The diagram shows a nest of 4 squares set one within another. The side of the outer most square is 20 cm. The midpoints of the sides are joined to give a second square, and the process is repeated to give the third and fourth squares. Find the length of a side of the smallest square.

Given, squares are formed joining mid-points of square ABCD.

AE = AH = $\dfrac{1}{2}$ × AB = $\dfrac{1}{2}$ × 20 = 10 cm.

In square ABCD,

∠A = 90°

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle AEH,

⇒ EH² = AE² + AH²

⇒ EH² = 10² + 10²

⇒ EH² = 100 + 100

⇒ EH² = 200

⇒ EH = $\sqrt{200} = \sqrt{2 \times 100}$

⇒ EH = $10\sqrt{2}$ cm.

In square EFGH,

∠E = 90°

EM = EN = $\dfrac{1}{2}$ × EH = $\dfrac{1}{2} \times 10 \sqrt{2} = 5\sqrt{2} \text{ cm.}$

By Pythagoras theorem,

In triangle EMN,

⇒ MN² = EN² + EM²

⇒ MN² = $(5\sqrt{2})^2 + (5\sqrt{2})^2$

⇒ MN² = 50 + 50

⇒ MN² = 100

⇒ MN = $\sqrt{100}$

⇒ MN = 10 cm.

In square MNOP,

∠M = 90°

IM = ML = $\dfrac{1}{2} × MN = \dfrac{1}{2} \times 10 = 5 \text{ cm.}$

By Pythagoras theorem,

In triangle IML,

⇒ IL² = IM² + ML²

⇒ IL² = 5² + 5²

⇒ IL² = 25 + 25

⇒ IL² = 50

⇒ IL = $\sqrt{25 \times 2}$

⇒ IL = $5\sqrt{2}$ cm.

Hence, the length of a side of the smallest square is $5\sqrt{2}$ cm.