Find all sides of a right angled triangle from given hypotenuse and area | Set 1

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.

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++

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// C++ program to get right angle triangle, given 
// hypotenuse and area of triangle 
#include <bits/stdc++.h> 
using namespace std; 
    
//  limit for float comparison 
#define eps 1e-6 
    
// Utility method to get area of right angle triangle, 
// given base and hypotenuse 
double getArea(double base, double hypotenuse) 
    double height = sqrt(hypotenuse*hypotenuse - base*base); 
    return 0.5 * base * height; 
    
// Prints base and height of triangle using hypotenuse 
// and area information 
void printRightAngleTriangle(int hypotenuse, int area) 
    int hsquare = hypotenuse * hypotenuse; 
    
    // maximum area will be obtained when base and height 
    // are equal (= sqrt(h*h/2)) 
    double sideForMaxArea = sqrt(hsquare / 2.0); 
    double maxArea = getArea(sideForMaxArea, hypotenuse); 
    
    // if given area itself is larger than maxArea then no 
    // solution is possible 
    if (area > maxArea) 
    
        cout << "Not possiblen"
        return
    
    
    double low = 0.0; 
    double high = sideForMaxArea; 
    double base; 
    
    // binary search for base 
    while (abs(high - low) > eps) 
    
        base = (low + high) / 2.0; 
        if (getArea(base, hypotenuse) >= area) 
            high = base; 
        else
            low = base; 
    
    
    // get height by pythagorean rule 
    double height = sqrt(hsquare - base*base); 
    cout << base << " " << height << endl; 
    
// Driver code to test above methods 
int main() 
    int hypotenuse = 5; 
    int area = 6; 
    
    printRightAngleTriangle(hypotenuse, area); 
    return 0; 

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Java

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// Java program to get right angle triangle, given 
// hypotenuse and area of triangle
public class GFG {
  
// limit for float comparison 
    final static double eps = (double) 1e-6;
  
// Utility method to get area of right angle triangle, 
// given base and hypotenuse 
    static double getArea(double base, double hypotenuse) {
        double height = Math.sqrt(hypotenuse * hypotenuse - base * base);
        return 0.5 * base * height;
    }
  
// Prints base and height of triangle using hypotenuse 
// and area information 
    static void printRightAngleTriangle(int hypotenuse, int area) {
        int hsquare = hypotenuse * hypotenuse;
  
        // maximum area will be obtained when base and height 
        // are equal (= sqrt(h*h/2)) 
        double sideForMaxArea = Math.sqrt(hsquare / 2.0);
        double maxArea = getArea(sideForMaxArea, hypotenuse);
  
        // if given area itself is larger than maxArea then no 
        // solution is possible 
        if (area > maxArea) {
            System.out.print("Not possible");
            return;
        }
  
        double low = 0.0;
        double high = sideForMaxArea;
        double base = 0;
  
        // binary search for base 
        while (Math.abs(high - low) > eps) {
            base = (low + high) / 2.0;
            if (getArea(base, hypotenuse) >= area) {
                high = base;
            } else {
                low = base;
            }
        }
  
        // get height by pythagorean rule 
        double height = Math.sqrt(hsquare - base * base);
        System.out.println(Math.round(base) + " " + Math.round(height));
    }
  
// Driver code to test above methods 
    static public void main(String[] args) {
        int hypotenuse = 5;
        int area = 6;
  
        printRightAngleTriangle(hypotenuse, area);
    }
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python 3 program to get right angle triangle, given 
# hypotenuse and area of triangle 
  
# limit for float comparison 
# define eps 1e-6 
import math
  
# Utility method to get area of right angle triangle, 
# given base and hypotenuse 
def getArea(base, hypotenuse):
    height = math.sqrt(hypotenuse*hypotenuse - base*base); 
    return 0.5 * base * height 
  
# Prints base and height of triangle using hypotenuse 
# and area information 
def printRightAngleTriangle(hypotenuse, area):
    hsquare = hypotenuse * hypotenuse 
  
    # maximum area will be obtained when base and height 
    # are equal (= sqrt(h*h/2)) 
    sideForMaxArea = math.sqrt(hsquare / 2.0
    maxArea = getArea(sideForMaxArea, hypotenuse)
  
