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++ program to find the area of // incircle of right angled triangle #include <bits/stdc++.h> using namespace std;
#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;
cout << area_inscribed(P, B, H) << endl;
return 0;
} // The code is contributed by Nidhi goel |
// 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 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 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# 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 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 ));
?> |
<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> |
3.141593
Time Complexity : O(1)
Auxiliary Space: O(1)