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Let $C_0$ be a circle of radius $I$ . For $n \geq 1$, let $C_n$ be a circle whose area equals the area of a square inscribed in $C_{n-1} .$ Then, $\sum \limits_{i=0}^{\infty}$ Area $\left(C_i\right)$ equals
$\pi^2$
$\frac{\pi-2}{\pi^2}$
$\frac{1}{\pi^2}$
$\frac{\pi^2}{\pi-2}$
Solution
(d)
We have, $C_0$ be a circle of radius $1.$ $C_n$ be a circle whose area equals the area of a square inscribed in $C_{n-1}$.
Let $a_0, a_1, a_2, a_3, \ldots, a_n$ be the length of sides of square inscribed in circle $C_0, C_1, C_2, \ldots, C_n$ and $r_0, r_1, r_2, \ldots, r_n$ be radius of circle.
$2 a_0^2 =4$
$a _0^2 =2$
$\pi r_1^2 =a_0^2$
$r_1^2 =\frac{2}{\pi}$
$2 a _1^2 =\left(2 r_1\right)^2=4 r_1^2$
$a _1^2 =\frac{4}{\pi}$
$\pi r_2^2 = c _1^2$
$\Rightarrow r_2^2=\frac{4}{\pi^2}$
Similarly, $r_n^2=\frac{2^2}{\pi^n}$
Now, $\sum \limits_{i=0}^{\infty}$ Area $\left(C_{i j}\right)$
$=\pi\left(-1+\frac{2}{\pi}+\frac{2^2}{\pi^2}+\frac{2^3}{\pi^3}+\ldots\right)$
$=\pi\left(\frac{1}{1-\frac{2}{\pi}}\right)$
$\left[S_{\infty}=1+r+r^2+\ldots=\frac{1}{1-r}\right]$
$=\frac{\pi^2}{\pi-2}$
