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++
// 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
// 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
Time Complexity : O(1)
Auxiliary Space: O(1)