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 = 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

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// 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;
}

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Java














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// 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)); 


    } 


} 



















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Python3














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# 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)) 



















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C#














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// 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)); 


    } 


} 



















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PHP














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<?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)); 


?> 



















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Output:


3.141593






          
          
          

          

    

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