循环四边形的对角线长度(使用边的长度)。

给定整数A ， B ， C和D ，它们表示循环四边形的边的长度，任务是找到循环四边形的对角线的长度。

例子：

Input: A = 10, B = 15, C = 20, D = 25 
Output: 22.06 26.07
Input: A = 10, B = 30, C =50, D = 20 
Output: 37.93 29.0 
 

方法：对角线的长度可以使用以下公式计算：

$Diagonal (p)=\sqrt{\frac{(ac+bd)(ad+bc)}{ab+cd}}$ [Tex]对角(q)= \ sqrt {\ frac {(ac + bd)(ab + cd)} {ad + bc}} [/ Tex]

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to calcualte the length of
// diagonals of a cyclic quadrilateral
vector Diagonals(int a, int b,
                        int c, int d)
{
    vector ans;
    ans.push_back(sqrt(((a * c) + (b * d)) *
                       ((a * d) + (b * c)) /
                       ((a * b) + (c * d))));
    ans.push_back(sqrt(((a * c) + (b * d)) *
                       ((a * b) + (c * d)) /
                       ((a * d) + (b * c))));
    return ans;
}
 
// Driver Code
int main()
{
    int A = 10;
    int B = 15;
    int C = 20;
    int D = 25;
 
    // Function Call
    vector ans = Diagonals(A, B, C, D);
 
    // Print the final answer
    printf("%.2f %.2f",
           (ans[0]) + .01,
            ans[1] + .01);
}
 
// This code is contributed by Amit Katiyar

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG{
 
// Function to calcualte the length of
// diagonals of a cyclic quadrilateral
static Vector Diagonals(int a, int b,
                               int c, int d)
{
    Vector ans = new Vector();
    ans.add((float) Math.sqrt(((a * c) + (b * d)) *
                              ((a * d) + (b * c)) /
                              ((a * b) + (c * d))));
    ans.add((float) Math.sqrt(((a * c) + (b * d)) *
                              ((a * b) + (c * d)) /
                              ((a * d) + (b * c))));
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    int A = 10;
    int B = 15;
    int C = 20;
    int D = 25;
 
    // Function Call
    Vector ans = Diagonals(A, B,
                                  C, D);
 
    // Print the final answer
    System.out.printf("%.2f %.2f",
                      (ans.get(0)) + .01,
                       ans.get(1) + .01);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement
# the above approach
 
import math
 
# Function to calcualte the length of
# diagonals of a cyclic quadrilateral
def Diagonals(a, b, c, d):
 
    p = math.sqrt(((a * c)+(b * d))*((a * d)+(b * c))
                  / ((a * b)+(c * d)))
    q = math.sqrt(((a * c)+(b * d))*((a * b)+(c * d))
                  / ((a * d)+(b * c)))
 
    return [p, q]
 
 
# Driver Code
A = 10
B = 15
C = 20
D = 25
 
# Function Call
ans = Diagonals(A, B, C, D)
 
# Print the final answer
print(round(ans[0], 2), round(ans[1], 2))

C#

// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Function to calcualte the length of
// diagonals of a cyclic quadrilateral
static List Diagonals(int a, int b,
                             int c, int d)
{
  List ans = new List();
  ans.Add((float) Math.Sqrt(((a * c) + (b * d)) *
                            ((a * d) + (b * c)) /
                            ((a * b) + (c * d))));
  ans.Add((float) Math.Sqrt(((a * c) + (b * d)) *
                            ((a * b) + (c * d)) /
                            ((a * d) + (b * c))));
  return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
  int A = 10;
  int B = 15;
  int C = 20;
  int D = 25;
 
  // Function Call
  List ans = Diagonals(A, B,
                              C, D);
 
  // Print the readonly answer
  Console.Write("{0:F2} {1:F2}",
                 (ans[0]) + .01,
                  ans[1] + .01);
}
}
 
// This code is contributed by 29AjayKumar

输出：

22.06 26.07

时间复杂度： O(1)
辅助空间： O(1)