Tutorialspoint.dev

Find other two sides of a right angle triangle

Given one side of right angle triangle, check if there exists a right angle triangle possible with any other two sides of the triangle. If possible print length of the other two sides.
Else print -1
Note: All the sides of triangle must be positive integers

Example 1:

Input : a = 3
Output : b = 4, c = 5
Explanation : a = 3, b = 4 and c = 5 form right 
angle triangle because 
32 + 42 = 9 + 16 = 25 = 52 => 32 + 42 = 52

Example 2:

Input : a = 11
Output : b = 60, c = 61
Explanation : a = 11, b = 60 and c = 61 form 
right angle triangle because
112 + 602 = 121 + 3600 = 3721 = 612 => 112 + 602 = 612



To solve this problem we first observe the Pythagoras equation. If a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.
This relationship is represented by the formula:

a2 + b2 = c2

  1. Case 1 – a is an odd number: Given a, find b and c
    c2 - b2 = a2
    OR
    c = (a2 + 1)/2;
    b = (a2 - 1)/2;
    

    Above solution works only for case when a is odd, because a2 + 1 is divisible by 2 only for odd a.

    /li>
  2. Case 2 – a is an even number: When c-b is 2 & c+b is (a2)/2
    c-b = 2 & c+b = (a2)/2
    Hence,
    c = (a2)/4 + 1;
    b = (a2)/4 - 1;
    

    This works when a is even.

This solution doesn’t work for case of a = 1 and a = 2 because there isn’t any right angle triangle with side 1 or 2 with all integer sides.

C++

// C++ program to print other two sides of right
// angle traingle given one side
#include <bits/stdc++.h>
using namespace std;
  
// Finds two sides of a right angle triangle
// if it exist.
void printOtherSides(int n)
{
    // if n is odd
    if (n & 1)
    {
        // case of n = 1 handled separately
        if (n == 1)
            cout << -1 << endl;
        else
        {
            int b = (n*n-1)/2;
            int c = (n*n+1)/2;
            cout << "b = " << b
                 << ", c = " << c << endl;
        }
    }
    else
    {
        // case of n = 2 handled separately
        if (n == 2)
            cout << -1 << endl;
        else
        {
            int b = n*n/4-1;
            int c = n*n/4+1;
            cout << "b = " << b
                 << ", c = " << c << endl;
        }
    }
}
  
// Driver program to test above function
int main()
{
    int a = 3;
    printOtherSides(a);
    return 0;
}

Java

// Java program to print other two 
// sides of right angle traingle 
// given one side
  
class GFG
{
    // Finds two sides of a right angle 
    // triangle if it they exist.
    static void printOtherSides(int n)
    {
        // if n is odd
        if (n % 2 != 0)
        {
            // case of n = 1 handled separately
            if (n == 1)
                System.out.println("-1");
            else
            {
                int b = (n * n -1) / 2;
                int c = (n *n  +1) / 2;
                System.out.println("b = "+ b +
                                   ", c = "+c);
            }
        }
        else
        {
            // case of n = 2 handled separately
            if (n == 2)
                System.out.println("-1");
            else
            {
                int b = n * n / 4 - 1;
                int c = n * n / 4 + 1;
                System.out.println("b = "+ b +
                                   ", c = "+c);
            }
        }
    
          
    // Driver code
    public static void main (String[] args)
    {
        int a = 3;
        printOtherSides(a);
    }
}
  
// This code is contributed by Anant Agarwal.

Python

# Python program to print other 
# two sides of right angle 
# triangle given one side
  
# Finds two sides of a right angle
# triangle if it they exist.
def printOtherSides(n):
      
    # if n is odd
    if(n & 1):
          
        # case of n = 1 handled
        # separately
        if(n == 1):
            print(-1)
        else:
            b = (n * n - 1) // 2
            c = (n * n + 1) // 2
            print("b =", b, ", c =", c)
    else:
          
        # case of n = 2 handled
        # separately
        if(n == 2):
            print(-1)
        else:
            b = n * n // 4 - 1
            c = n * n // 4 + 1
            print("b =", b", c =", c)
  
# Driver Code
a = 3
printOtherSides(a)
  
# This code is contributed by
# Sanjit_Prasad

C#

// C# program to print other two 
// sides of right angle traingle 
// given one side
using System;
  
class GFG {
      
    // Finds two sides of a right angle 
    // triangle if it they exist.
    static void printOtherSides(int n)
    {
          
        // if n is odd
        if (n % 2 != 0)
        {
              
            // case of n = 1 handled
            // separately
            if (n == 1)
            Console.WriteLine("-1");
            else
            {
                int b = (n * n - 1) / 2;
                int c = (n * n + 1) / 2;
                Console.Write("b = "+ b +
                              ", c = "+ c);
            }
        }
        else
        {
              
            // case of n = 2 handled
            // separately
            if (n == 2)
            Console.Write("-1");
            else
            {
                int b = n * n / 4 - 1;
                int c = n * n / 4 + 1;
                Console.Write("b = "+ b +
                              ", c = "+ c);
            }
        }
    
          
    // Driver code
    public static void Main ()
    {
        int a = 3;
        printOtherSides(a);
    }
}
  
// This code is contributed by Nitin Mittal.

PHP

<?php
// PHP program to print other two 
// sides of right angle traingle 
// given one side
  
// Finds two sides of a right angle
// triangle if it they exist.
function printOtherSides($n)
{
      
    // if n is odd
    if ($n & 1)
    {
        // case of n = 1 
        // handled separately
        if ($n == 1)
            echo -1 ;
        else
        {
            $b = ($n * $n - 1) / 2;
            $c = ($n * $n + 1) / 2;
            echo "b = " ,$b,", c = " ,$c ;
        }
    }
    else
    {
          
        // case of n = 2 
        // handled separately
        if ($n == 2)
            echo -1 ;
        else
        {
            $b = $n * $n / 4 - 1;
            $c = $n * $n / 4 + 1;
            echo "b = " ,$b, ", c = ", $c ;
        }
    }
}
  
    // Driver Code
    $a = 3;
    printOtherSides($a);
    return 0;
  
// This code is contributed by nitin mittal.
?>

Output:

b = 4, c = 5

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



This article is attributed to GeeksforGeeks.org

You Might Also Like

leave a comment

code

0 Comments

load comments

Subscribe to Our Newsletter