Respuesta:
[tex]25cm[/tex]
Explicación paso a paso:
[tex]Largo: L[/tex]
[tex]Ancho: a[/tex]
[tex]Alto: h = 15cm[/tex]
Diagonal de la cara de la base: [tex]d = 20cm[/tex]
Diagonal del prisma: D = ?
Por Pitágoras:
[tex]a^{2} +L^{2} = d^{2}[/tex]
[tex]a^{2} +L^{2} = 20^{2}[/tex]
[tex]a^{2} +L^{2} = 400[/tex] [tex]ecuac.1[/tex]
[tex]a^{2} +L^{2} +h^{2} = D^{2}[/tex]
[tex]a^{2} +L^{2} + 15^{2} = D^{2}[/tex] [tex]ecuac.2[/tex]
Restando ecuac.2 y ecuac.1
[tex]a^{2} +L^{2} +15^{2} = D^{2}[/tex]
[tex]-a^{2} -L^{2} = -400[/tex]
________________
[tex]+15^{2} = D^{2} -400[/tex]
[tex]+225 +400 = D^{2}[/tex]
[tex]625 = D^{2}[/tex]
[tex]D = \sqrt{625}[/tex]
[tex]D = 25 cm[/tex]
La medida de la diagonal del prisma es: [tex]25cm[/tex]
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Respuesta:
[tex]25cm[/tex]
Explicación paso a paso:
[tex]Largo: L[/tex]
[tex]Ancho: a[/tex]
[tex]Alto: h = 15cm[/tex]
Diagonal de la cara de la base: [tex]d = 20cm[/tex]
Diagonal del prisma: D = ?
Por Pitágoras:
[tex]a^{2} +L^{2} = d^{2}[/tex]
[tex]a^{2} +L^{2} = 20^{2}[/tex]
[tex]a^{2} +L^{2} = 400[/tex] [tex]ecuac.1[/tex]
[tex]a^{2} +L^{2} +h^{2} = D^{2}[/tex]
[tex]a^{2} +L^{2} + 15^{2} = D^{2}[/tex] [tex]ecuac.2[/tex]
Restando ecuac.2 y ecuac.1
[tex]a^{2} +L^{2} +15^{2} = D^{2}[/tex]
[tex]-a^{2} -L^{2} = -400[/tex]
________________
[tex]+15^{2} = D^{2} -400[/tex]
[tex]+225 +400 = D^{2}[/tex]
[tex]625 = D^{2}[/tex]
[tex]D = \sqrt{625}[/tex]
[tex]D = 25 cm[/tex]
La medida de la diagonal del prisma es: [tex]25cm[/tex]