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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