Respuestas
Respuesta dada por:
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A = 9450
P = 390
Para el area utilizamos la sgte formula
A = b * h
Y para el permimetro
P = 2b + 2h
Entonces reemplazamos con los datos que tenemos
*A = 9450
b * h = A
b * h = 9450
*P = 390
2b + 2h = P
2b + 2h = 390
Entonces nos queda un sistema de ecuaciones
![\left \{ {{b * h = 9450 } \atop {2b + 2h=390}} \right.
\left \{ {{b * h = 9450 } \atop {2b + 2h=390}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bb+%2A+h+%3D+9450+%7D+%5Catop+%7B2b+%2B+2h%3D390%7D%7D+%5Cright.+%0A)
Resolvemos, método: sustitución
2b + 2h = 390
![2b = 390 - 2h
\\ b = \frac{390-2h}{2}
2b = 390 - 2h
\\ b = \frac{390-2h}{2}](https://tex.z-dn.net/?f=2b+%3D+390+-+2h%0A+%5C%5C++b+%3D++%5Cfrac%7B390-2h%7D%7B2%7D%0A+)
b * h = 9450
![\frac{390-2h}{2} * h = 9450
\\\\(390-2h) * h = 9450*2
\\\\ 390h-2h^2 =18900
\\\\0 = 18900 +2h^2-390h
\\\\2h^2-390h+18900=0
\frac{390-2h}{2} * h = 9450
\\\\(390-2h) * h = 9450*2
\\\\ 390h-2h^2 =18900
\\\\0 = 18900 +2h^2-390h
\\\\2h^2-390h+18900=0](https://tex.z-dn.net/?f=+%5Cfrac%7B390-2h%7D%7B2%7D+%2A+h+%3D+9450+%0A%5C%5C%5C%5C%28390-2h%29+%2A+h+%3D+9450%2A2%0A%5C%5C%5C%5C+390h-2h%5E2+%3D18900%0A%5C%5C%5C%5C0+%3D+18900+%2B2h%5E2-390h%0A%5C%5C%5C%5C2h%5E2-390h%2B18900%3D0%0A%0A)
Utilizando formula general
donde
a = 2 ; b = -390 ; c = 18900
![x_{1,\:2}=\frac{-(-390)\pm \sqrt{(-390)^2-4(2)(18900)}}{2(2)}
\\\\x_{1,\:2}=\frac{390\pm \sqrt{152100-151200}}{4}
\\\\x_{1,\:2}=\frac{390\pm \sqrt{900}}{4}
\\\\x_{1,\:2}=\frac{390\pm 30}{4}
\\ ---------------------------\\x_{1}=\frac{390+30}{4}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{2}=\frac{390-30}{4}
\\\\x_{1}=\frac{420}{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{2}=\frac{360}{4}
\\\\x_{1}=105 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_2=90
x_{1,\:2}=\frac{-(-390)\pm \sqrt{(-390)^2-4(2)(18900)}}{2(2)}
\\\\x_{1,\:2}=\frac{390\pm \sqrt{152100-151200}}{4}
\\\\x_{1,\:2}=\frac{390\pm \sqrt{900}}{4}
\\\\x_{1,\:2}=\frac{390\pm 30}{4}
\\ ---------------------------\\x_{1}=\frac{390+30}{4}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{2}=\frac{390-30}{4}
\\\\x_{1}=\frac{420}{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{2}=\frac{360}{4}
\\\\x_{1}=105 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_2=90](https://tex.z-dn.net/?f=x_%7B1%2C%5C%3A2%7D%3D%5Cfrac%7B-%28-390%29%5Cpm+%5Csqrt%7B%28-390%29%5E2-4%282%29%2818900%29%7D%7D%7B2%282%29%7D%0A%5C%5C%5C%5Cx_%7B1%2C%5C%3A2%7D%3D%5Cfrac%7B390%5Cpm+%5Csqrt%7B152100-151200%7D%7D%7B4%7D%0A%5C%5C%5C%5Cx_%7B1%2C%5C%3A2%7D%3D%5Cfrac%7B390%5Cpm+%5Csqrt%7B900%7D%7D%7B4%7D%0A%5C%5C%5C%5Cx_%7B1%2C%5C%3A2%7D%3D%5Cfrac%7B390%5Cpm+30%7D%7B4%7D%0A%5C%5C+---------------------------%5C%5Cx_%7B1%7D%3D%5Cfrac%7B390%2B30%7D%7B4%7D%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+x_%7B2%7D%3D%5Cfrac%7B390-30%7D%7B4%7D%0A%5C%5C%5C%5Cx_%7B1%7D%3D%5Cfrac%7B420%7D%7B4%7D+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C++%5C++%5C++%5C++%5C++%5C+x_%7B2%7D%3D%5Cfrac%7B360%7D%7B4%7D+%0A%5C%5C%5C%5Cx_%7B1%7D%3D105+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C++%5C++%5C+%5C+%5C+%5C++x_2%3D90%0A%0A)
Las longitudes del terreno son de 105 y 90 metros respectivamente
P = 390
Para el area utilizamos la sgte formula
A = b * h
Y para el permimetro
P = 2b + 2h
Entonces reemplazamos con los datos que tenemos
*A = 9450
b * h = A
b * h = 9450
*P = 390
2b + 2h = P
2b + 2h = 390
Entonces nos queda un sistema de ecuaciones
Resolvemos, método: sustitución
2b + 2h = 390
b * h = 9450
Utilizando formula general
donde
a = 2 ; b = -390 ; c = 18900
Las longitudes del terreno son de 105 y 90 metros respectivamente
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