Question

Determina el volumen del tronco de pirámide
Por favor ayuda es para mi examen

Answer 1

Respuesta:

         [tex]1.09m^{3}[/tex]

Explicación paso a paso:

[tex]Lado: L= 1.4m[/tex]

[tex]Lado: L' = 1m[/tex]

[tex]ap = \sqrt{0.8^{2}-(\frac{1.4-1}{2} )^{2} } =\sqrt{0.64-(0.2)^{2 } }=\sqrt{0.64-0.04} =\sqrt{0.6m^{2} } = 0.77m[/tex]

[tex]ap^{2} = h^{2} + (\frac{L-L'}{2} )^{2}[/tex]

[tex](0.77)^{2} = h^{2} + (\frac{1.4-1}{2} )^{2}[/tex]

[tex](0.77)^{2} = h^{2} + (0.2)^{2}[/tex]

[tex]0.59 =h^{2} +0.04[/tex]

[tex]0.59-0.04 = h^{2}[/tex]

[tex]0.55 = h^{2}[/tex]

[tex]h = \sqrt{0.55m^{2} }[/tex]

[tex]h = 0.74m[/tex]

[tex]Area-base-mayor: AB = L^{2} = (1.4m)^{2} =1.96m^{2}[/tex]

[tex]Area-de-la-menor:Ab = (L')^{2} =(1m)^{2} = 1m^{2}[/tex]

[tex]Volumen: V = ?[/tex]

[tex]V = \frac{h}{3} (AB +Ab+ \sqrt{AB*Ab} )[/tex]

[tex]V= \frac{0.74m}{3} (1.96m^{2}+1m^{2} +\sqrt{1.96m^{2} +1m^{2} } )[/tex]

[tex]V= 0.25m ( 2.96m^{2} +\sqrt{1.96m^{2} } ) = 0.25m( 2.96m^{2} +1.4m^{2} )[/tex]

[tex]V = 0.25m(4.36m^{2} ) = 1.09m^{3}[/tex]

[tex]V = 1.09m^{3}[/tex]