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:

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

edit
close

play_arrow

link
brightness_4
code

// 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

edit
close

play_arrow

link
brightness_4
code

# 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

edit
close

play_arrow

link
brightness_4
code

// 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

edit
close

play_arrow

link
brightness_4
code

<?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

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.





Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.


Article Tags :
Practice Tags :