1. A square is drawn by joining the midpoints of the sides of a given square. A third square is drawn inside the second square in the same way and this process is continued indefinitely. If a side of the first square is 8 cm, the sum of the areas of all the squares such formed (in sq.cm.)is
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By: anil on 05 May 2019 02.34 am
Side of first square = 8cm.
Side of second square made by joining mid-points of first square = $$frac{8}{sqrt2}$$
Similarly side of third square = $$frac{8}{sqrt2 imessqrt2}$$ and so on.
Now summation of areas will be $$8^2+(frac{8}{sqrt2})^2+(frac{8}{sqrt2 imessqrt2})^2$$ .........
or $$8^2 ( 1+frac{1}{2}+frac{1}{4}+frac{1}{8}.....)$$
or $$64 imes (frac{1}{1-frac{1}{2}})$$ (As we know sum of an infinite G.P. is $$frac{a}{1-r}$$ where a is first term and r is common ratio)
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Side of second square made by joining mid-points of first square = $$frac{8}{sqrt2}$$
Similarly side of third square = $$frac{8}{sqrt2 imessqrt2}$$ and so on.
Now summation of areas will be $$8^2+(frac{8}{sqrt2})^2+(frac{8}{sqrt2 imessqrt2})^2$$ .........
or $$8^2 ( 1+frac{1}{2}+frac{1}{4}+frac{1}{8}.....)$$
or $$64 imes (frac{1}{1-frac{1}{2}})$$ (As we know sum of an infinite G.P. is $$frac{a}{1-r}$$ where a is first term and r is common ratio)
or 128 sq. cm.