# Area of Circumcircle of a Right Angled Triangle

Given an integer C which is the length of the hypotenuse of a right angled triangle of a circumcircle passing through the centre of the circumcircle. The task is to find the area of the circumcircle. Examples:

Input: C = 8
Output: 50.26
Input: C = 10
Output: 78.53

Approach: Since the hypotenuse C passes through the center of the circle, the radius of the circle will be C / 2
And we know that the area of a circle is PI * r2 where PI = 22 / 7 and r is the radius of the circle.
Hence the area of the circumcircle will be PI * (C / 2)2 i.e. PI * C2 / 4.
Below is the implementation of the above approach:

## C++

 `// C++ program to find the area of Circumscribed` `// circle of right angled triangle` `#include ` `#define PI 3.14159265` `using` `namespace` `std;`   `// Function to find area of` `// circumscribed circle` `float` `area_circumscribed(``float` `c)` `{` `    ``return` `(c * c * (PI / 4));` `}`   `// Driver code` `int` `main()` `{` `    ``float` `c = 8;` `    ``cout << area_circumscribed(c);` `    ``return` `0;` `}`   `// This code is contributed by Shivi_Aggarwal`

## C

 `// C program to find the area of Circumscribed` `// circle of right angled triangle` `#include ` `#define PI 3.14159265`   `// Function to find area of` `// circumscribed circle` `float` `area_circumscribed(``float` `c)` `{` `    ``return` `(c * c * (PI / 4));` `}`   `// Driver code` `int` `main()` `{` `    ``float` `c = 8;` `    ``printf``(``"%f"``,` `           ``area_circumscribed(c));` `    ``return` `0;` `}`

## Java

 `// Java code to find the area of circumscribed` `// circle of right angled triangle` `import` `java.lang.*;`   `class` `GFG {`   `    ``static` `double` `PI = ``3.14159265``;`   `    ``// Function to find the area of` `    ``// circumscribed circle` `    ``public` `static` `double` `area_cicumscribed(``double` `c)` `    ``{` `        ``return` `(c * c * (PI / ``4``));` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``double` `c = ``8.0``;` `        ``System.out.println(area_cicumscribed(c));` `    ``}` `}`

## Python3

 `# Python3 code to find the area of circumscribed ` `# circle of right angled triangle` `PI ``=` `3.14159265` `    `  `# Function to find the area of ` `# circumscribed circle` `def` `area_cicumscribed(c):` `    ``return` `(c ``*` `c ``*` `(PI ``/` `4``))` `    `  `# Driver code` `c ``=` `8.0` `print``(area_cicumscribed(c))`

## C#

 `// C# code to find the area of` `// circumscribed circle` `// of right angled triangle` `using` `System;`   `class` `GFG {` `    ``static` `double` `PI = 3.14159265;`   `    ``// Function to find the area of` `    ``// circumscribed circle` `    ``public` `static` `double` `area_cicumscribed(``double` `c)` `    ``{` `        ``return` `(c * c * (PI / 4));` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``double` `c = 8.0;` `        ``Console.Write(area_cicumscribed(c));` `    ``}` `}`

## PHP

 ``

## Javascript

 ``

Output:

`50.265484`

Time complexity: O(1), since there is no loop or recursion.

Auxiliary Space: O(1), since no extra space has been taken.

Another Approach:

To find the area of a circumcircle, we need to know its radius first. In a right-angled triangle, the circumcenter is the midpoint of the hypotenuse. Therefore, the radius of the circumcircle can be calculated as half of the length of the hypotenuse.

Once we have the radius, we can use the formula for the area of a circle, which is:

Area = ? * r^2

In this problem, we are given the length of the hypotenuse, which is also the diameter of the circumcircle. Therefore, we can calculate the radius as C/2. Then, we can use the above formula to find the area of the circumcircle.

## C++

 `#include ` `#include `   `using` `namespace` `std;`   `double` `areaOfCircumcircle(``double` `c) {` `    ``double` `a = c/2.0;` `    ``double` `b = ``sqrt``(``pow``(c, 2) - ``pow``(a, 2));` `    ``double` `r = c/2.0;` `    ``double` `area = M_PI*``pow``(r, 2);` `    ``return` `area;` `}`   `int` `main() {` `    ``double` `c = 8;` `    ``cout << areaOfCircumcircle(c) << endl;` `}`

## Java

 `import` `java.lang.Math;`   `public` `class` `Main {` `    ``public` `static` `double` `areaOfCircumcircle(``double` `c) {` `        ``double` `a = c / ``2.0``;` `        ``double` `b = Math.sqrt(Math.pow(c, ``2``) - Math.pow(a, ``2``));` `        ``double` `r = c / ``2.0``;` `        ``double` `area = Math.PI * Math.pow(r, ``2``);` `        ``return` `area;` `    ``}`   `    ``public` `static` `void` `main(String[] args) {` `        ``double` `c = ``8``;` `        ``System.out.println(areaOfCircumcircle(c));` `    ``}` `}`   `// Contributed by sdeadityasharma`

## Python3

 `import` `math`   `def` `areaOfCircumcircle(c):` `    ``a ``=` `c``/``2.0` `    ``b ``=` `math.sqrt(``pow``(c, ``2``) ``-` `pow``(a, ``2``))` `    ``r ``=` `c``/``2.0` `    ``area ``=` `math.pi``*``pow``(r, ``2``)` `    ``return` `area`   `if` `__name__ ``=``=` `'__main__'``:` `    ``c ``=` `8` `    ``print``(areaOfCircumcircle(c))` `# This code is contributed by Prajwal Kandekar`

## C#

 `using` `System;`   `public` `class` `MainClass {` `public` `static` `double` `areaOfCircumcircle(``double` `c) {` `double` `a = c / 2.0;` `double` `b = Math.Sqrt(Math.Pow(c, 2) - Math.Pow(a, 2));` `double` `r = c / 2.0;` `double` `area = Math.PI * Math.Pow(r, 2);` `return` `area;` `}``public` `static` `void` `Main() {` `    ``double` `c = 8;` `    ``Console.WriteLine(areaOfCircumcircle(c));` `}` `}`

## Javascript

 `function` `areaOfCircumcircle(c) {` `    ``const a = c/2.0;` `    ``const b = Math.sqrt(Math.pow(c, 2) - Math.pow(a, 2));` `    ``const r = c/2.0;` `    ``const area = Math.PI*Math.pow(r, 2);` `    ``return` `area;` `}`   `const c = 8;` `console.log(areaOfCircumcircle(c));`

Output

`50.2655`

Time Complexity: O(1)

Space Complexity: O(1)

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