Check if right triangle possible from given area and hypotenuse
Last Updated :
27 Aug, 2022
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++
#include <bits/stdc++.h>
using namespace std;
void findRightAngle(int A, int H)
{
long D = pow(H, 4) - 16 * A * A;
if (D >= 0)
{
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";
}
int main()
{
findRightAngle(6, 5);
}
|
Java
class GFG {
static void findRightAngle(double A, double H)
{
double D = Math.pow(H, 4) - 16 * A * A;
if (D >= 0)
{
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");
}
public static void main(String arg[])
{
findRightAngle(6, 5);
}
}
|
Python3
from math import sqrt
def findRightAngle(A, H):
D = pow(H,4) - 16 * A * A
if D >= 0:
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")
findRightAngle(6, 5)
|
C#
using System;
class GFG {
static void findRightAngle(double A, double H)
{
double D = Math.Pow(H, 4) - 16 * A * A;
if (D >= 0) {
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");
}
public static void Main()
{
findRightAngle(6, 5);
}
}
|
PHP
<?php
function findRightAngle($A, $H)
{
$D = pow($H, 4) - 16 * $A * $A;
if ($D >= 0)
{
$root1 = ($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";
}
findRightAngle(6, 5);
?>
|
Javascript
<script>
function findRightAngle(A,H)
{
let D = Math.pow(H, 4) - 16 * A * A;
if (D >= 0)
{
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");
}
findRightAngle(6, 5);
</script>
|
Output:
3 4 5
Time complexity: O(log(n)) since using inbuilt sqrt functions
Auxiliary Space: O(1)
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