Question

The side of 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 aquare are joined to form another square. The process is continued indefinitely. Find (i) the sum of the areas and (ii) the sum of the perimeters of the squares.

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.
`:. BE=BF=5` cm
`rArr EF=sqrt(BE^(2)+BF^(2))=sqrt(25+25) cm`
`=sqrt(50)` cm `=5sqrt(2)` cm.
`:. FQ=FP=1/2 EF =(5sqrt(2))/2 cm=5/sqrt(2)` cm
`rArr PQ=sqrt(FP^(2)+FQ^(2))=sqrt(25/2+25/2) cm=sqrt(25) cm= 5cm`
Thus, the sides of the squres are 10 cm, `5sqrt(2)` cm, 5 cm, ...
(i) Sum of the areas of the squares formed
`={(10)^(2)+(5sqrt(2))^(2)+5^(2)+...oo} cm^(2)`
`=(100+50+25+...oo) cm^(2)`
`=100/((1-1/2))cm^(2)=200 cm^(2) [("taking the sum of infinite GP"),("With "a=100 and r=1/2)]`.
(ii) Sum of perimeters of the squares formed
`=(40+20sqrt(2)+20+...) cm`
`=40/((1-1/sqrt(2))) cm=(40sqrt(2))/((sqrt(2)-1))xx((sqrt(2)+1))/((sqrt(2)+1)) cm=(80+40sqrt(2)) cm`.