Respuestas
Respuesta dada por:
1
¡Hola!
1) Tenemos los siguientes datos del paralelepipedo:
a = 4,5 cm
b = 3,2 cm
c = 1,8 cm
Vamos a encontrar el volumen del paralelepipedo
![V_{par} = a*b*c V_{par} = a*b*c](https://tex.z-dn.net/?f=V_%7Bpar%7D+%3D+a%2Ab%2Ac)
![V_{par} = 4,5*3,2*1,8 V_{par} = 4,5*3,2*1,8](https://tex.z-dn.net/?f=V_%7Bpar%7D+%3D+4%2C5%2A3%2C2%2A1%2C8)
![\boxed{\boxed{V_{par} = 25,92\:cm^3}}\end{array}}\qquad\checkmark \boxed{\boxed{V_{par} = 25,92\:cm^3}}\end{array}}\qquad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7BV_%7Bpar%7D+%3D+25%2C92%5C%3Acm%5E3%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Ccheckmark)
Vamos a encontrar el área lateral del paralelepipedo:
![AL_{par} = 2*(a)*(b) + 2*(a)*(c) AL_{par} = 2*(a)*(b) + 2*(a)*(c)](https://tex.z-dn.net/?f=AL_%7Bpar%7D+%3D+2%2A%28a%29%2A%28b%29+%2B+2%2A%28a%29%2A%28c%29)
![AL_{par} = 2*(4,5)*(3,2) + 2*(4,5)*(1,8) AL_{par} = 2*(4,5)*(3,2) + 2*(4,5)*(1,8)](https://tex.z-dn.net/?f=AL_%7Bpar%7D+%3D+2%2A%284%2C5%29%2A%283%2C2%29+%2B+2%2A%284%2C5%29%2A%281%2C8%29)
![AL_{par} = 28,8 + 16,2 AL_{par} = 28,8 + 16,2](https://tex.z-dn.net/?f=AL_%7Bpar%7D+%3D+28%2C8+%2B+16%2C2)
![\boxed{\boxed{AL_{par} = 45\:cm^2}}\end{array}}\qquad\checkmark \boxed{\boxed{AL_{par} = 45\:cm^2}}\end{array}}\qquad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7BAL_%7Bpar%7D+%3D+45%5C%3Acm%5E2%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Ccheckmark)
Vamos a encontrar el Área Total del paralelepipedo:
![AT_{par} = 2*(a*b+a*c+b*c) AT_{par} = 2*(a*b+a*c+b*c)](https://tex.z-dn.net/?f=AT_%7Bpar%7D+%3D+2%2A%28a%2Ab%2Ba%2Ac%2Bb%2Ac%29)
![AT_{par} = 2*(4,5*3,2+4,5*1,8+3,2*1,8) AT_{par} = 2*(4,5*3,2+4,5*1,8+3,2*1,8)](https://tex.z-dn.net/?f=AT_%7Bpar%7D+%3D+2%2A%284%2C5%2A3%2C2%2B4%2C5%2A1%2C8%2B3%2C2%2A1%2C8%29)
![AT_{par} = 2*(14,4+8,1+5,76) AT_{par} = 2*(14,4+8,1+5,76)](https://tex.z-dn.net/?f=AT_%7Bpar%7D+%3D+2%2A%2814%2C4%2B8%2C1%2B5%2C76%29)
![AT_{par} = 2*(28,26) AT_{par} = 2*(28,26)](https://tex.z-dn.net/?f=AT_%7Bpar%7D+%3D+2%2A%2828%2C26%29)
![\boxed{\boxed{AT_{par} = 56,52\:cm^2}} \end{array}}\qquad\checkmark \boxed{\boxed{AT_{par} = 56,52\:cm^2}} \end{array}}\qquad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7BAT_%7Bpar%7D+%3D+56%2C52%5C%3Acm%5E2%7D%7D+%5Cend%7Barray%7D%7D%5Cqquad%5Ccheckmark)
2) Tenemos los siguientes datos del prisma triangular:
altura (h) = 5 m
lado de la base (l) = 3 m
perímetro de la base = 3 + 3 + 3 + 3 = 12 m
utilizar: √3 ≈ 1,73
Vamos a encontrar el área lateral del prisma triangular:
![AL_{pt} = Perimetro\:de\:la\:base\:\:*\:\:\:altura AL_{pt} = Perimetro\:de\:la\:base\:\:*\:\:\:altura](https://tex.z-dn.net/?f=AL_%7Bpt%7D+%3D+Perimetro%5C%3Ade%5C%3Ala%5C%3Abase%5C%3A%5C%3A%2A%5C%3A%5C%3A%5C%3Aaltura)
![AL_{pt} = 12 *5 AL_{pt} = 12 *5](https://tex.z-dn.net/?f=AL_%7Bpt%7D+%3D+12+%2A5)
![\boxed{\boxed{AL_{pt} = 60\:m^2}}\end{array}}\qquad\checkmark \boxed{\boxed{AL_{pt} = 60\:m^2}}\end{array}}\qquad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7BAL_%7Bpt%7D+%3D+60%5C%3Am%5E2%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Ccheckmark)
Vamos a encontrar el área de la base del prisma triangular:
![A_{base} = \frac{l^2* \sqrt{3} }{4} A_{base} = \frac{l^2* \sqrt{3} }{4}](https://tex.z-dn.net/?f=A_%7Bbase%7D+%3D++%5Cfrac%7Bl%5E2%2A+%5Csqrt%7B3%7D+%7D%7B4%7D+)
![A_{base} = \frac{3^2* \sqrt{3} }{4} A_{base} = \frac{3^2* \sqrt{3} }{4}](https://tex.z-dn.net/?f=A_%7Bbase%7D+%3D+%5Cfrac%7B3%5E2%2A+%5Csqrt%7B3%7D+%7D%7B4%7D)
![A_{base} = \frac{9* 1,73 }{4} A_{base} = \frac{9* 1,73 }{4}](https://tex.z-dn.net/?f=A_%7Bbase%7D+%3D+%5Cfrac%7B9%2A+1%2C73+%7D%7B4%7D)
![A_{base} = \frac{15,57}{4} A_{base} = \frac{15,57}{4}](https://tex.z-dn.net/?f=A_%7Bbase%7D+%3D+%5Cfrac%7B15%2C57%7D%7B4%7D)
![A_{base} = 3,8925\to \boxed{A_{base} \approx 3,90\:m^2} A_{base} = 3,8925\to \boxed{A_{base} \approx 3,90\:m^2}](https://tex.z-dn.net/?f=A_%7Bbase%7D+%3D+3%2C8925%5Cto+%5Cboxed%7BA_%7Bbase%7D+%5Capprox+3%2C90%5C%3Am%5E2%7D)
Vamos a encontrar el área total del prisma triangular:
![AT_{pt} = AL_{pt} + 2*A_{base} AT_{pt} = AL_{pt} + 2*A_{base}](https://tex.z-dn.net/?f=AT_%7Bpt%7D+%3D+AL_%7Bpt%7D+%2B+2%2AA_%7Bbase%7D)
![AT_{pt} = 60 + 2*3,90 AT_{pt} = 60 + 2*3,90](https://tex.z-dn.net/?f=AT_%7Bpt%7D+%3D+60+%2B+2%2A3%2C90)
![AT_{pt} = 60 + 7,8 AT_{pt} = 60 + 7,8](https://tex.z-dn.net/?f=AT_%7Bpt%7D+%3D+60+%2B+7%2C8)
![\boxed{\boxed{AT_{pt} = 67,8\:m^2}}\end{array}}\qquad\checkmark \boxed{\boxed{AT_{pt} = 67,8\:m^2}}\end{array}}\qquad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7BAT_%7Bpt%7D+%3D+67%2C8%5C%3Am%5E2%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Ccheckmark)
Vamos a encontrar el volumen del prisma triangular
![V_{pt} = A_{base}*altura V_{pt} = A_{base}*altura](https://tex.z-dn.net/?f=V_%7Bpt%7D+%3D+A_%7Bbase%7D%2Aaltura)
![V_{pt} = 3,90*5 V_{pt} = 3,90*5](https://tex.z-dn.net/?f=V_%7Bpt%7D+%3D+3%2C90%2A5)
![\boxed{\boxed{V_{pt} = 19,5\:m^3}}\end{array}}\qquad\checkmark \boxed{\boxed{V_{pt} = 19,5\:m^3}}\end{array}}\qquad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7BV_%7Bpt%7D+%3D+19%2C5%5C%3Am%5E3%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Ccheckmark)
3) Tenemos los siguientes datos del cilindro:
