A square is drawn inside another square such that each vertex of the inner square lies at the midpoint of a side of the outer square. If the perimeter of the inner square is 40 cm, find the perimeter of the outer square.
\(40\sqrt{2}\) cm
The inner square's perimeter is 40 cm, so its side = \(\frac{40}{4} = 10\) cm.
Each side of the inner square joins the midpoints of two adjacent sides of the outer square, forming the hypotenuse of a right triangle whose legs are each half the outer side \(\frac{a}{2}\).
So inner side \(= \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2} = \frac{a}{\sqrt{2}}\), giving \(\frac{a}{\sqrt{2}} = 10\).
Thus outer side \(a = 10\sqrt{2}\) cm, and outer perimeter \(= 4 \times 10\sqrt{2} = 40\sqrt{2}\) cm.
Hence, the perimeter of the outer square is \(40\sqrt{2}\) cm.
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