Question

As medidas dos catetos de um triângulo retângulo são ( x + 5) cm e ( x + 1) cm e a
hipotenusa ( x + 9) cm. Determine o perímetro desse triângulo.

Answer 1

$(x + 9)^2=(x + 5)^2 + (x +1)^2\\x^2 +18x + 81 = x^2 + 10x + 25 + x^2 +2x + 1 \\x^2 - 2x^2 +18x -12x + 81 -26 = 0\\-x^2 + 6x + 55 = 0\\x^2 -6x -55=0\\\\ x = \frac{-(-6) +/- \sqrt{256} }{2}\\\\x = \frac{(6 +/-16}{2} \left \{ {{x'= 22/2} \atop {x''=-10/2}} \left \{ {{x'=11} \atop {x''=-5}} \right. \right. \\\\ (11 +9)^2 = (11 +5)^2 + (11+1)^2\\ 20^2 = 16^2 + 12^2 \\400 = 256 + 144\\ 400 = 400 \\\\ (-5+9)^2 = (-5 + 5)^2 + (5+1)^2\\4^2 = 0^2 + 6^2\\16 \neq 36$


$P_{t} = (x + 5) + (x+1) + (x+9) \\P_{t} = (11+5) +(11+1)+(11+9)\\P_{t} = 16 + 12 + 20\\\\P_{t} = 48 cm$