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Ray Of Light [21]
3 years ago
5

Tenemos un rectángulo cuya área es (15n + 9nx) cm 2 y su base mide (3n) cm. Existe también un cuadrado con la misma altura del r

ectángulo, tal que si a su área se le disminuye (9x 2 ) cm 2, , ésta medirá 1825 cm 2 . ¿Cuánto mide el área original del cuadrado?
Mathematics
1 answer:
LekaFEV [45]3 years ago
3 0

Answer:

El área original del cuadrado es de 1825 centímetros cuadrados.

Step-by-step explanation:

De acuerdo con el enunciado, tenemos que el área del cuadrilátero es:

(15\cdot n +9\cdot n\cdot x)\,cm^{2} = (3\cdot n\,cm)\cdot h (1)

Donde h es la altura del cuadrilátero, medida en centímetros.

De ahí sacamos que la altura es igual a:

h = (15+9\cdot x)\,cm

También se sabe que el cuadrado con la misma altura del cuadrilátero menos un área determinada es igual a lo siguiente:

(h^{2}-9\cdot x^{2})\,cm^{2} = 1825\,cm^{2}

[(15+9\cdot x)^{2}-9\cdot x^{2}]\,cm^{2} = 1825\,cm^{2}

225+270\cdot x+81\cdot x^{2}-9\cdot x^{2} = 1825

72\cdot x^{2}+270\cdot x +225 = 1825

72\cdot x^{2}+270\cdot x -1600 = 0 (2)

Esta es una ecuación cuadrática, el cual puede ser resuelto por la Fórmula de la Cuadrática, cuyas soluciones son, respectivamente:

x_{1} \approx 3.198 y x_{2}\approx -6.948

Sabemos que las longitudes son variables positivas, por tanto, solo se debe escoger aquellas soluciones de x que cumplan tal condición. Empleamos la altura del cuadrilátero para la evaluación:

x_{1} \approx 3.198

h_{1} = 15+9\cdot (3.198)

h_{1}\approx 43.782\,cm

x_{2}\approx -6.948

h_{2} = 15+9\cdot (-6.948)

h_{2}\approx -47.532\,cm

Entonces, el valor apropiado de x es aproximadamente 3.198.

Por último el área original del cuadrado es:

A = (43.782\,cm)^{2}

A = 1916.864\,cm^{2}

El área original del cuadrado es de 1825 centímetros cuadrados.

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