Question

determine o valor da incognita x nos triangulos retangulos

Answer 1

Utilize o teorema de Pitágoras: $a^2=b^2+c^2$.

a) $x^2=(\sqrt{6})^2+(\sqrt{19})^2~~\Rightarrow~~x^2=6+19~~\Rightarrow~~x^2=25$

$x=\sqrt{25}~~\Rightarrow~~\boxed{x=5}$.


b) $(3\sqrt{5})^2=x^2+3^2~~\Rightarrow~~45=x^2+9~~\Rightarrow~~x^2=36$

$x=\sqrt{36}~~\Rightarrow~~\boxed{x=6}$.


c) $(7x)^2=(5x)^2+(\sqrt{6})^2~~\Rightarrow~~49x^2=25x^2+6~~\Rightarrow~~24x^2=6$

$x^2=\dfrac{1}{4}~~\Rightarrow~~x=\sqrt{\dfrac{1}{4}}~~\Rightarrow~~\boxed{x=\dfrac{1}{2}}$.


d) $(x+4)^2=x^2+12^2~~\Rightarrow~~x^2+8x+16=x^2+144~~\Rightarrow~~8x=128$

$x=\dfrac{128}{8}~~\Rightarrow~~\boxed{x=16}$.


e) $(x+6)^2=x^2+(x+3)^2~~\Rightarrow~~x^2+12x+36=x^2+x^2+6x+9~~\Rightarrow~~x^2-6x-27=0$

$\Delta=(-6)^2-4\cdot1\cdot(-27)=36+108=144$

$x=\dfrac{-(-6)\pm\sqrt{144}}{2}=\dfrac{6\pm12}{2}=3\pm6$

$x'=3+6=9$ e $x"=3-6=-3$ (não serve).