Check if right triangle possible from given area and hypotenuse

Given area and hypotenuse, the aim is to print all sides if right triangle can exist, else print -1. We need to print all sides in ascending order.

Examples: 

Input  : 6 5
Output : 3 4 5

Input  : 10 6
Output : -1 

We have discussed a solution of this problem in below post. 
Find all sides of a right angled triangle from given hypotenuse and area | Set 1
In this post, a new solution with below logic is discussed.
Let the two unknown sides be a and b 
Area : A = 0.5 * a * b 
Hypotenuse Square : H^2 = a^2 + b^2 
Substituting b, we get H² = a² + (4 * A²)/a²
On re-arranging, we get the equation a⁴ – (H²)(a²) + 4*(A²)
The discriminant D of this equation would be D = H⁴ – 16*(A²) 
If D = 0, then roots are given by the linear equation formula, roots = (-b +- sqrt(D) )/2*a 
these roots would be equal to the square of the sides, finding the square roots would give us the sides. 

C++

// C++ program to check existence of
// right triangle.
#include <bits/stdc++.h>
using namespace std;

// Prints three sides of a right triangle
// from given area and hypotenuse if triangle
// is possible, else prints -1.
void findRightAngle(int A, int H)
{
    // Descriminant of the equation
    long D = pow(H, 4) - 16 * A * A;

    if (D >= 0)
    {
        // applying the linear equation
        // formula to find both the roots
        long root1 = (H * H + sqrt(D)) / 2;
        long root2 = (H * H - sqrt(D)) / 2;

        long a = sqrt(root1);
        long b = sqrt(root2);

        if (b >= a)
        cout << a << " " << b << " " << H;
        else
        cout << b << " " << a << " " << H;
    }
    else
        cout << "-1";
}

// Driver code
int main()
{
    findRightAngle(6, 5);

}

// This code is contributed By Anant Agarwal.

Java

// Java program to check existence of
// right triangle.

class GFG {

    // Prints three sides of a right triangle
    // from given area and hypotenuse if triangle
    // is possible, else prints -1.
    static void findRightAngle(double A, double H)
    {
        // Descriminant of the equation
        double D = Math.pow(H, 4) - 16 * A * A;

        if (D >= 0)
        {
            // applying the linear equation
            // formula to find both the roots
            double root1 = (H * H + Math.sqrt(D)) / 2;
            double root2 = (H * H - Math.sqrt(D)) / 2;

            double a = Math.sqrt(root1);
            double b = Math.sqrt(root2);
            if (b >= a)
                System.out.print(a + " " + b + " " + H);
            else
                System.out.print(b + " " + a + " " + H);
        }
        else
            System.out.print("-1");
    }

    // Driver code
    public static void main(String arg[])
    {
        findRightAngle(6, 5);
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python program to check existence of
# right triangle.
from math import sqrt

# Prints three sides of a right triangle
# from given area and hypotenuse if triangle
# is possible, else prints -1.
def findRightAngle(A, H):

    # Descriminant of the equation
    D = pow(H,4) - 16 * A * A
    if D >= 0:

        # applying the linear equation
        # formula to find both the roots
        root1 = (H * H + sqrt(D))//2
        root2 = (H * H - sqrt(D))//2

        a = int(sqrt(root1))
        b = int(sqrt(root2))
        if b >= a:
            print (a, b, H)
        else:
            print (b, a, H)
    else:
        print ("-1")

# Driver code
# Area is 6 and hypotenuse is 5.
findRightAngle(6, 5)

C#

// C# program to check existence of
// right triangle.
using System;

class GFG {

    // Prints three sides of a right triangle
    // from given area and hypotenuse if triangle
    // is possible, else prints -1.
    static void findRightAngle(double A, double H)
    {

        // Descriminant of the equation
        double D = Math.Pow(H, 4) - 16 * A * A;

        if (D >= 0) {

            // applying the linear equation
            // formula to find both the roots
            double root1 = (H * H + Math.Sqrt(D)) / 2;
            double root2 = (H * H - Math.Sqrt(D)) / 2;

            double a = Math.Sqrt(root1);
            double b = Math.Sqrt(root2);

            if (b >= a)
                Console.WriteLine(a + " " + b + " " + H);
            else
                Console.WriteLine(b + " " + a + " " + H);
        }
        else
            Console.WriteLine("-1");
    }

    // Driver code
    public static void Main()
    {
        findRightAngle(6, 5);
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to check existence of
// right triangle.

// Prints three sides of a right triangle
// from given area and hypotenuse if
// triangle is possible, else prints -1.
function findRightAngle($A, $H)
{

    // Descriminant of the equation
    $D = pow($H, 4) - 16 * $A * $A;

    if ($D >= 0)
    {

        // applying the linear equation
        // formula to find both the roots
        $root1 = ($H * $H + sqrt($D)) / 2;
        $root2 = ($H * $H - sqrt($D)) / 2;

        $a = sqrt($root1);
        $b = sqrt($root2);

        if ($b >= $a)
            echo $a , " ", $b , " " , $H;
        else
        echo $b , " " , $a , " " , $H;
    }
    else
        echo "-1";
}

    // Driver code
    findRightAngle(6, 5);

// This code is contributed By Anuj_67
?>

Javascript

<script>

// Javascript program to check existence of
// right triangle.

// Prints three sides of a right triangle
// from given area and hypotenuse if triangle
// is possible, else prints -1.
function findRightAngle(A,H)
{
    // Descriminant of the equation
    let D = Math.pow(H, 4) - 16 * A * A;

    if (D >= 0)
    {
        // applying the linear equation
        // formula to find both the roots
        let root1 = (H * H + Math.sqrt(D)) / 2;
        let root2 = (H * H - Math.sqrt(D)) / 2;

        let a = Math.sqrt(root1);
        let b = Math.sqrt(root2);

        if (b >= a)
       document.write(a + " " + b + " " + H+"<br/>");
        else
       document.write(b + " " + a + " " + H+"<br/>");
    }
    else
        document.write("-1");
}

// Driver code

    findRightAngle(6, 5);

// This code contributed by Rajput-Ji

</script>

Output: 

3 4 5

Time complexity: O(log(n)) since using inbuilt sqrt functions

Auxiliary Space: O(1)

This article is contributed by Harshit Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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