### $ABCD$ is a square with each side measuring $144 \space cm$. $M$ is a point on $CB$ such that $CM = 36 \space cm$. If $N$ is a variable point on the diagonal $DB$, find the least value of $CN + MN$. D C B A M N

Answer:

$180 \space cm$

Step by Step Explanation:
1. Given, $BC = 144 \space cm$ and $CM = 36 \space cm$
$\implies BM = CB - CM = 144 - 36 = 108$
Let's join $A$ to $N$
2. \begin{align} & \text{Since } \triangle ADN \cong \triangle CDN && [\text{By SAS criterion}] \\ & \therefore AN = CN && [\text{Corresponding sides of congruent triangles}] \space\space\space\space \end{align}
$\implies \space AN + MN = CN + NM$
Observe that the value of $AN + NM$ is least when $ANM$ is a straight line.
3. Now, if $ANM$ is a straight line, then $\triangle AMB$ is a right-angled triangle.
$\therefore$ by Pythagoras theorem,
 Least value of $AN + NM$ \begin{align} & = \sqrt{ AB^2 + BM^2 } \\ & = \sqrt{ 144^2 + 108^2 } \\ & = 180 \end{align}
From step 2, we have $AN + MN = CN + NM$.
4. Hence, the least value of $CN + MN$ is $180 \space cm$.

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