Find all sides of a right angled triangle from given hypotenuse and area - GeeksforGeeks

Find all sides of a right angled triangle from given hypotenuse and area - GeeksforGeeks
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.

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.

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 possible\n";

        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;

}

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