h (altura) = 0,4 dm → 4 cm
r (rayo) = 2 cm
utilizar: π ≈ 3,14
Vamos a encontrar el área lateral del cilindro:
![AL_{cil} = 2* \pi *r*h AL_{cil} = 2* \pi *r*h](https://tex.z-dn.net/?f=AL_%7Bcil%7D+%3D+2%2A+%5Cpi+%2Ar%2Ah)
![AL_{cil} = 2* 3,14*2*4 AL_{cil} = 2* 3,14*2*4](https://tex.z-dn.net/?f=AL_%7Bcil%7D+%3D+2%2A+3%2C14%2A2%2A4)
![\boxed{\boxed{AL_{cil} = 50,24\:cm^2}}\end{array}}\qquad\checkmark \boxed{\boxed{AL_{cil} = 50,24\:cm^2}}\end{array}}\qquad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7BAL_%7Bcil%7D+%3D+50%2C24%5C%3Acm%5E2%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Ccheckmark)
Vamos a encontrar el volumen del cilindro:
![V_{cil} = \pi *r^2*h V_{cil} = \pi *r^2*h](https://tex.z-dn.net/?f=V_%7Bcil%7D+%3D++%5Cpi+%2Ar%5E2%2Ah)
![V_{cil} = 3,14 *2^2*4 V_{cil} = 3,14 *2^2*4](https://tex.z-dn.net/?f=V_%7Bcil%7D+%3D+3%2C14+%2A2%5E2%2A4)
![V_{cil} = 3,14 *4*4 V_{cil} = 3,14 *4*4](https://tex.z-dn.net/?f=V_%7Bcil%7D+%3D+3%2C14+%2A4%2A4)
![\boxed{\boxed{V_{cil} = 50,24\:cm^3}}\end{array}}\qquad\checkmark \boxed{\boxed{V_{cil} = 50,24\:cm^3}}\end{array}}\qquad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7BV_%7Bcil%7D+%3D+50%2C24%5C%3Acm%5E3%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Ccheckmark)
Vamos a encontrar el área de la base del cilindro:
![A_{base\:cilindro} = \pi *r^2 A_{base\:cilindro} = \pi *r^2](https://tex.z-dn.net/?f=A_%7Bbase%5C%3Acilindro%7D+%3D++%5Cpi+%2Ar%5E2)
![A_{base\:cilindro} = 3,14 *2^2 A_{base\:cilindro} = 3,14 *2^2](https://tex.z-dn.net/?f=A_%7Bbase%5C%3Acilindro%7D+%3D+3%2C14+%2A2%5E2)
![A_{base\:cilindro} = 3,14 *4 A_{base\:cilindro} = 3,14 *4](https://tex.z-dn.net/?f=A_%7Bbase%5C%3Acilindro%7D+%3D+3%2C14+%2A4)
![\boxed{A_{base\:cilindro} = 12,56\:cm^2} \boxed{A_{base\:cilindro} = 12,56\:cm^2}](https://tex.z-dn.net/?f=%5Cboxed%7BA_%7Bbase%5C%3Acilindro%7D+%3D+12%2C56%5C%3Acm%5E2%7D)
Usando la información arriba, vamos a encontrar el Área total del cilindro:
![AT_{cilindro} = AL_{cil} + 2*(A_{base\:cil}) AT_{cilindro} = AL_{cil} + 2*(A_{base\:cil})](https://tex.z-dn.net/?f=AT_%7Bcilindro%7D+%3D+AL_%7Bcil%7D+%2B+2%2A%28A_%7Bbase%5C%3Acil%7D%29)
![AT_{cilindro} = 2* \pi *r*h + 2*( \pi *r^2) AT_{cilindro} = 2* \pi *r*h + 2*( \pi *r^2)](https://tex.z-dn.net/?f=AT_%7Bcilindro%7D+%3D+2%2A+%5Cpi+%2Ar%2Ah+%2B+2%2A%28+%5Cpi+%2Ar%5E2%29)
![AT_{cilindro} = 2* \pi *r*(h+r) AT_{cilindro} = 2* \pi *r*(h+r)](https://tex.z-dn.net/?f=AT_%7Bcilindro%7D+%3D+2%2A+%5Cpi+%2Ar%2A%28h%2Br%29)
![AT_{cilindro} = 2* 3,14 *2*(4+2) AT_{cilindro} = 2* 3,14 *2*(4+2)](https://tex.z-dn.net/?f=AT_%7Bcilindro%7D+%3D+2%2A+3%2C14+%2A2%2A%284%2B2%29)
