Question

Seg PM is a median of ∆PQR. If PQ = 40, PR = 42 and PM = 29, find QR.

Answer 1

In ∆PQR, point M is the midpoint of side QR.

$\mathrm{QM}=\mathrm{MR}=\frac{1}{2}\mathrm{QR}$

$$\begin{gathered} {\mathrm{PQ}}^2+{\mathrm{PR}}^2=2{\mathrm{PM}}^2+2{\mathrm{QM}}^2\left(\mathrm{by}\mathrm{Apollonius}\mathrm{theorem}\right) \\ \Rightarrow {\left(40\right)}^2+{\left(42\right)}^2=2{\left(29\right)}^2+2{\mathrm{QM}}^2 \\ \Rightarrow 1600+1764=1682+2{\mathrm{QM}}^2 \\ \Rightarrow 3364-1682=2{\mathrm{QM}}^2 \\ \Rightarrow 1682=2{\mathrm{QM}}^2 \\ \Rightarrow {\mathrm{QM}}^2=841 \\ \Rightarrow \mathrm{QM}=29 \\ \Rightarrow \mathrm{QR}=2\times 29 \\ \Rightarrow \mathrm{QR}=58 \end{gathered}$$

Hence, QR = 58.