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3 years ago

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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 al área del cuadrado se le disminuye (9x 2 ) cm 2 , ésta medirá 1825 cm 2 . ¿Cuánto mide el área original del cuadrado?

1 answer:

lana66690 [7]3 years ago

4 0

Answer:

El área original del cuadrado es 34225 cm²

Step-by-step explanation:

Los parámetros dados son;

El área del rectángulo = (15·n + 9·n·x) cm²

La longitud de la base del rectángulo = (3·n) cm

La altura del cuadrado = La altura del rectángulo

El área original del cuadrado - 9·x² cm² = 1825 cm²

La altura del rectángulo = (El área del rectángulo) / (La longitud de la base del rectángulo)

∴ La altura del rectángulo = (15·n + 9·n·x) / (3·n) = 5 + 3·x

La altura del rectángulo = 5 + 3 · x = La altura del cuadrado

El área del cuadrado = Lado × Lado = Altura × Altura = (5 + 3·x) × (5 + 3·x)

∴ El área del cuadrado = (5 + 3·x) × (5 + 3·x) = 9·x² + 30·x + 25

Del área original del cuadrado - 9·x² cm² = 1825 cm², tenemos;

9·x² + 30·x + 25 - 9·x² = 1825

30·x + 25 = 1825

30·x = 1825 - 25 = 1800

x = 1800/30 = 60

x = 60

El área original del cuadrado - 9·x² cm² = 1825 cm²

El área original del cuadrado = 1825 cm² + 9·x² cm² = 1825 cm² + 9 × 60² cm² = 34225 cm²

El área original del cuadrado también se puede encontrar de la siguiente manera;

El área original del cuadrado = (5 + 3 × 60) × (5 + 3 × 60) = 34225 cm²

El área original del cuadrado = 34225 cm².

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