Question

Tema
relaciones métricas
procedimiento​

Answer 1

Verified answer

Respuesta:

Explicación paso a paso:

1.

[tex]AB= 2u ; BC = 7u ; AC = 8u ; AH = x ; HC = 8-x[/tex]

Por Pitágoras:

[tex]Como: (HB)^{2} = (HB)^{2}[/tex]

[tex]2^{2} -x^{2} =7^{2} -(8-x)^{2}[/tex]

[tex]4-x^{2} =49-64+16x-x^{2}[/tex]

[tex]-x^{2} +x^{2} -16x=49-64-4[/tex]

[tex]-16x = -19[/tex]

[tex]x = \frac{-19}{-16} = \frac{19}{16}[/tex]

[tex]Luego: x = \frac{19}{16}[/tex]

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2 )

[tex]AB= 2m ; BC = 5m; AC = 6m[/tex]

Trazando la altura BD del vértice B a un Punto D del lado AC,entonces:

[tex]AD = x ; DC = 6-x[/tex]

[tex]Como: ( BD)^{2} = (BD)^{2}[/tex]

Por Pitágoras.

[tex]2^{2} -x^{2} =5^{2} - (6-x)^{2}[/tex]

[tex]4-x^{2} =25-36+12x-x^{2}[/tex]

[tex]-12x = 25-36-4[/tex]

[tex]-12x =-15[/tex]

[tex]x = \frac{-15}{-12} = \frac{5}{4}[/tex]

[tex]x = 1.25[/tex]

[tex]Altura: BD = ?[/tex]

[tex]BD = \sqrt{(AB)^{2} -x^{2} } = \sqrt{2^{2} -(1.25)^{2} } =\sqrt{4-1.5625} =\sqrt{2.4375}[/tex]

[tex]BD = 1.56m[/tex]

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3)

[tex]AB= 5cm; BC= 12cm; AC = 10cm; BM = MC; AM = ?[/tex]

[tex](AB)^{2} +(AC)^{2} =2(AM)^{2} +\frac{(BC)^{2} }{2}[/tex]

[tex]5^{2} +(10)^{2} = 2(AM)^{2} +\frac{(12)^{2} }{2}[/tex]

[tex]25+100= 2 (AM)^{2} +\frac{144}{2}[/tex]

[tex]125 = 2(AM)^{2} +72[/tex]

[tex]125-72 = 2(AM)^{2}[/tex]

[tex]53 = 2 (AM)^{2}[/tex]

[tex](AM)^{2} = \frac{53}{2}[/tex]

[tex](AM)^{2} = 26.5[/tex]

[tex]AM = \sqrt{26.5}[/tex]

[tex]AM = 5.15m[/tex]

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4 )

[tex]BC = 10m ; AC = 5m ; AB = 6m ; BH = ? ; AH= ?[/tex]

[tex]( BH)^{2} = (BH)^{2}[/tex]

[tex](AC)^{2}- y^{2} =(AB)^{2} -(CB-y)^{2}[/tex]

[tex]5^{2} -y^{2} = 6^{2} -(10-y)^{2}[/tex]

[tex]25-y^{2} = 36-100+20y-y^{2}[/tex]

[tex]-20y =36-100-25[/tex]

[tex]-20y = -89[/tex]

[tex]y = \frac{-89}{-20 } = 4.45m[/tex]

[tex](AB)^{2} =x^{2} +y^{2}[/tex]

[tex]6 ^{2} = x^{2} + (4.45)^{2}[/tex]

[tex]36 =x^{2} +19.8[/tex]

[tex]36-19.8 = x^{2}[/tex]

[tex]16.2 = x^{2}[/tex]

[tex]x = \sqrt{16.2}[/tex]

[tex]x = 4.02m[/tex]