Related Articles

# Area of Incircle of a Right Angled Triangle

• Difficulty Level : Medium
• Last Updated : 25 Mar, 2021

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 ``#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

 ``

## Javascript

 ``
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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up