Question

aplicando as relações métricas nos triângulos abaixo, determine o Valor de X. PRECISO DA RESPOSTA AGORA POR FAVOR!!!!​

Answer 1

$\large\boxed{\begin{array}{l}\sf a)~\rm 12\cdot n=6^2\\\rm 12n=36\\\rm n=\dfrac{36}{12}\\\\\rm n=3\end{array}}$

$\large\boxed{\begin{array}{l}\sf b)~\rm b^2=(3+9)\cdot3\\\rm b^2=12\cdot3\\\rm b^2=36\\\rm b=\sqrt{36}\\\rm b=6\end{array}}$

$\large\boxed{\begin{array}{l}\sf c)~\rm3\cdot x=(2\sqrt{6})^2\\\rm 3x=4\cdot6\\\rm 3x=24\\\rm x=\dfrac{24}{3}\\\\\rm x=8\\\rm y^2=3\cdot(8-3)\\\rm y^2=3\cdot5\\\rm y^2=15\\\rm y=\sqrt{15} \end{array}}$

$\large\boxed{\begin{array}{l}\sf d)~\rm a=2+4=6\\\rm h^2=2\cdot4\\\rm h=\sqrt{2\cdot4}\\\rm h=4\sqrt{2}\\\rm c^2=6\cdot2\\\rm c^2=12\\\rm c=\sqrt{12}\\\rm c=\sqrt{2^2\cdot3}\\\rm c=2\sqrt{3}\\\rm a\cdot h=b\cdot c\\\rm b=\dfrac{a\cdot h}{c}\\\\\rm b=\dfrac{6\cdot4\sqrt{2}}{2\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}\\\\\rm b=\dfrac{12\sqrt{6}}{2\cdot3}=\dfrac{12\sqrt{6}}{6}\\\\\rm b=2\sqrt{6}\end{array}}$