Question

In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS =13, find QR.

Answer 1

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

$\mathrm{QS}=\mathrm{SR}=\frac{1}{2}\mathrm{QR}$

$$\begin{gathered} {\mathrm{PQ}}^2+{\mathrm{PR}}^2=2{\mathrm{PS}}^2+2{\mathrm{QS}}^2\left(\mathrm{by}\mathrm{Apollonius}\mathrm{theorem}\right) \\ \Rightarrow {11}^2+{17}^2=2{\left(13\right)}^2+2{\mathrm{QS}}^2 \\ \Rightarrow 121+289=2\left(169\right)+2{\mathrm{QS}}^2 \\ \Rightarrow 410=338+2{\mathrm{QS}}^2 \\ \Rightarrow 2{\mathrm{QS}}^2=410-338 \\ \Rightarrow 2{\mathrm{QS}}^2=72 \\ \Rightarrow {\mathrm{QS}}^2=36 \\ \Rightarrow \mathrm{QS}=6 \\ \therefore \mathrm{QR}=2\times \mathrm{QS} \\ =2\times 6 \\ =12 \end{gathered}$$


Hence, QR = 12.