给定两个整数r和d ,其中r是较小圆的半径, d是该圆与某个较大半径圆的面积之差。任务是找到较大圆的面积。
例子:
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
方法:设较小圆和较大圆的半径分别为r和R ,面积差为d即PI * R 2 – PI * r 2 = d其中PI = 3.14
或者, R 2 = (d / PI) + r 2 。
现在,大圆的面积可以计算为PI * R 2 。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
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
Javascript
输出:
55.24