The third proportional to $$(\frac{x}{y}+\frac{y}{x}) and \sqrt{x^{2}+y^{2}}$$ is
Let the third proportion be x.
$$(\frac{x}{y}+\frac{y}{x}) : \sqrt{x^{2}+y^{2}}$$ :: $$ \sqrt{x^{2}+y^{2}} : z$$
$$(\frac{x^2+y^2}{xy}) : \sqrt{x^{2}+y^{2}}$$ :: $$ \sqrt{x^{2}+y^{2}}: z$$
$$(\frac{\sqrt{x^2+y^2}}{xy}) : 1$$ :: $$\frac{ \sqrt{x^{2}+y^{2}}}{z}$$
$$(\frac{\sqrt{x^2+y^2}}{xy}) $$ :: $$\frac{ \sqrt{x^{2}+y^{2}}}{z}$$
z=xy
Option A is the correct answer.
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