Question

Del Grafico, Si BP = 6, el valor de BC es:

a) 6√6
b)9√ 2
c)6√ 3
d)7√ 3

Answer 1

Respuesta:

$\mathbf{c).-6\sqrt{3}}$

Explicación paso a paso:

$\texttt{\sl{Usemos el Teorema de Pit\'agoras, para encontrar}}\\\texttt{\sl{el valor de m, donde:}}\\\mathbf{a=2\sqrt{m^{2}+9}\,\,, b=m+2m=3m\,\,, c=\sqrt{m^{2}+46}}\\\texttt{\sl{Entonces aplicando el Teorema nos queda:}}\\a^{2}+c^{2}=b^{2}\\\\(2\sqrt{m^{2}+9})^{2}+(\sqtt{m^{2}+36})^{2}=(3m)^{2}\\\\\texttt{\sl{Desarrollando}}\\\\\mathbf{4(m^{2}+9)+m^{2}+36=9m^{2}}\\\\\texttt{\sl{Simplificando}}\\\\\mathbf{4m^{2}+36+m^{2}+36=9m^{2}}\\\\\mathbf{5m^{2}-9m^{2}+72=0}\\\\\mathbf{-4m^{2}+72=0}\\\\\mathbf{-4m^{2}=-72}\\\\\mathbf{4m^{2}=72}\\\\\mathbf{m^{2}=\frac{72}{4}=18}\\\\\mathbf{m=\pm3\sqrt{2}}\\\\\texttt{\sl{En consecuencia, descartando la}}\\\texttt{\sl{la ra\'iz negativa, los lados buscados son:}}\\\\\mathbf{b=3m=3(3\sqrt{2})=9\sqrt{2}}\\\\\mathbf{m=3\sqrt{2}}\\\\\mathbf{2m=2(3\sqrt{2})=6\sqrt{2}}$

$\texttt{\sl{Por lo tanto, aplicando nuevamente}}\\\texttt{\sl{el Teorema de Pit\'agoras para hallar BC nos queda:}}\\\\\mathbf{BC=\sqrt{(6\sqrt{2})^{2}+(6)^{2}}}\\\\\mathbf{BC=\sqrt{36(2)+ +36}}\\\\\mathbf{BC=\sqrt{72+36}}\\\\\mathbf{BC=\sqrt{108}=6\sqrt{3}}$