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Differential Calculus

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7 Sep 2008 12:09:32 IST

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Limit:

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A square is inscribed in a circle of radius R, a circle is inscribed in this square, then a square and so on n times. Find the limit of the sum of areas of all the squares as .

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Nikhil Bhattacharya

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Joined: 2 Jul 2008

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7 Sep 2008 12:26:44 IST

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$\large{\text{Side of Ist square=$\sqrt{2}{R}$}}}$

$\large{\text{Side of IInd square $={R}$}}}$

$\large{\text{Side of IIIrd square $=\frac{R}{\sqrt{2}}$}}}$

$\large{\text{Hence total Area is = $2R^2+R^2+\frac{R^2}{2}...\infty$}}}$

$\large{\text{Hence total Area is = $4R^2$}}}$

rohan sen sharma

Cool goIITian

Joined: 3 Feb 2007

Posts: 50

7 Sep 2008 12:37:41 IST

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ist square will have side sqrt2*r

2nd square will have side r

3rd will have r/(sqrt2)

and so on

therfore sum of areas=2r²+r²+r²/2...........upto infinity

as this is in gp....wit common diff 1/2 and 1st term 2r²

therfore applying sum of infinite gp=a/(1-r)

we get - 2r²/(1-1/2)

therefore total area=4r²

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