# Area of Incircle of a Right Angled Triangle

• Difficulty Level : Medium
• Last Updated : 06 Jul, 2022

Given the P, B and H are the perpendicular, base and hypotenuse respectively of a right angled triangle. The task is to find the area of the incircle of radius r as shown below:

Examples:

Input: P = 3, B = 4, H = 5
Output: 3.14
Input: P = 5, B = 12, H = 13
Output: 12.56

Approach: Formula for calculating the inradius of a right angled triangle can be given as r = ( P + B – H ) / 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 incircle will be PI * ((P + B – H) / 2)2.
Below is the implementation of the above approach:

## C++

 `// C++ program to find the area of``// incircle of right angled triangle``#include ``using` `namespace` `std;``#define PI 3.14159265` `// Function to find area of``// incircle``float` `area_inscribed(``float` `P, ``float` `B, ``float` `H)``{``    ``return` `((P + B - H) * (P + B - H) * (PI / 4));``}` `// Driver code``int` `main()``{``    ``float` `P = 3, B = 4, H = 5;``    ``cout << area_inscribed(P, B, H) << endl;` `    ``return` `0;``}` `// The code is contributed by Nidhi goel`

## C

 `// C program to find the area of``// incircle of right angled triangle``#include ``#define PI 3.14159265` `// Function to find area of``// incircle``float` `area_inscribed(``float` `P, ``float` `B, ``float` `H)``{``    ``return` `((P + B - H) * (P + B - H) * (PI / 4));``}` `// Driver code``int` `main()``{``    ``float` `P = 3, B = 4, H = 5;``    ``printf``(``"%f"``, area_inscribed(P, B, H));``    ``return` `0;``}`

## Java

 `// Java code to find the area of inscribed``// circle of right angled triangle``import` `java.lang.*;` `class` `GFG {` `    ``static` `double` `PI = ``3.14159265``;` `    ``// Function to find the area of``    ``// inscribed circle``    ``public` `static` `double` `area_inscribed(``double` `P, ``double` `B,``                                        ``double` `H)``    ``{``        ``return` `((P + B - H) * (P + B - H) * (PI / ``4``));``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``double` `P = ``3``, B = ``4``, H = ``5``;``        ``System.out.println(area_inscribed(P, B, H));``    ``}``}`

## Python3

 `# Python3 code to find the area of inscribed``# circle of right angled triangle``PI ``=` `3.14159265` `# Function to find the area of``# inscribed circle`  `def` `area_inscribed(P, B, H):``    ``return` `((P ``+` `B ``-` `H)``*``(P ``+` `B ``-` `H)``*``(PI ``/` `4``))`  `# Driver code``P ``=` `3``B ``=` `4``H ``=` `5``print``(area_inscribed(P, B, H))`

## C#

 `// C# code to find the area of``// inscribed circle``// of right angled triangle``using` `System;` `class` `GFG {``    ``static` `double` `PI = 3.14159265;` `    ``// Function to find the area of``    ``// inscribed circle``    ``public` `static` `double` `area_inscribed(``double` `P, ``double` `B,``                                        ``double` `H)``    ``{``        ``return` `((P + B - H) * (P + B - H) * (PI / 4));``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``double` `P = 3.0, B = 4.0, H = 5.0;``        ``Console.Write(area_inscribed(P, B, H));``    ``}``}`

## PHP

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## Javascript

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Output:

`3.141593`

Time Complexity : O(1)

Auxiliary Space: O(1)

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