Question

Concentric circles of radii $1,2,3,.......,100$ cm are drawn. The interior of the smallest circle is coloured red and the angular regions are coloured alternately green and red so that no two adjacent regions are of the same colour. Then, the total area of the green regions is sq.cm is equal to

• A. $1000\pi$
• B. $5050\pi$
• C. $4950\pi$
• D. $5151\pi$

Answer 1

Answer: (B)

$5050\pi$

Given,

Angular regions of the circle with radii $2,4,6,8,....100$ have green colour

$\therefore$ Total area of green-coloured regions

$A_g=\pi (2^2-1^2)+\pi (4^2-3^2)+\pi (6^2-5^2)+...+\pi ({100}^2-{99}^2)$

$=\pi \left[\right(2^2-1^2)+(4^2-3^2)+(6^2-5^2)+...+({100}^2-{99}^2\left)\right]$

$=\pi \left[\right(2-1\left)\right(2+1)+(4-3\left)\right(4+3)+(6-5\left)\right(6+5)+...+(100-99\left)\right(100+99\left)\right]$

$=\pi [3+7+11+...+199]$

Now, the above series is an AP, where

$a=3,l=199,d=4$ and

$n=\frac{199-3}{4}+1=\frac{196}{4}+1=50$

$\therefore A_g=\pi \left[\frac{n}{2}\right(a+l\left)\right]=\pi \left[\frac{50}{2}\right(3+199\left)\right]$

$=\pi \left[25\right(202\left)\right]=5050\pi$

Hence, option (B) is correct.