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.

$\mathrm{BP}=\mathrm{PC}=\frac{1}{2}\mathrm{BC}=9$

$$\begin{gathered} {\mathrm{AB}}^2+{\mathrm{AC}}^2=2{\mathrm{AP}}^2+2{\mathrm{BP}}^2\left(\mathrm{by}\mathrm{Apollonius}\mathrm{theorem}\right) \\ \Rightarrow 260=2{\mathrm{AP}}^2+2\left(9^2\right) \\ \Rightarrow 260=2{\mathrm{AP}}^2+2\left(81\right) \\ \Rightarrow 260=2{\mathrm{AP}}^2+162 \\ \Rightarrow 2{\mathrm{AP}}^2=260-162 \\ \Rightarrow 2{\mathrm{AP}}^2=98 \\ \Rightarrow {\mathrm{AP}}^2=49 \\ \Rightarrow \mathrm{AP}=7 \end{gathered}$$

Hence, AP = 7.