+62d=16.643cm
a = 20 cmb = 10 cmc = 5 cm
d=√202+102+52d=22.912cm
PIRAMIDE FORMULAS
V= Ah3
34
![Page 35: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/35.jpg)
PROBLEMASCALCULAR LOS SIGUIENTES VOLUMENES UTILIZANDO LOS VALORES DADOS
A= 480 cm2
h= 25 cm
V= 480∗253
V=4000 cm3
A= 600 cm2
h= 53 cm
V=600∗533
V=10600 cm3
ESFERA FORMULAS
V= 43πr3=1
6πd3
≅ 4.189 r3
A=4 πr2=πd2
35
![Page 36: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/36.jpg)
PROBLEMASCALCULAR LOS SIGUIENTES VOLUMENES UTILIZANDO LOS VALORES DADOS
d= 1050 cm
r= 525 cm
A= ?
V= ?
V= 43πr3 A=π∗10502
V=6.06 X108 cm3 A=3.46 X 106 cm2
CILINDRO HUECO FORMULAS
V= π4h (D2−d2)
PROBLEMASCALCULAR LOS SIGUIENTES VOLUMENES UTILIZANDO LOS VALORES DADOS
V= ? cm3
d= 3 cmD= 5 cm
36
![Page 37: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/37.jpg)
h= ? cm
V= π432(452−402)
V=10681.4 cm3
PIRAMIDE TRUNCA FORMULAS
V=h3
¿
PROBLEMASCALCULAR LOS SIGUIENTES VOLUMENES UTILIZANDO LOS VALORES DADOS
V= ? cm3
A1= 774 cm2
A2= 1400 cm2
h= 103 cm
V=1033
¿
V=110380 cm3
37
![Page 38: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/38.jpg)
PRISMA RECTANGULAR
FORMULAS
V=a∗b∗c
A=2 (ab+ac+bc )
d=√a2+b2+c2
PROBLEMASHACER UNA TABLA CON LOS VOLUMENES Y AREAS CORRESPONDIENTES PARA LOS SIGUIENTES VALORES DANDO UN 12% DE MARGEN PARA EL LIQUIDO.
# V A d +12%1 127 164.38 10.04 142.242 312.15 300 14.36 349.60
83 170 202 12.03 190.44 16.005 40.01 4.9 17.925
65 61.041 99.388 8 68.365
9
# a b c1 8 3 5.292 12.5 5 53 10.625 4 4
38
![Page 39: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/39.jpg)
4 4.00125
2 2
5 6.78233
3 3
PROBLEMA
DETERMINAR:
DATOS
I II III IV
a / 28 / /b / 10 / /c / 7 / /V 240 1960 14137.
2581.60
3d / 30.54 30 6.6A 40 1092 2827.4
32403.3
2h 18 / / 17r / / 15 22.5
I. PIRAMIDE RECTANGULARV= Ah
3=40∗18
3=240
II. PRISMA RECTANGULAR
V=a∗b∗c=28∗10∗7=1096
A=2 (ab+ac+bc )=2 (28∗10+28∗7+10∗7 )=1092
d=√a2+b2+c2=√282+102+72=30.545III. ESFERA
V=16πd3=1
6π 303=14137.2
A=πd2=π 302=2827.43IV. CILINDRO
V= π4d2h= π
46.62∗17=581.603
A1=2πrh=2 π∗22.5∗17=2403.32
39
![Page 40: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/40.jpg)
PROBLEMA
ENCONTRAR LOS DATOS FALTANTES PARA LOS TIPOS DE CONTENEDORES. PIRAMIDE RECTA
A= 25 cm2
h= 15 cm
V= Ah3
=25∗153
=125 cm3
PIRAMIDE TRUNCAh= 18 cmA1= 30 cm2
A2= 20 cm2
V=h3
¿
ESFERAd= 26 cmr= 13 cmA=V=
V=16πd3=1
6π∗263=9202.77c m3
A=πd2=π 262=2123.72c m2
PROBLEMA
40
![Page 41: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/41.jpg)
VISTA FRONTAL
VTOTAL= ? VTOTAL= 6154.62 cm3=61.54 m3
PIRAMIDE h=7 A=16V= Ah
3=16∗7
3=37.333 cm3
ESFERA r= 10V=1
6πd3=1
6π∗203=4188.7902 cm3
CUBO a= 4V=a3=43=64c m3
PRISMA RECTANGULAR a= 4 b=16 c=4V=a∗b∗c=4∗16∗4=256 c m3
CILINDRO r=8 h=16V= π
4d2h= π
4162∗16=61.54 cm3
PROBLEMAS
FORMULAS V=a∗b∗c
41
![Page 42: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/42.jpg)
A=2 (ab+ac+bc )
V 1=a∗b∗c=4.5∗14.5∗22.4=1461.6V 2=a∗b∗c=4.5∗14.5∗22.4=1461.6V 3=a∗b∗c=16.3∗12∗12=2347.2
V TOTAL=5270.4 cm3
A1=2 (ab+ac+bc )=2 (4.5∗14.5+4.5∗22.4+14.5∗22.4 )=533.7A2=2 (ab+ac+bc )=2 (4.5∗14.5+4.5∗22.4+14.5∗22.4 )=533.7A3=2 (ab+ac+bc )=2 (12∗12+12∗16.3∗12∗16.3 )=1070.4
ATOTAL=2141.8 cm2
PROBLEMAS
FORMULAS
V=16πd3
A=πd2
V= π4d2h
42
![Page 43: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/43.jpg)
A1=2πrh
V 1=16πd3=1
6π∗323=17157.3 c m3
V 2=16πd3=1
6π∗323=17157.3 c m3
V 3=16πd3=1
6π∗323=17157.3 c m3
V 4=π4d2h=π
424.42∗98.2=45917.8 cm3
V TOTAL=97389.7 cm3
A1=πd2=π 322=3216.99 cm2
A2=πd2=π 322=3216.99 cm2
A3=πd2=π 322=3216.99 c m2
A4=2πrh=2π∗12.2∗98.2=7527.51 cm2
ATOTAL=17178.5 cm2
PROBLEMAS
FORMULAS
V=h3
¿
43
![Page 44: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/44.jpg)
V 1=h3
¿
V 2=h3
¿
V TOTAL=850.133 cm3
PROBLEMAS
FORMULAS V= π
4d2h
V=a3
44
![Page 45: Mate Desarrollo](https://reader033.fdocumento.com/reader033/viewer/2022061122/546fa654b4af9f8c398b45a3/html5/thumbnails/45.jpg)
V 1=π4d2h= π
4202∗22=6911.5038c m3
V 2=a3=73=343cm3
V TOTAL=7254.5038 c m3
45