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)
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