Skip to content
Related Articles

Related Articles

Find area of the larger circle when radius of the smaller circle and difference in the area is given
  • Last Updated : 16 Mar, 2021

Given two integers r and d where r is the radius of the smaller circle and d is the difference of the area of this circle with some larger radius circle. The task is to find the area of the larger circle.
Examples: 
 

Input: r = 4, d = 5 
Output: 55.24 
Area of the smaller circle = 3.14 * 4 * 4 = 50.24 
55.24 – 50.24 = 5
Input: r = 12, d = 3 
Output: 455.16 
 

 

Approach: Let radius of the smaller and the larger circles be r and R respectively and the difference in the areas is given to be d i.e. PI * R2 – PI * r2 = d where PI = 3.14 
Or, R2 = (d / PI) + r2
Now, area of the bigger circle can be calculated as PI * R2.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
const double PI = 3.14;
 
// Function to return the area
// of the bigger circle
double find_area(int r, int d)
{
    // Find the radius of
    // the bigger circle
    double R = d / PI;
    R += pow(r, 2);
    R = sqrt(R);
 
    // Calculate the area of
    // the bigger circle
    double area = PI * pow(R, 2);
    return area;
}
 
// Driver code
int main()
{
    int r = 4, d = 5;
 
    cout << find_area(r, d);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
    static double PI = 3.14;
 
    // Function to return the area
    // of the bigger circle
    static double find_area(int r, int d)
    {
        // Find the radius of
        // the bigger circle
        double R = d / PI;
        R += Math.pow(r, 2);
        R = Math.sqrt(R);
 
        // Calculate the area of
        // the bigger circle
        double area = PI * Math.pow(R, 2);
        return area;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int r = 4, d = 5;
 
        System.out.println(find_area(r, d));
    }
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python 3 implementation of the approach
PI = 3.14
from math import pow, sqrt
 
# Function to return the area
# of the bigger circle
def find_area(r, d):
     
    # Find the radius of
    # the bigger circle
    R = d / PI
    R += pow(r, 2)
    R = sqrt(R)
 
    # Calculate the area of
    # the bigger circle
    area = PI * pow(R, 2)
    return area
 
# Driver code
if __name__ == '__main__':
    r = 4
    d = 5
 
    print(find_area(r, d))
 
# This code is contributed by
# Surendra_Gangwar

C#




// C# implementation of the approach
using System;
 
public class GFG
{
    static double PI = 3.14;
 
    // Function to return the area
    // of the bigger circle
    static double find_area(int r, int d)
    {
        // Find the radius of
        // the bigger circle
        double R = d / PI;
        R += Math.Pow(r, 2);
        R = Math.Sqrt(R);
 
        // Calculate the area of
        // the bigger circle
        double area = PI * Math.Pow(R, 2);
        return area;
    }
 
    // Driver code
    static public void Main ()
    {
     
        int r = 4, d = 5;
        Console.Write(find_area(r, d));
    }
}
 
// This code is contributed by ajit.

PHP




<?php
// PHP implementation of the approach
const PI = 3.14;
 
// Function to return the area
// of the bigger circle
function find_area($r, $d)
{
     
    // Find the radius of
    // the bigger circle
    $R = $d / PI;
    $R += pow($r, 2);
    $R = sqrt($R);
 
    // Calculate the area of
    // the bigger circle
    $area = PI * pow($R, 2);
    return $area;
}
 
// Driver Code
$r = 4;
$d = 5;
 
echo (find_area($r, $d));
 
// This code is contributed by Naman_Garg
?>

Javascript




<script>
 
// Javascript implementation of the approach
 
 let PI = 3.14;
 
// Function to return the area
// of the bigger circle
function find_area(r, d)
{
    // Find the radius of
    // the bigger circle
    let R = d / PI;
    R += Math.pow(r, 2);
    R = Math.sqrt(R);
 
    // Calculate the area of
    // the bigger circle
    let area = PI * Math.pow(R, 2);
    return area;
}
 
// Driver code
let r = 4, d = 5;
    document.write( find_area(r,d).toFixed(2));
 
// This code contributed by Rajput-Ji
 
</script>
Output: 
55.24

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :