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In ∆ABC, seg AP is a median. If BC = 18, AB^{2}^{ }+ AC^{2}^{ }= 260, Find AP.

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#### Solution

In ∆ABC, point P is the midpoint of side BC.

\[ \Rightarrow 260 = 2 {AP}^2 + 2\left( 9^2 \right)\]

\[ \Rightarrow 260 = 2 {AP}^2 + 2\left( 81 \right)\]

\[ \Rightarrow 260 = 2 {AP}^2 + 162\]

\[ \Rightarrow 2 {AP}^2 = 260 - 162\]

\[ \Rightarrow 2 {AP}^2 = 98\]

\[ \Rightarrow {AP}^2 = 49\]

\[ \Rightarrow AP = 7\]

Hence, AP = 7.

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