Find other two sides of a right angle triangle
Last Updated :
16 Feb, 2023
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
- 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.
- 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++
#include <bits/stdc++.h>
using namespace std;
void printOtherSides( int n)
{
if (n & 1)
{
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
{
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;
}
}
}
int main()
{
int a = 3;
printOtherSides(a);
return 0;
}
|
Java
class GFG
{
static void printOtherSides( int n)
{
if (n % 2 != 0 )
{
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
{
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);
}
}
}
public static void main (String[] args)
{
int a = 3 ;
printOtherSides(a);
}
}
|
Python3
def printOtherSides(n):
if (n & 1 ):
if (n = = 1 ):
print ( - 1 )
else :
b = (n * n - 1 ) / / 2
c = (n * n + 1 ) / / 2
print ( "b =" , b, ", c =" , c)
else :
if (n = = 2 ):
print ( - 1 )
else :
b = n * n / / 4 - 1
c = n * n / / 4 + 1
print ( "b =" , b ", c =" , c)
a = 3
printOtherSides(a)
|
C#
using System;
class GFG {
static void printOtherSides( int n)
{
if (n % 2 != 0)
{
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
{
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);
}
}
}
public static void Main ()
{
int a = 3;
printOtherSides(a);
}
}
|
PHP
<?php
function printOtherSides( $n )
{
if ( $n & 1)
{
if ( $n == 1)
echo -1 ;
else
{
$b = ( $n * $n - 1) / 2;
$c = ( $n * $n + 1) / 2;
echo "b = " , $b , ", c = " , $c ;
}
}
else
{
if ( $n == 2)
echo -1 ;
else
{
$b = $n * $n / 4 - 1;
$c = $n * $n / 4 + 1;
echo "b = " , $b , ", c = " , $c ;
}
}
}
$a = 3;
printOtherSides( $a );
return 0;
?>
|
Javascript
<script>
function printOtherSides(n)
{
if (n % 2 != 0)
{
if (n == 1)
document.write( "-1" );
else
{
var b = (n * n -1) / 2;
var c = (n *n +1) / 2;
document.write( "b = " + b +
", c = " +c);
}
}
else
{
if (n == 2)
document.write( "-1" );
else
{
var b = n * n / 4 - 1;
var c = n * n / 4 + 1;
document.write( "b = " + b +
", c = " +c);
}
}
}
var a = 3;
printOtherSides(a);
</script>
|
Output:
b = 4, c = 5
Time complexity: O(1)
Auxiliary Space: O(1)
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