![AT_{cilindro} = 2* 3,14 *2*(6) AT_{cilindro} = 2* 3,14 *2*(6)](https://tex.z-dn.net/?f=AT_%7Bcilindro%7D+%3D+2%2A+3%2C14+%2A2%2A%286%29)
![AT_{cilindro} = 2* 3,14 *12 AT_{cilindro} = 2* 3,14 *12](https://tex.z-dn.net/?f=AT_%7Bcilindro%7D+%3D+2%2A+3%2C14+%2A12)
![\boxed{\boxed{AT_{cilindro} = 75,36\:cm^2}}\end{array}}\qquad\checkmark \boxed{\boxed{AT_{cilindro} = 75,36\:cm^2}}\end{array}}\qquad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7BAT_%7Bcilindro%7D+%3D+75%2C36%5C%3Acm%5E2%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Ccheckmark)
o si desea resolver directamente sustituyendo los resultados encontrados, tenemos:
![AT_{cilindro} = AL_{cil} + 2*(A_{base\:cil}) AT_{cilindro} = AL_{cil} + 2*(A_{base\:cil})](https://tex.z-dn.net/?f=AT_%7Bcilindro%7D+%3D+AL_%7Bcil%7D+%2B+2%2A%28A_%7Bbase%5C%3Acil%7D%29)
![AT_{cilindro} = 50,24 + 2*(12,56) AT_{cilindro} = 50,24 + 2*(12,56)](https://tex.z-dn.net/?f=AT_%7Bcilindro%7D+%3D+50%2C24+%2B+2%2A%2812%2C56%29)
![AT_{cilindro} = 50,24 + 25,12 AT_{cilindro} = 50,24 + 25,12](https://tex.z-dn.net/?f=AT_%7Bcilindro%7D+%3D+50%2C24+%2B+25%2C12)
![\boxed{\boxed{AT_{cilindro} = 75,36\:cm^2}}\end{array}}\qquad\checkmark \boxed{\boxed{AT_{cilindro} = 75,36\:cm^2}}\end{array}}\qquad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7BAT_%7Bcilindro%7D+%3D+75%2C36%5C%3Acm%5E2%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Ccheckmark)
¡Espero haber ayudado!
1) Tenemos los siguientes datos del paralelepipedo:
a = 4,5 cm
b = 3,2 cm
c = 1,8 cm
Vamos a encontrar el volumen del paralelepipedo
Vamos a encontrar el área lateral del paralelepipedo:
Vamos a encontrar el Área Total del paralelepipedo:
2) Tenemos los siguientes datos del prisma triangular:
altura (h) = 5 m
lado de la base (l) = 3 m
perímetro de la base = 3 + 3 + 3 + 3 = 12 m
utilizar: √3 ≈ 1,73
Vamos a encontrar el área lateral del prisma triangular:
Vamos a encontrar el área de la base del prisma triangular:
Vamos a encontrar el área total del prisma triangular:
Vamos a encontrar el volumen del prisma triangular
3) Tenemos los siguientes datos del cilindro:
h (altura) = 0,4 dm → 4 cm
r (rayo) = 2 cm
utilizar: π ≈ 3,14
Vamos a encontrar el área lateral del cilindro:
Vamos a encontrar el volumen del cilindro:
Vamos a encontrar el área de la base del cilindro:
Usando la información arriba, vamos a encontrar el Área total del cilindro:
o si desea resolver directamente sustituyendo los resultados encontrados, tenemos:
¡Espero haber ayudado!
Anónimo:
Para una mejor visualización, actualice la página
Preguntas similares
hace 6 años
hace 9 años
hace 9 años