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 = 30Output:Area = 235.57Input:a = 20, b = 36Output:Area = 565.38

**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 * R

^{2}= ( 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 ` |

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## 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 ` |

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## 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)) ` |

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## 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 ` |

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**Output:**

235.575

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