^@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


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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^@
    D C B A M 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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