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 * 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; ` `} ` |

*chevron_right*

*filter_none*

## 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)); ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

## 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)) ` |

*chevron_right*

*filter_none*

## 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)); ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

## 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` `)); ` `?> ` |

*chevron_right*

*filter_none*

**Output:**

3.141593

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.