# 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 H^{2} = a^{2} + (4 * A^{2})/a^{2}

On re-arranging, we get the equation a^{4} – (H^{2})(a^{2}) + 4*(A^{2})

The discriminant D of this equation would be D = H^{4} – 16*(A^{2})

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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.