Question

Determine o valor de X nos triângulos abaixo.​

Answer 1

$\boxed{\begin{array}{l}\boxed{\rm 1}\\\tt 1)~\sf x^2+4^2=5^2 \\\sf x^2+16=25\\\sf x^2=25-16\\\sf x^2=9\\\sf x=\sqrt{9}\\\sf x=3\\\tt b)~\sf x^2=9^2+12^2\\\sf x^2=81+144\\\sf x^2=225\\\sf x=\sqrt{225}\\\sf x=15\\\tt c)~\sf x^2+8^2=10^2\\\sf x^2+64=100\\\sf x^2=100-64\\\sf x^2=36\\\sf x=\sqrt{36}\\\sf x=6\\\tt d)~\sf(3x)^2+(4x)^2=20^2\\\sf 9x^2+16x^2=400\\\sf 25x^2=400\\\sf x^2=\dfrac{400}{25}\\\sf x^2=16\\\sf x=\sqrt{16}\\\sf x=4\end{array}}$

$\boxed{\begin{array}{l}\boxed{\rm 2}\\\sf x^2=12^2+16^2\\\sf x^2=144+256\\\sf x^2=400\\\sf x=\sqrt{400}\\\sf x=20\\\sf y^2=15^2+x^2\\\sf y^2=225+400\\\sf y^2=625\\\sf y=\sqrt{625}\\\sf y=25\end{array}}$

$\boxed{\begin{array}{l}\boxed{\rm 3}\\\sf h^2+3^2=5^2\\\sf h^2+9=25\\\sf h^2=25-9\\\sf h^2=16\\\sf h=\sqrt{16}\\\sf h=4~m\end{array}}$