Question

In ∆ABC, seg AP is a median. If BC = 18, AB² + AC² = 260, Find AP.

Answer 1

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

\[BP = PC = \frac{1}{2}BC = 9\]

\[{AB}^2 + {AC}^2 = 2 {AP}^2 + 2 {BP}^2\] ...(by Apollonius theorem)
\[ \Rightarrow 260 = 2 {AP}^2 + 2\left( 9^2 \right)\]
\[ \Rightarrow 260 = 2 {AP}^2 + 2\left( 81 \right)\]
\[ \Rightarrow 260 = 2 {AP}^2 + 162\]
\[ \Rightarrow 2 {AP}^2 = 260 - 162\]
\[ \Rightarrow 2 {AP}^2 = 98\]
\[ \Rightarrow {AP}^2 = 49\]
\[ \Rightarrow AP = 7\]

Hence, AP = 7.