Sum of area of alternative concentric circle with radius 1,2,3,4..........N

Last Updated : 20 Jan, 2022

Given that Concentric circle of radii 1,2,3,4..........N cm are drawn. The interior of the smallest circle is colored white and angular regions are colored alternately green and white so that no two adjacent regions are of the same color. The task is to find the total area of the green region in sq cm.

Examples:

Input: N=100
Output: 15865.042900628456

Input: N=10
Output: 172.78759594743863

Approach: Area of the first green region would be-

[?(R2²- R1²)]

Similarly, the area of all alternative green circles would be-

[?(R4²- R3²)], [?(R6²- R5²)].....

The total area of the green region is the sum of the area of the green region is-

A = [ ? { (R2² - R1²) + (R4² - R3²) + (R6² - R5²) ............. (R(N)² - R(N-1)²}]

A = [ ? { ((R2 - R1)(R2 + R1)) + ((R4 - R3)(R4 + R3)) + ((R6 - R5)(R6 + R5)) ............. ((RN - R(N - 1)(RN + R(N - 1)) } ]

Since the difference between the radius of 2 concentric circles is only 1 cm so R(N) - R(N - 1) = 1. Hence,

A = [ ? {R1 + R2 + R3 + R4 + R5 + R6 .............R(N-1) + R(N) } ]

(R1 + R2 + R3 + R4 + R5 + R6 .............R(N-1) + R(N) forms an arithmetic progression so we have to find the sum of arithmetic progression with starting radius as 1 and last radius as N with a common difference of 1.

Sum of A.P. is-

S_N = N * (2 * a +(N - 1) * d) / 2

or

S_N = N * (a + l) / 2

Thus, the area of the green region is-

A = Sn * ?

Below is the implementation of the above approach:

C++

// C++ code for the above approach #include <bits/stdc++.h> using namespace std;  // Function to calculate area // of concentric circles double area(int N, int a, int l) {      // Formula to find sum of     // Arithmetic Progression     double S = N * (a + l) / 2;      // return area     return S * M_PI; }  // Driver code int main() {        // N is number of concentric circles     int N = 100;      // Radius of interior circle     int a = 1;      // l radius of exterior circle     int l = N;      // r is result     double r = area(N, a, l);      // Print result     cout << fixed << setprecision(12) << r; }  // This code is contributed by Samim Hossain Mondal.

Java

// Java code for the above approach class GFG {    // Function to calculate area // of concentric circles static double area(int N, int a, int l) {      // Formula to find sum of     // Arithmetic Progression     double S = N * (a + l) / 2;      // return area     return S * Math.PI; }  // Driver code public static void main(String args[]) {        // N is number of concentric circles     int N = 100;      // Radius of interior circle     int a = 1;      // l radius of exterior circle     int l = N;      // r is result     double r = area(N, a, l);      // Print result     System.out.println(r); } }  // This code is contributed by gfgking

Python3

# import pi from math module from math import pi  # Function to calculate area  # of concentric circles def area(N,a):            # Formula to find sum of      # Arithmetic Progression     S = N * (a + l) / 2          # return area     return S * pi  # N is number of concentric circles N = 100  # Radius of interior circle a = 1  # l radius of exterior circle l = N  # r is result  r = area(N, a)  # Print result print(r)

// C# program for the above approach using System; class GFG {    // Function to calculate area   // of concentric circles   static double area(int N, int a, int l)   {      // Formula to find sum of     // Arithmetic Progression     double S = N * (a + l) / 2;      // return area     return S * Math.PI;   }    // Driver Code:   public static void Main()   {     // N is number of concentric circles     int N = 100;      // Radius of interior circle     int a = 1;      // l radius of exterior circle     int l = N;      // r is result     double r = area(N, a, l);      // Print result     Console.WriteLine(r);   } }  // This code is contributed by Samim Hossain Mondal.

JavaScript

 <script>         // JavaScript code for the above approach           // Function to calculate area          // of concentric circles         function area(N, a) {              // Formula to find sum of              // Arithmetic Progression             S = N * (a + l) / 2              // return area             return S * Math.PI         }                  // N is number of concentric circles         N = 100          // Radius of interior circle         a = 1          // l radius of exterior circle         l = N          // r is result          r = area(N, a)          // Print result         document.write(r)            // This code is contributed by Potta Lokesh     </script>

Output

15865.042900628456

Time Complexity: O(1)
Auxiliary Space: O(1)

Sum of area of alternative concentric circle with radius 1,2,3,4..........N

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Sum of area of alternative concentric circle with radius 1,2,3,4..........N

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