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$$\triangle$$ABC and $$\triangle$$XYZ are equilateral triangles of 54 cm sides. All smaller triangles like $$\triangle$$ANM, $$\triangle$$OCP, $$\triangle$$QPX etc. are also equilateral triangles. Find the area of the shape MNOPQRM.
AB = 54 cm and $$\triangle$$ANM , $$\triangle$$OCP , $$\triangle$$OPX are equilateral triangles.
=> MN = MR = NO = OP = PQ = QR = $$\frac{54}{3} = 18$$ cm
Thus, MNOPQRM is a regular hexagon with side 18 cm
$$\therefore$$ Area of MNOPQRM = $$\frac{3 \sqrt{3}}{2} (side)^2$$
= $$\frac{3 \sqrt{3}}{2} \times 18^2 = 486 \sqrt{3} cm^2$$
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