    # if given area itself is larger than maxArea then no 
    # solution is possible 
    if (area > maxArea):
        print("Not possiblen"
        return
      
    low = 0.0
    high = sideForMaxArea 
      
    # binary search for base 
    while (abs(high - low) > 1e-6):
        base = (low + high) / 2.0
        if (getArea(base, hypotenuse) >= area):
            high =base
        else:
            low = base
      
    # get height by pythagorean rule 
    height = math.ceil(math.sqrt(hsquare - base*base))
    base = math.floor(base)
    print(base,height) 
  
# Driver code to test above methods 
if __name__ == '__main__':
    hypotenuse = 5
    area = 6
  
    printRightAngleTriangle(hypotenuse, area) 
  
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# program to get right angle triangle, given 
// hypotenuse and area of triangle 
  
using System;
public class GFG{
  
  
// limit for float comparison 
     static double eps = (double) 1e-6; 
  
// Utility method to get area of right angle triangle, 
// given base and hypotenuse 
    static double getArea(double base1, double hypotenuse) { 
        double height = Math.Sqrt(hypotenuse * hypotenuse - base1 * base1); 
        return 0.5 * base1 * height; 
    
  
// Prints base and height of triangle using hypotenuse 
// and area information 
    static void printRightAngleTriangle(int hypotenuse, int area) { 
        int hsquare = hypotenuse * hypotenuse; 
  
        // maximum area will be obtained when base and height 
        // are equal (= sqrt(h*h/2)) 
        double sideForMaxArea = Math.Sqrt(hsquare / 2.0); 
        double maxArea = getArea(sideForMaxArea, hypotenuse); 
  
        // if given area itself is larger than maxArea then no 
        // solution is possible 
        if (area > maxArea) { 
            Console.Write("Not possible"); 
            return
        
  
        double low = 0.0; 
        double high = sideForMaxArea; 
        double base1 = 0; 
  
        // binary search for base 
        while (Math.Abs(high - low) > eps) { 
            base1 = (low + high) / 2.0; 
            if (getArea(base1, hypotenuse) >= area) { 
                high = base1; 
            } else
                low = base1; 
            
        
  
        // get height by pythagorean rule 
        double height = Math.Sqrt(hsquare - base1 * base1); 
        Console.WriteLine(Math.Round(base1) + " " + Math.Round(height)); 
    
  
// Driver code to test above methods 
    static public void Main() { 
        int hypotenuse = 5; 
        int area = 6; 
  
        printRightAngleTriangle(hypotenuse, area); 
    
  
// This code is contributed by 29AjayKumar 

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PHP

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<?php
// PHP program to get right angle triangle, 
// given hypotenuse and area of triangle 
  
// limit for float comparison 
$eps =.0000001;
  
// Utility method to get area of right 
// angle triangle, given base and hypotenuse 
function getArea($base, $hypotenuse
    $height = sqrt($hypotenuse * $hypotenuse
                                 $base * $base); 
    return 0.5 * $base * $height
  
// Prints base and height of triangle 
// using hypotenuse and area information 
function printRightAngleTriangle($hypotenuse
                                 $area
    global $eps;
    $hsquare = $hypotenuse * $hypotenuse
  
    // maximum area will be obtained when base 
    // and height are equal (= sqrt(h*h/2)) 
    $sideForMaxArea = sqrt($hsquare / 2.0); 
    $maxArea = getArea($sideForMaxArea
                       $hypotenuse); 
  
    // if given area itself is larger than 
    // maxArea then no solution is possible 
    if ($area > $maxArea
    
        echo "Not possiblen"
        return
    
  
    $low = 0.0; 
    $high = $sideForMaxArea
    $base
  
    // binary search for base 
    while (abs($high - $low) > $eps
    
        $base = ($low + $high) / 2.0; 
        if (getArea($base, $hypotenuse) >= $area
            $high = $base
        else
            $low = $base
    
  
    // get height by pythagorean rule 
    $height = sqrt($hsquare - $base * $base); 
        echo (ceil($base)) ," "
             (floor($height)), "\n"
  
// Driver Code 
$hypotenuse = 5; 
$area = 6; 
  
printRightAngleTriangle($hypotenuse, $area); 
  
// This code is contributed by Sachin
?>

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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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