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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 <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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