Question

Calcule o valor de X aplicando o Teorema de Pitágoras

Answer 1

$A) A^{2} = B^{2} + C^{2} 15^{2} = 12^{2} + X^{2} 225 = 144 + X^{2} 225 - 144 = X^{2} \sqrt{81} = X^{2} 9 = X^{2}$

$B) A^{2} = B^{2} + C^{2} 10^{2} = X^{2} + X^{2} 100 = 2x 2x = 100/2 = X^{2} x= \sqrt{50} x= 5\sqrt{2}$

$C) A^{2} = B^{2} + C^{2} 14^{2} = (5 \sqrt{3}) ^{2} + X^{2} 196 = 25.3 + X^{2} 196 = 75 + X^{2} x^{2} =196-75 x^{2} = 121 x = \sqrt{121} x = 11$

$D) A^{2} = B^{2} + C^{2} (x+1)^{2} = (\sqrt{7}) ^{2} + X^{2} (x+1).(x+1) = (\sqrt{7})^{2} + X^{2} X^{2} + X + X + 1 = 7 + X^{2} X^{2} + X^{2} + 2X = 7-1 2X = 6 X = 6/2 X = 3$

Answer 2

Explicação passo-a-passo:

$A) </p><p></p><p>A^{2} = B^{2} + C^{2} </p><p></p><p> 15^{2} = 12^{2} + X^{2} </p><p></p><p>225 = 144 + X^{2} </p><p></p><p>225 - 144 = X^{2} </p><p></p><p> \sqrt{81} = X^{2} </p><p></p><p>9 = X^{2} </p><p>B)</p><p></p><p> A^{2} = B^{2} + C^{2} </p><p></p><p> 10^{2} = X^{2} + X^{2} </p><p></p><p>100 = 2x</p><p></p><p>2x = 100/2 = X^{2} </p><p></p><p>x= \sqrt{50} </p><p></p><p>x= 5\sqrt{2} C) </p><p></p><p> A^{2} = B^{2} + C^{2} </p><p></p><p> 14^{2} = (5 \sqrt{3}) ^{2} + X^{2} </p><p></p><p>196 = 25.$