Question

neste triangulo retângulo a hipotenuza BC mede 20cm calcule. ajudem aii por favor agradeço

Answer 1

$h^2=m\cdot{n}\\\\(6)^2=(x)\cdot(20-x)\\\\36=20x-x^2\\\\x^2-20x+36=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4\cdot{a}\cdot{c}}}{2\cdot{a}}\\\\\\x=\dfrac{-(-20)\pm\sqrt{(-20)^2-4\cdot1\cdot36}}{2\cdot1}\\\\\\x=\dfrac{20\pm\sqrt{400-144}}{2}\\\\\\x=\dfrac{20\pm\sqrt{256}}{2}\\\\\\x=\dfrac{20\pm16}{2}\left\langle\begin{matrix}x'=18\\\\\boxed{x''=2}\end{matrix}\right$


Tendo em vista que o comprimento 'x' é o menor, seu valor corresponde à x''=2. O valor de 'x' é 2 cm.