Question

The side fo a given square is 10 cm. The midpoints of its sides are joined to form a new square. Again, the midpoints of the sides of this new square are joined to form another square. This process is continued indefinitely. Find (i) The sum of the areas and (ii) The sum of the perimeters of the square.

Answer 1

Let ABCD be the given square with each side equal to 10 cm. Let E,F,G,H be the midpoints of the sides AB, BC, CD and DA respectively. Let P,Q,R,S be the midpoints of the sides EF,FG,GH, and HE Respectively.
$\therefore BE=BF=5cm\Rightarrow EF=\sqrt{B{E}^2+B{F}^2}=\sqrt{25+25}cm=\sqrt{50}cm=5\sqrt{2}cm.\therefore FQ=FP=\frac{1}{2}EF=\frac{5\sqrt{5}}{2}c,=\frac{5}{\sqrt{2}}cm\Rightarrow PQ=\sqrt{F{P}^2+F{Q}^2}=\sqrt{\frac{25}{2}+\frac{25}{2}}cm=\sqrt{25}cm=5cm.\text{Thus, the sides of the squares are 10 cm,}5\sqrt{2}cm,5cm,\cdots \left(i\right)\text{Sum of the areas of the squares formed}=\left\{\right(10{)}^2+(5\sqrt{2}{)}^2+5^2+\cdots \infty \}c{m}^2=(100+50+25+\cdots \infty )c{m}^2=\frac{100}{(1-\frac{1}{2})}c{m}^2=200c{m}^2\text{(ii) Sum of perimeters of the squares formed}=(40+20\sqrt{2}+20+\cdots )cm=\frac{40}{(1-\frac{1}{\sqrt{2}})}cm=\frac{40\sqrt{2}}{(\sqrt{2}-1)}\times \frac{(\sqrt{2}+1)}{(\sqrt{2}+1)}cm=(80+40\sqrt{2})cm.$