Question

$$\begin{gathered} \mathrm{In}\mathrm{the}\text{given}\mathrm{figure}, \\ {\mathrm{AB}}^2+{\mathrm{AC}}^2=122\text{and}\mathrm{BC}=10. \\ \mathrm{Find}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{median}\mathrm{on}\mathrm{side}\mathrm{BC}. \end{gathered}$$

Answer 1

In ΔABC, AQ is the median.
Since the median AQ bisects the side BC of Δ ABC, BQ = QC =$\frac{\mathrm{BC}}{2}=\frac{10}{2}=5$.
Now, by Appollonius' Theorem, we get:
AB² + AC² = 2AQ² + 2BQ²
AB² + AC² = 122 and BQ = 5. So,
122 = 2AQ² + 2(5)²
$\Rightarrow$ 122 = 2AQ² + 50
$\Rightarrow$ 2AQ² = 122 – 50 = 72
$\Rightarrow$ AQ² = $\frac{72}{2}$=36
$\Rightarrow$ AQ =$\sqrt{36}=6$