Solve this following

Question:

Let $\mathrm{AD}$ and $\mathrm{BC}$ be two vertical poles at $\mathrm{A}$ and B respectively on a horizontal ground. If $\mathrm{AD}=8 \mathrm{~m}, \mathrm{BC}=11 \mathrm{~m}$ and $\mathrm{AB}=10 \mathrm{~m}$; then the distance (in meters) of a point $\mathrm{M}$ on $\mathrm{AB}$ from the point $A$ such that $M D^{2}+M C^{2}$ is minimum is.

Solution:

$\left(\mathrm{MD}^{2}+(\mathrm{MC})^{2}=\mathrm{h}^{2}+64+(\mathrm{h}-10)^{2}+121\right.$

$=2 \mathrm{~h}^{2}-20 \mathrm{~h}+64+100+121$

$=2\left(\mathrm{~h}^{2}-10 \mathrm{~h}\right)+285$

$=2(\mathrm{~h}-5)^{2}+235$

it is minimum if $\mathrm{h}=5$

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