Program to calculate the length of the diagonal in a rectangle

Write a program to calculate the length of the diagonal in a rectangle with the given length and breadth.

Examples:

Input: Length = 5, Width = 3
Output: Diagonal Length = 5.83095

Input: Length = 8, Width = 6
Output: Diagonal Length = 10

Approach: To solve the problem, follow the below idea:

The diagonal of a rectangle forms a right-angled triangle with the length and width of the rectangle. We can use the Pythagoras theorem to calculate the length of the diagonal. So, we can use this formula Diagonal² = Length² + Breadth² to calculate the length of the diagonal.

Step-by-step algorithm:

• Square the length and width.
• Sum the squares.
• Take the square root of the sum.

Below is the implementation of the algorithm:

#include <cmath>
#include <iostream>

using namespace std;

int main()
{
    double L = 5, W = 3;
    double diagonal = sqrt(L * L + W * W);
    cout << "Diagonal Length: " << diagonal << endl;
    return 0;
}

#include <math.h>
#include <stdio.h>

int main()
{
    double L = 5, W = 3;
    double diagonal = sqrt(L * L + W * W);
    printf("Diagonal Length: %lf\n", diagonal);
    return 0;
}

public class DiagonalLength {
    public static void main(String[] args)
    {
        double L = 5, W = 3;
        double diagonal
            = Math.sqrt(Math.pow(L, 2) + Math.pow(W, 2));
        System.out.println("Diagonal Length: " + diagonal);
    }
}

import math

L, W = 5, 3
diagonal = math.sqrt(L**2 + W**2)
print(f"Diagonal Length: {diagonal}")

using System;

class Program {
    static void Main()
    {
        double L = 5, W = 3;
        double diagonal
            = Math.Sqrt(Math.Pow(L, 2) + Math.Pow(W, 2));
        Console.WriteLine("Diagonal Length: " + diagonal);
    }
}

let L = 5, W = 3;
let diagonal = Math.sqrt(L ** 2 + W ** 2);
console.log("Diagonal Length: " + diagonal);

Diagonal Length: 5.83095

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