使用半径和周长的三角形的内心和外心之间的距离

给定两个整数r和R分别表示半径和圆周半径的长度，任务是计算中心距和中心距之间的距离d 。

Inradius正三角形的inradius(R)(ABC)是(具有中心为L)的内切圆，其是将配合在三角形内的最大圆的半径。
外接半径：三角形的外接半径(R)是该三角形的外接圆的半径(以O为中心)。

例子：

Input: r = 2, R = 5 
Output: 2.24

Input: r = 5, R = 12 
Output: 4.9

方法：
可以使用几何学中的欧拉定理解决该问题，该定理指出，三角形的中心和外心之间的距离可以通过以下公式计算：

$Distance = \sqrt{R^2 - 2rR}$

下面是上述方法的实现：

C++14

// C++14 program for the above approach
#include 
using namespace std;
 
// Function returns the required distance
double distance(int r, int R)
{
    double d = sqrt(pow(R, 2) -
                       (2 * r * R));
                             
    return d;
}
 
// Driver code
int main()
{
     
    // Length of Inradius
    int r = 2;
     
    // Length of Circumradius
    int R = 5;
 
    cout << (round(distance(r, R) * 100.0) / 100.0);
}
 
// This code is contributed by sanjoy_62

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function returns the required distance
static double distance(int r,int R)
{
    double d = Math.sqrt(Math.pow(R, 2) -
                         (2 * r * R));
                          
    return d;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Length of Inradius
    int r = 2;
     
    // Length of Circumradius
    int R = 5;
 
    System.out.println(Math.round(
        distance(r, R) * 100.0) / 100.0);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
import math
 
# Function returns the required distance
def distance(r,R):
 
    d = math.sqrt( (R**2) - (2 * r * R))
     
    return d
 
# Driver Code
 
# Length of Inradius
r = 2
 
# Length of Circumradius
R = 5
 
print(round(distance(r,R),2))

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function returns the required distance
static double distance(int r, int R)
{
    double d = Math.Sqrt(Math.Pow(R, 2) -
                         (2 * r * R));
                         
    return d;
}
 
// Driver code
public static void Main(string[] args)
{
     
    // Length of Inradius
    int r = 2;
     
    // Length of Circumradius
    int R = 5;
     
    Console.Write(Math.Round(
        distance(r, R) * 100.0) / 100.0);
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// Javascript program for
// the above approach
 
// Function returns the required distance
function distance(r, R)
{
    let d = Math.sqrt(Math.pow(R, 2) -
                         (2 * r * R));
                            
    return d;
}
 
// Driver code
 
    // Length of Inradius
    let r = 2;
       
    // Length of Circumradius
    let R = 5;
   
    document.write(Math.round(
        distance(r, R) * 100.0) / 100.0);
      
     // This code is contributed by susmitakundugoaldanga.

输出：

2.24

时间复杂度： O(1)
辅助空间： O(1)