Question

In $△ABC$, $AP$ is a median. If $AP=7$, ${AB}^2+{AC}^2=260$, then find $BC$.

Answer 1

In $△ABC$,
$AP$ is a median
$\therefore PB=PC$, and $AP=7$
In $△APB$ by Pythagoras theorem
${AB}^2={AP}^2+{PB}^2$ ...(i)
In $△APC$ by Pythagoras theorem,
${AC}^2={AP}^2+{PC}^2$ ...(ii)
Adding (i) and (ii), we get
${AB}^2+{AC}^2=2{AP}^2+{PB}^2+{PC}^2$ ...(iii)
${AB}^2+{AC}^2=260$ (Given) ...(iv)
From (iii) and (iv),
$260=2\times 49+{PB}^2+{PB}^2$ (Using iii)
$\Rightarrow 260-98=2{PB}^2$
$\Rightarrow {PB}^2=\frac{162}{2}$
$\Rightarrow {PB}^2=81$
$\Rightarrow PB=9$
$\Rightarrow PB=PC=9$
$\therefore BC=PB+PC$
$\therefore =9+9=18$.