How To Find Sides Of Right Triangle Given Hypotenuse

Given hypotenuse and area of a right angle triangle, get its base and height and if any triangle with given hypotenuse and area is not possible, print not possible.
Examples:

Input  : hypotenuse = 5,    area = 6 Output : base = 3, height = 4  Input : hypotenuse = 5, area = 7  Output : No triangle possible with above specification.

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hypotenuse

We can use a property of right angle triangle for solving this problem, which can be stated as follows,

A right angle triangle with fixed hypotenuse attains maximum area, when it is isosceles i.e. both height and base becomes equal so if hypotenuse if H, then  by pythagorean theorem, Base²          + Height²          = H²          For maximum area both base and height should be equal,  b²          + b²          = H²          b = sqrt(H²/2)  Above is the length of base at which triangle attains maximum area, given area must be less than this maximum area, otherwise no such triangle will possible.

Now if given area is less than this maximum area, we can do a binary search for length of base, as increasing base will increases area, it is a monotonically increasing function where binary search can be applied easily.
In below code, a method is written for getting area of right angle triangle, recall that for right angle triangle area is ½*base*height and height can be calculated from base and hypotenuse using pythagorean theorem.
Below is the implementation of above approach:

C++

#include <bits/stdc++.h>

using namespace std;

#define eps 1e-6

double getArea( double base, double hypotenuse)

{

double height = sqrt (hypotenuse*hypotenuse - base*base);

return 0.5 * base * height;

}

void printRightAngleTriangle( int hypotenuse, int area)

{

int hsquare = hypotenuse * hypotenuse;

double sideForMaxArea = sqrt (hsquare / 2.0);

double maxArea = getArea(sideForMaxArea, hypotenuse);

if (area > maxArea)

{

cout << "Not possiblen" ;

return ;

}

double low = 0.0;

double high = sideForMaxArea;

double base;

while ( abs (high - low) > eps)

{

base = (low + high) / 2.0;

if (getArea(base, hypotenuse) >= area)

high = base;

else

low = base;

}

double height = sqrt (hsquare - base*base);

cout << base << " " << height << endl;

}

int main()

{

int hypotenuse = 5;

int area = 6;

printRightAngleTriangle(hypotenuse, area);

return 0;

}

Java

public class GFG {

final static double eps = ( double ) 1e- 6 ;

static double getArea( double base, double hypotenuse) {

double height = Math.sqrt(hypotenuse * hypotenuse - base * base);

return 0.5 * base * height;

}

static void printRightAngleTriangle( int hypotenuse, int area) {

int hsquare = hypotenuse * hypotenuse;

double sideForMaxArea = Math.sqrt(hsquare / 2.0 );

double maxArea = getArea(sideForMaxArea, hypotenuse);

if (area > maxArea) {

System.out.print( "Not possible" );

return ;

}

double low = 0.0 ;

double high = sideForMaxArea;

double base = 0 ;

while (Math.abs(high - low) > eps) {

base = (low + high) / 2.0 ;

if (getArea(base, hypotenuse) >= area) {

high = base;

} else {

low = base;

}

double height = Math.sqrt(hsquare - base * base);

System.out.println(Math.round(base) + " " + Math.round(height));

}

static public void main(String[] args) {

int hypotenuse = 5 ;

int area = 6 ;

printRightAngleTriangle(hypotenuse, area);

}

Python3

import math

def getArea(base, hypotenuse):

height = math.sqrt(hypotenuse * hypotenuse - base * base);

return 0.5 * base * height

def printRightAngleTriangle(hypotenuse, area):

hsquare = hypotenuse * hypotenuse

sideForMaxArea = math.sqrt(hsquare / 2.0 )

maxArea = getArea(sideForMaxArea, hypotenuse)

if (area > maxArea):

print ( "Not possiblen" )

return

low = 0.0

high = sideForMaxArea

while ( abs (high - low) > 1e - 6 ):

base = (low + high) / 2.0

if (getArea(base, hypotenuse) > = area):

high = base

else :

low = base

height = math.ceil(math.sqrt(hsquare - base * base))

base = math.floor(base)

print (base,height)

if __name__ = = '__main__' :

hypotenuse = 5

area = 6

printRightAngleTriangle(hypotenuse, area)

C#

using System;

public class GFG{

static double eps = ( double ) 1e-6;

static double getArea( double base1, double hypotenuse) {

double height = Math.Sqrt(hypotenuse * hypotenuse - base1 * base1);

return 0.5 * base1 * height;

}

static void printRightAngleTriangle( int hypotenuse, int area) {

int hsquare = hypotenuse * hypotenuse;

double sideForMaxArea = Math.Sqrt(hsquare / 2.0);

double maxArea = getArea(sideForMaxArea, hypotenuse);

if (area > maxArea) {

Console.Write( "Not possible" );

return ;

}

double low = 0.0;

double high = sideForMaxArea;

double base1 = 0;

while (Math.Abs(high - low) > eps) {

base1 = (low + high) / 2.0;

if (getArea(base1, hypotenuse) >= area) {

high = base1;

} else {

low = base1;

}

double height = Math.Sqrt(hsquare - base1 * base1);

Console.WriteLine(Math.Round(base1) + " " + Math.Round(height));

}

static public void Main() {

int hypotenuse = 5;

int area = 6;

printRightAngleTriangle(hypotenuse, area);

}

PHP

<?php

$eps =.0000001;

function getArea( $base , $hypotenuse )

{

$height = sqrt( $hypotenuse * $hypotenuse -

$base * $base );

return 0.5 * $base * $height ;

}

function printRightAngleTriangle( $hypotenuse ,

$area )

{

global $eps ;

$hsquare = $hypotenuse * $hypotenuse ;

$sideForMaxArea = sqrt( $hsquare / 2.0);

$maxArea = getArea( $sideForMaxArea ,

$hypotenuse );

if ( $area > $maxArea )

{

echo "Not possiblen" ;

return ;

}

$low = 0.0;

$high = $sideForMaxArea ;

$base ;

while ( abs ( $high - $low ) > $eps )

{

$base = ( $low + $high ) / 2.0;

if (getArea( $base , $hypotenuse ) >= $area )

$high = $base ;

else

$low = $base ;

}

$height = sqrt( $hsquare - $base * $base );

echo ( ceil ( $base )) , " " ,

( floor ( $height )), "\n" ;

}

$hypotenuse = 5;

$area = 6;

printRightAngleTriangle( $hypotenuse , $area );

?>

Javascript

<script>

let eps = 1e-6;

function getArea(base, hypotenuse) {

let height = Math.sqrt(hypotenuse * hypotenuse - base * base);

return 0.5 * base * height;

}

function printRightAngleTriangle(hypotenuse, area) {

let hsquare = hypotenuse * hypotenuse;

let sideForMaxArea = Math.sqrt(hsquare / 2.0);

let maxArea = getArea(sideForMaxArea, hypotenuse);

if (area > maxArea) {

document.write( "Not possible" );

return ;

}

let low = 0.0;

let high = sideForMaxArea;

let base = 0;

while (Math.abs(high - low) > eps) {

base = (low + high) / 2.0;

if (getArea(base, hypotenuse) >= area) {

high = base;

} else {

low = base;

}

let height = Math.sqrt(hsquare - base * base);

document.write(Math.round(base) + " " + Math.round(height));

}

let hypotenuse = 5;

let area = 6;

printRightAngleTriangle(hypotenuse, area);

</script>

Output:

3 4

One more solution is discussed in below post.
Check if right angles possible from given area and hypotenuse
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