等腰梯形内切圆的面积

给定等腰梯形 ABCD 的两个底为a和b ，任务是找到该梯形内切圆的面积

例子：

Input: a = 10, b = 30 
Output: Area = 235.57

Input: a = 20, b = 36 
Output: Area = 565.38

推导：给定一个圆内切于梯形ABCD（边AB = n，CD = m），我们需要找出梯形的高度，即(AL)，它是圆半径的一半，以求出其面积圈子。

为了找到圆的高度，我们执行以下操作。

1. 圆总是在它们的中点处接触梯形的边，假设AB，BD，CD，AC的中点是G，F，H，E，并与圆的中心连接。
2. 现在从对称性，我们可以看到

AG = AE = n/2, 
EC = CG = m/2, 
HD = DF = n/2,
GB = FB = m/2

1. 现在在三角形 ACL 中应用勾股定理。

Hypotenuse AC = m/2 + n/2
Base CL = CH - AG = m/2 - n/2

we get 
Perpendicular AL = Square_root(m * n)

2. 因此梯形的高度 = AL = Square_Root（给定边的乘积）
3. 现在圆的半径只是高度的一半，因此可以轻松计算面积。

方法：

1. 求梯形的高度为(square_root( m * n )) 。
2. 求内圆的半径

R = height / 2 
  = square_root(m * n) / 2

1. 现在找到圆的面积

= Pi * R2 
= ( 3.141 * m * n ) / 4

下面是上述方法的实现：

C++

// CPP implementation to find
// the rea of the circle
// inscribed in a trapezoid
// having non- parallel sides m, n
#include
using namespace std;
 
// Function to find area of circle
// inscribed in a trapezoid
// having non- parallel sides m, n
double area_of_circle(int m, int n)
{
    // radius of circle by the
    // formula i.e. root( m * n) / 2
    // area of circle = (3.141 ) * ( R ** 2 )
 
    int square_of_radius = ( m * n ) / 4;
    double area = ( 3.141 * square_of_radius );
    return area;
}
 
// Driver Code
int main(){
    int n = 10;
    int m = 30;
    cout << (area_of_circle(m, n));
}
 
// This code is contributed by mohit kumar 29

Java

// Java Program to find
// the rea of the circle
// inscribed in a trapezoid
// having non- parallel sides m, n
class GFG
{
     
    // Function to find area of circle
    // inscribed in a trapezoid
    // having non- parallel sides m, n
    static double area_of_circle(int m, int n)
    {
        // radius of circle by the
        // formula i.e. root( m * n) / 2
        // area of circle = (3.141 ) * ( R ** 2 )
     
        int square_of_radius = ( m * n ) / 4;
        double area = ( 3.141 * square_of_radius );
        return area;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 10;
        int m = 30;
        System.out.println(area_of_circle(m, n));
    }
}
 
// This code is contributed by Yash_R

Python3

# Python 3 implementation to find
# the rea of the circle
# inscribed in a trapezoid
# having non- parallel sides  m, n
 
 
# Function to find area of circle
# inscribed in a trapezoid
# having non- parallel sides  m, n
def area_of_circle(m, n):
    # radius of circle by the
    # formula i.e. root( m * n) / 2
    # area of circle = (3.141 ) * ( R ** 2 )
     
    square_of_radius = ( m * n ) / 4
    area = ( 3.141 * square_of_radius )
    return area
 
# Driver Code
if __name__=='__main__':
    n = 10
    m = 30
    print(area_of_circle(m, n))

C#

// C# Program to find
// the rea of the circle
// inscribed in a trapezoid
// having non- parallel sides m, n
using System;
 
class GFG
{
     
// Function to find area of circle
// inscribed in a trapezoid
// having non- parallel sides m, n
static double area_of_circle(int m, int n)
{
    // radius of circle by the
    // formula i.e. root( m * n) / 2
    // area of circle = (3.141 ) * ( R ** 2 )
 
    int square_of_radius = ( m * n ) / 4;
    double area = ( 3.141 * square_of_radius );
    return area;
}
 
// Driver code
public static void Main ()
{
    int n = 10;
    int m = 30;
    Console.WriteLine(area_of_circle(m, n));
}
}
 
// This code is contributed by Sanjit_Prasad

Javascript

 
// Javascript Program to find
// the rea of the circle
// inscribed in a trapezoid
// having non- parallel sides m, n
 
// Function to find area of circle
// inscribed in a trapezoid
// having non- parallel sides m, n
function area_of_circle(m, n)
{
     
    // Radius of circle by the
    // formula i.e. root( m * n) / 2
    // area of circle = (3.141 ) * ( R ** 2 )
 
    var square_of_radius = ( m * n ) / 4;
    var area = ( 3.141 * square_of_radius );
    return area;
}
 
// Driver Code
var n = 10;
var m = 30;
 
document.write(area_of_circle(m, n));
 
// This code is contributed by Khushboogoyal499
     

输出：

235.575

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