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π r² + 33 π = π ( r + x )²
sustituimos r
16 π + 33 π = π ( 16 + 8 x + x² )
49 π = 16 π + 8 π x + π x² igualamos a cero
π x² + 8 π x - 33 π = 0 ecuación de 2o. grado
a = π ; b = 8 π ; c = - 33 π
x = - 8π +- √ 64π² +132π² / 2π
x = - 8π +- √ 196π² / 2π
x = - 8π +- 14π / 2π podemos eliminar π y no queda
x₁ = - 8 + 14 / 2 = 3 la otra solución es negativa y se descarta
entonces hay que agregarle 3 cm al radio