$\triangle ABC$ and $\triangle XYZ$ are equilateral triangles of side $54\ \text{cm}$. All smaller triangles like $\triangle ANM,\ \triangle OCP,\ \triangle QPX$ etc.\ are also equilateral. Find the area of the shape $MNOPQRM$.
Show Hint
In the standard hexagram made from two congruent equilateral triangles, the side of the central regular hexagon is \(\tfrac{1}{3}\) of the big triangle’s side \(\Rightarrow\) use \(A_{\text{hex}}=\tfrac{3\sqrt{3}}{2}s^2\) quickly.
Step 1: Recognize the central figure.
Two congruent equilateral triangles arranged as a hexagram (Star of David) produce a {central regular hexagon} $MNOPQR$. The six small corner triangles (e.g., $\triangle ANM$) are given equilateral and similar to the big ones. Their sides are parallel to the sides of the big triangles, so the sides of the big triangle are cut into equal thirds.
Step 2: Side of the small equilateral triangles and the hexagon.
Each corner triangle is an equilateral with side equal to one–third of the large side:
\[
\text{small side}=\frac{54}{3}=18\ \text{cm}.
\]
The side of the central regular hexagon equals this length (adjacent edges are collinear with sides of those small equilateral triangles), hence
\[
s_{\text{hex}}=18\ \text{cm}.
\]
Step 3: Area of the regular hexagon.
For a regular hexagon of side $s$,
\[
[\text{Hexagon}] = \frac{3\sqrt{3}}{2}\,s^2.
\]
With $s=18$,
\[
[\text{Hexagon }MNOPQRM]=\frac{3\sqrt{3}}{2}\times 18^{2}
=\frac{3\sqrt{3}}{2}\times 324
=486\sqrt{3}\ \text{sq.\ cm}.
\]
\[
\boxed{486\sqrt{3}\ \text{sq.\ cm}}
\]
