Area of circle inscribed in a Isosceles Trapezoid

Given two bases of the isoceles trapezoid ABCD as a and b, the task is to find the area of circle inscribed in this trapezoid

Examples:

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

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Derivation: Given a circle inscribed in trapezium ABCD (sides AB = n and CD = m), we need to find out the height of the trapezium i.e., (AL), which is half of the radius of the circle to find the area of the circle.

For finding the height of circle we do following operation.

The circle will always touch the sides of trapezium at their midpoints, Say the midpoints of AB, BD, CD, AC are G, F, H, E and join them with the centre of the circle.

Now from Symmetry, we can see that

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

Now in Triangle ACL apply the Pythagoras theorem.

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

we get 
Perpendicular AL = Square_root(m * n)

Therfore the height of the Trapezium = AL = Square_Root(Product of given sides)
Now the radius of the circle is simple half of the height and hence the area can be calculated easily.

Appriach:

Find the height of the trapezoid as (square_root( m * n )).

Find the radius of the incircle

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

Now find the area of the circle
```
= Pi * R² 
= ( 3.141 * m * n ) / 4
```

Below is the implementation of the above approach:

C++

// CPP implementation to find 
// the rea of the circle 
// inscribed in a trapezoid 
// having non- parllel sides m, n 
#include<bits/stdc++.h> 
using namespace std; 
  
// Function to find area of circle 
// inscribed in a trapezoid 
// having non- parllel 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- parllel sides m, n  
class GFG  
{  
      
    // Function to find area of circle  
    // inscribed in a trapezoid  
    // having non- parllel 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- parllel sides  m, n 
  
  
# Function to find area of circle  
# inscribed in a trapezoid 
# having non- parllel 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- parllel sides m, n  
using System;  
  
class GFG  
{  
      
// Function to find area of circle  
// inscribed in a trapezoid  
// having non- parllel 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 

Output:

235.575

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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