Q.If in triangle \( \triangle ABC \), the sum of the medians AD, BE, and CF is \(x\), and the sum of the sides is \(y\), then the relationship between \(x\) and \(y\) is — (a) \(x\gt y\) (b) \(x\lt y\) (c) \(x= y\) (d) None of the above
Answer: B
AD < AB + BD BE < BC + EC CF < AC + AF ∴ AD + BE + CF < AB + BC + AC + \(\frac{1}{2}\)(BC + AC + AB) ⇒ AD + BE + CF < \(\frac{3}{2}\)(BC + AC + AB) ∴ \(x < \frac{3}{2}y\) That is, \(x < y\)
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