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Find all sides of a right angled triangle from given hypotenuse and area | Set 1
• Difficulty Level : Medium
• Last Updated : 26 Mar, 2021

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.``` 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++

 `// C++ program to get right angle triangle, given``// hypotenuse and area of triangle``#include ``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;``}`

## Java

 `// 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`

## Python3

 `# 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`

## C#

 `// 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`

## PHP

 ` ``\$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``?>`

## Javascript

 ``

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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