计算具有给定边长的循环四边形的面积

给定四个正整数A 、 B 、 C和D表示循环四边形的边长，任务是找到循环四边形的面积。

例子：

Input: A = 10, B = 15, C = 20, D = 25
Output: 273.861

Input: A = 10, B = 30, C = 50, D = 20
Output: 443.706

方法：根据以下观察可以解决给定的问题：

• 循环四边形是顶点都在一个圆上的四边形。圆称为外接圆或外接圆，顶点称为同环。

• 在上图中， r是外接圆的半径， A 、 B 、 C和D分别是边PQ 、 QR 、 RS和SP的长度。
• 四边形的面积由Bretschneider 公式给出：

$Area = \sqrt(s - A)*(s - B)*(s - C)*(s - D) - A*B*C*D *\cos\frac{(\alpha + \gamma)}{2}$

where, A, B, C, and D are the sides of the triangle and
α and γ are the opposite angles of the quadrilateral.

Since, the sum of opposite angles of the quadrilateral is 180 degree. Therefore, the value of cos(180/2) = cos(90) = 0.

Therefore, the formula for finding the area reduces to $\sqrt(s - A)*(s - B)*(s - C)*(s - D)$.

因此，想法是打印的值$\sqrt(s - A)*(s - B)*(s - C)*(s - D)$作为给定四边形的合成面积。

下面是上述方法的实现：

C++

// C++ programm for the above approach
 
#include 
using namespace std;
 
// Function to find the area
// of cyclic quadrilateral
float calculateArea(float A, float B,
                    float C, float D)
{
    // Stores the value of
    // half of the perimeter
    float S = (A + B + C + D) / 2;
 
    // Stores area of cyclic quadrilteral
    float area = sqrt((S - A) * (S - B)
                      * (S - C) * (S - D));
 
    // Return the resultant area
    return area;
}
 
// Driver Code
int main()
{
    float A = 10;
    float B = 15;
    float C = 20;
    float D = 25;
    cout << calculateArea(A, B, C, D);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
     
// Function to find the area
// of cyclic quadrilateral
static float calculateArea(float A, float B,
                           float C, float D)
{
     
    // Stores the value of
    // half of the perimeter
    float S = (A + B + C + D) / 2;
 
    // Stores area of cyclic quadrilteral
    float area = (float)Math.sqrt((S - A) * (S - B) *
                                  (S - C) * (S - D));
 
    // Return the resultant area
    return area;
}
 
// Driver code
public static void main (String[] args)
{
    float A = 10;
    float B = 15;
    float C = 20;
    float D = 25;
     
    System.out.println(calculateArea(A, B, C, D));
 
}
}
 
// This code is contributed by Ankita saini

Python3

# Python3 programm for the above approach
from math import sqrt
 
# Function to find the area
# of cyclic quadrilateral
def calculateArea(A, B, C, D):
     
    # Stores the value of
    # half of the perimeter
    S = (A + B + C + D) // 2
 
    # Stores area of cyclic quadrilteral
    area = sqrt((S - A) * (S - B) *
                (S - C) * (S - D))
 
    # Return the resultant area
    return area
 
# Driver Code
if __name__ == '__main__':
     
    A = 10
    B = 15
    C = 20
    D = 25
     
    print(round(calculateArea(A, B, C, D), 3))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the area
// of cyclic quadrilateral
static float calculateArea(float A, float B,
                           float C, float D)
{
     
    // Stores the value of
    // half of the perimeter
    float S = (A + B + C + D) / 2;
 
    // Stores area of cyclic quadrilteral
    float area = (float)Math.Sqrt((S - A) * (S - B) *
                                  (S - C) * (S - D));
 
    // Return the resultant area
    return area;
}
 
// Driver Code
static public void Main()
{
    float A = 10;
    float B = 15;
    float C = 20;
    float D = 25;
     
    Console.Write(calculateArea(A, B, C, D));
}
}
 
// This code is contributed by code_hunt

Javascript

// java script  programm for the above approach
 
// Function to find the area
//of cyclic quadrilateral
function calculateArea(A, B, C, D){
     
    //Stores the value of
    // half of the perimeter
    let S = (A + B + C + D) /2
 
    // Stores area of cyclic quadrilteral
    let area = Math.sqrt((S - A) * (S - B) *
                (S - C) * (S - D))
 
    //Return the resultant area
    return area;
    }
 
// Driver Code
 
     
    let  A = 10;
    let B = 15;
    let C = 20;
    let D = 25;
     
    document.write(calculateArea(A, B, C, D).toFixed(3))
 
//this code is contributed by sravan kumar

输出：

273.861

时间复杂度： O(1)
辅助空间： O(1)

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