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Given the radius of a circle, write a program to find the diameter of a circle.
Examples:
Input: r=4
Output: 8
Input: r=5
Output:10
Approach:
The diameter of a circle is simply twice the radius.
Here’s how you could express this formula in a mathematical sense:
D= 2* r
r is radius and D is diameter of a circle
Below is the implementation of the code:
C++
#include <iostream>
using namespace std;
int main()
{
int radius = 4, diameter;
cout << "The Given radius of the circle is: " << radius
<< endl;
diameter = 2 * radius;
cout << "The diameter of the circle is : " << diameter
<< endl;
return 0;
}
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Java
import java.util.Scanner;
public class CircleDiameter {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int radius = 4;
int diameter;
System.out.println("The Given radius of the circle is: " + radius);
diameter = 2 * radius;
System.out.println("The diameter of the circle is: " + diameter);
scanner.close();
}
}
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Python3
radius = 4
print("The Given radius of the circle is:", radius)
diameter = 2 * radius
print("The diameter of the circle is:", diameter)
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C#
using System;
public class GFG {
static public void Main()
{
int radius = 4, diameter;
Console.WriteLine(
"The Given radius of the circle is: " + radius);
diameter = 2 * radius;
Console.WriteLine("The diameter of the circle is : "
+ diameter);
}
}
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Javascript
let radius = 4;
let diameter;
console.log("The Given radius of the circle is: " + radius);
diameter = 2 * radius;
console.log("The diameter of the circle is : " + diameter);
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Output
The Given radius of the circle is: 4
The diameter of the circle is : 8
Time Complexity: O(1), since there is one multiplication operation which takes constant time.
Auxiliary Space: O(1), since no extra space has been taken.
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