Respuesta:
[tex]A_{L} = 48\sqrt{2} u^{2}[/tex]
Explicación paso a paso:
Datos:
Area de la sección recta: [tex]A_{SR} = 12\sqrt{3} u^{2}[/tex]
Altura: [tex]h = 2\sqrt{3} u[/tex]
Area Lateral: [tex]A_{L} =?[/tex]
Arista: a = ?
[tex]Sen 60 = \frac{h}{a}[/tex]
[tex]\frac{\sqrt{3} }{2} = \frac{2\sqrt{3} }{a}[/tex]
[tex]( a ) ( \sqrt{3} ) = ( 2) ( 2\sqrt{3 } )[/tex]
[tex](a)(\sqrt{3}) = 4\sqrt{3}[/tex]
[tex]a = \frac{4\sqrt{3} }{\sqrt{3} }[/tex]
[tex]a = 4u[/tex]
Area del sección hexagonal:
[tex]A_{SR} = \frac{3L^{2}\sqrt{3} }{2}[/tex]
[tex]12\sqrt{3} = \frac{3L^{2} \sqrt{3} }{2}[/tex]
[tex]3L^{2} \sqrt{3} = ( 2 ) ( 12\sqrt{3} )[/tex]
[tex]( L^{2}) ( 3\sqrt{3} ) = ( 24\sqrt{3})[/tex]
[tex]L^{2} =\frac{24\sqrt{3} }{3\sqrt{3} } =\frac{24}{3}[/tex]
[tex]L^{2} = 8[/tex]
[tex]L = \sqrt{8} = \sqrt{4 * 2} = 2\sqrt{2} u[/tex]
Perímetro de la base:
[tex]P_{B} = 6L = 6 ( 2\sqrt{2} ) = 12\sqrt{2}[/tex] u
Area Lateral:
[tex]A_{L} = P_{B} * a = ( 12\sqrt{2}u )( 4u)[/tex]
[tex]A_{L} = 48\sqrt{2} u^{2}[/tex]
