Check if right triangle possible from given area and hypotenuse - GeeksforGeeks

Check if right triangle possible from given area and hypotenuse - GeeksforGeeks
Given area and hypotenuse, the aim is to print all sides if right triangle can exist, else print -1. We need to print all sides in ascending order.

In this post, a new solution with below logic is discussed.

Let the two unknown sides be a and b
Area : A = 0.5 * a * b
Hypotenuse Square : H^2 = a^2 + b^2
Substituting b, we get H2 = a2 + (4 * A2)/a2
On re-arranging, we get the equation a4 – (H2)(a2) + 4*(A2)

The discriminant D of this equation would be D = H⁴ – 16*(A²)
If D = 0, then roots are given by the linear equation formula, roots = (-b +- sqrt(D) )/2*a
these roots would be equal to the square of the sides, finding the square roots would give us the sides.

def findRightAngle(A, H):

    # Descriminant of the equation

    D = pow(H,4) - 16 * A * A

    if D >= 0:

        # applying the linear equation

        # formula to find both the roots

        root1 = (H * H + sqrt(D))/2

        root2 = (H * H - sqrt(D))/2

        a = sqrt(root1)

        b = sqrt(root2)

        if b >= a:

            print a, b, H

        else:

            print b, a, H

    else:

        print "-1"

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

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.

// 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 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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