#### Question

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

#### Solution

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

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

\[{AB}^2 + {AC}^2 = 2 {AP}^2 + 2 {BP}^2 \left( \text{by Apollonius theorem} \right)\]

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

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

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

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

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

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

\[ \Rightarrow AP = 7\]

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

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

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

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

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

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

\[ \Rightarrow AP = 7\]

Hence, AP = 7.

Is there an error in this question or solution?

Solution In ∆Abc Seg Ap is a Median. If Bc = 18, Ab2 + Ac2 = 260 Find Ap. Concept: Apollonius Theorem.