Given an integer C which is the length of the hypotenuse of a right angled triangle of a circumcircle passing through the centre of the circumcircle. The task is to find the area of the circumcircle.
Examples:
Input: C = 8
Output: 50.26
Input: C = 10
Output: 78.53
Approach: Since the hypotenuse C passes through the center of the circle, the radius of the circle will be C / 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 circumcircle will be PI * (C / 2)2 i.e. PI * C2 / 4.
Below is the implementation of the above approach:
C++
// C++ program to find the area of Circumscribed// circle of right angled triangle#include <bits/stdc++.h>#define PI 3.14159265using namespace std;// Function to find area of// circumscribed circlefloat area_circumscribed(float c){ return (c * c * (PI / 4));}// Driver codeint main(){ float c = 8; cout << area_circumscribed(c); return 0;}// This code is contributed by Shivi_Aggarwal |
C
// C program to find the area of Circumscribed// circle of right angled triangle#include <stdio.h>#define PI 3.14159265// Function to find area of// circumscribed circlefloat area_circumscribed(float c){ return (c * c * (PI / 4));}// Driver codeint main(){ float c = 8; printf("%f", area_circumscribed(c)); return 0;} |
Java
// Java code to find the area of circumscribed// circle of right angled triangleimport java.lang.*;class GFG { static double PI = 3.14159265; // Function to find the area of // circumscribed circle public static double area_cicumscribed(double c) { return (c * c * (PI / 4)); } // Driver code public static void main(String[] args) { double c = 8.0; System.out.println(area_cicumscribed(c)); }} |
Python3
# Python3 code to find the area of circumscribed # circle of right angled trianglePI = 3.14159265 # Function to find the area of # circumscribed circledef area_cicumscribed(c): return (c * c * (PI / 4)) # Driver codec = 8.0print(area_cicumscribed(c)) |
C#
// C# code to find the area of// circumscribed circle// of right angled triangleusing System;class GFG { static double PI = 3.14159265; // Function to find the area of // circumscribed circle public static double area_cicumscribed(double c) { return (c * c * (PI / 4)); } // Driver code public static void Main() { double c = 8.0; Console.Write(area_cicumscribed(c)); }} |
PHP
<?php// PHP program to find the // area of Circumscribed // circle of right angled triangle$PI = 3.14159265;// Function to find area of // circumscribed circlefunction area_circumscribed($c){ global $PI; return ($c * $c * ($PI / 4));}// Driver code$c = 8;echo(area_circumscribed($c));?> |
Javascript
<script>// javascript code to find the area of circumscribed// circle of right angled triangle let PI = 3.14159265; // Function to find the area of // circumscribed circle function area_cicumscribed(c) { return (c * c * (PI / 4)); } // Driver code var c = 8.0; document.write(area_cicumscribed(c).toFixed(6));// This code is contributed by Rajput-Ji</script> |
50.265484
Time complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.
Another Approach:
To find the area of a circumcircle, we need to know its radius first. In a right-angled triangle, the circumcenter is the midpoint of the hypotenuse. Therefore, the radius of the circumcircle can be calculated as half of the length of the hypotenuse.
Once we have the radius, we can use the formula for the area of a circle, which is:
Area = ? * r^2
where r is the radius.
In this problem, we are given the length of the hypotenuse, which is also the diameter of the circumcircle. Therefore, we can calculate the radius as C/2. Then, we can use the above formula to find the area of the circumcircle.
C++
#include <iostream>#include <cmath>using namespace std;double areaOfCircumcircle(double c) { double a = c/2.0; double b = sqrt(pow(c, 2) - pow(a, 2)); double r = c/2.0; double area = M_PI*pow(r, 2); return area;}int main() { double c = 8; cout << areaOfCircumcircle(c) << endl;} |
Java
import java.lang.Math;public class Main { public static double areaOfCircumcircle(double c) { double a = c / 2.0; double b = Math.sqrt(Math.pow(c, 2) - Math.pow(a, 2)); double r = c / 2.0; double area = Math.PI * Math.pow(r, 2); return area; } public static void main(String[] args) { double c = 8; System.out.println(areaOfCircumcircle(c)); }}// Contributed by sdeadityasharma |
Python3
import mathdef areaOfCircumcircle(c): a = c/2.0 b = math.sqrt(pow(c, 2) - pow(a, 2)) r = c/2.0 area = math.pi*pow(r, 2) return areaif __name__ == '__main__': c = 8 print(areaOfCircumcircle(c))# This code is contributed by Prajwal Kandekar |
C#
using System;public class MainClass {public static double areaOfCircumcircle(double c) {double a = c / 2.0;double b = Math.Sqrt(Math.Pow(c, 2) - Math.Pow(a, 2));double r = c / 2.0;double area = Math.PI * Math.Pow(r, 2);return area;}public static void Main() { double c = 8; Console.WriteLine(areaOfCircumcircle(c));}} |
Javascript
function areaOfCircumcircle(c) { const a = c/2.0; const b = Math.sqrt(Math.pow(c, 2) - Math.pow(a, 2)); const r = c/2.0; const area = Math.PI*Math.pow(r, 2); return area;}const c = 8;console.log(areaOfCircumcircle(c)); |
50.2655
Time Complexity: O(1)
Space Complexity: O(1)
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