Respuesta:
[tex]\frac{9\sqrt{3} }{2}[/tex]
Explicación paso a paso:
Hallando el otro cateto (x)
[tex]6^{2}[/tex] = [tex]3^{2}[/tex] + [tex]x^{2}[/tex]
36 = 9 + [tex]x^{2}[/tex]
[tex]x^{2}[/tex] = 27
x =[tex]\sqrt{27}[/tex]
Área del triangulo: [tex](\frac{b . h}{2})[/tex]
[tex]\frac{3. \sqrt{27} }{{2} }[/tex] = [tex]\frac{9\sqrt{3} }{2}[/tex]
[tex] {6}^{2} = {3}^{2} + {x}^{2} [/tex]
[tex]36 = 9 + {x}^{2} [/tex]
[tex]36 - 9 = {x}^{2} [/tex]
[tex]27 = {x}^{2} [/tex]
[tex] \sqrt{27} = x[/tex]
[tex]5.19 = x[/tex]
A = b×h/2
[tex]a = \frac{5.19 \times 3}{2} [/tex]
[tex]a = \frac{15.57}{2} [/tex]
[tex]a = 7.785m[/tex]
espero te sirva saludos✓
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Verified answer
Respuesta:
[tex]\frac{9\sqrt{3} }{2}[/tex]
Explicación paso a paso:
Hallando el otro cateto (x)
[tex]6^{2}[/tex] = [tex]3^{2}[/tex] + [tex]x^{2}[/tex]
36 = 9 + [tex]x^{2}[/tex]
[tex]x^{2}[/tex] = 27
x =[tex]\sqrt{27}[/tex]
Área del triangulo: [tex](\frac{b . h}{2})[/tex]
[tex]\frac{3. \sqrt{27} }{{2} }[/tex] = [tex]\frac{9\sqrt{3} }{2}[/tex]
Explicación paso a paso:
hola ✓
Para poder sacar el área primero tenemos que saber el valor del otro cateto ✓
[tex] {6}^{2} = {3}^{2} + {x}^{2} [/tex]
[tex]36 = 9 + {x}^{2} [/tex]
[tex]36 - 9 = {x}^{2} [/tex]
[tex]27 = {x}^{2} [/tex]
[tex] \sqrt{27} = x[/tex]
[tex]5.19 = x[/tex]
ya sabiendo el valor del otro cateto remplazamos los datos a la formula !
A = b×h/2
[tex]a = \frac{5.19 \times 3}{2} [/tex]
[tex]a = \frac{15.57}{2} [/tex]
[tex]a = 7.785m[/tex]
espero te sirva saludos✓