# Area of Incircle of a Right Angled Triangle

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 = 5Output:3.14Input:P = 5, B = 12, H = 13Output: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 * r ^{2}** 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 <stdio.h>` `#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

`<?php` `// PHP program to find the` `// area of inscribed` `// circle of right angled triangle` `$PI` `= 3.14159265;` `// Function to find area of` `// inscribed circle` `function` `area_inscribed(` `$P` `, ` `$B` `, ` `$H` `)` `{` ` ` `global` `$PI` `;` ` ` `return` `((` `$P` `+ ` `$B` `- ` `$H` `)*(` `$P` `+ ` `$B` `- ` `$H` `)* (` `$PI` `/ 4));` `}` `// Driver code` `$P` `=3;` `$B` `=4;` `$H` `=5;` `echo` `(area_inscribed(` `$P` `, ` `$B` `, ` `$H` `));` `?>` |

## Javascript

`<script>` `// javascript code to find the area of inscribed` `// circle of right angled triangle` ` ` `let PI = 3.14159265;` ` ` `// Function to find the area of` ` ` `// inscribed circle` ` ` `function` `area_inscribed(P , B , H) {` ` ` `return` `((P + B - H) * (P + B - H) * (PI / 4));` ` ` `}` ` ` `// Driver code` ` ` `var` `P = 3, B = 4, H = 5;` ` ` `document.write(area_inscribed(P, B, H).toFixed(6));` `// This code is contributed by Rajput-Ji` `</script>` |

**Output:**

3.141593

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