Area of a triangle with two vertices at midpoints of opposite sides of a square and the other vertex lying on vertex of a square

Given a positive integer N representing the side of a square, the task is to find the area of a triangle formed by connecting the midpoints of two adjacent sides and vertex opposite to the two sides.

Examples:

Input: N = 10
Output: 37.5

Input: N = 1
Output: 0.375

Approach: The given problem can be solved based on the following observations:

• The one side of the triangle will be the hypotenuse of the triangle formed with the vertices as two middle point and one vertex of the square at the intersection of the sides whose length of the side is given by .
• The length of the other two sides of the triangle is given by .
• Now, the sides of the triangle are known, therefore, the area of the triangle can be calculated using the Heron’s Formula.

Follow the steps below to solve the problem:

• Find the side of the triangle as discussed above formula and store it in the variables say a, b, and c respectively.
• After completing the above steps, print the value of (s*(s – a)*(s – b)*(s – c))^1/2 where s = (a + b + c) / 2.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the area of the
// triangle that inscribed in square
double areaOftriangle(int side)
{
    // Stores the length of the first
    // side of triangle
    double a = sqrt(pow(side / 2, 2)
                    + pow(side / 2, 2));

    // Stores the length of the second
    // side of triangle
    double b = sqrt(pow(side, 2)
                    + pow(side / 2, 2));

    // Stores the length of the third
    // side of triangle
    double c = sqrt(pow(side, 2)
                    + pow(side / 2, 2));

    double s = (a + b + c) / 2;

    // Stores the area of the triangle
    double area = sqrt(s * (s - a)
                       * (s - b) * (s - c));

    // Return the resultant area
    return area;
}

// Driver Code
int main()
{
    int N = 10;
    cout << areaOftriangle(N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the area of the
// triangle that inscribed in square
static double areaOftriangle(int side)
{

    // Stores the length of the first
    // side of triangle
    double a = Math.sqrt(Math.pow(side / 2, 2) +
                         Math.pow(side / 2, 2));

    // Stores the length of the second
    // side of triangle
    double b = Math.sqrt(Math.pow(side, 2) +
                         Math.pow(side / 2, 2));

    // Stores the length of the third
    // side of triangle
    double c = Math.sqrt(Math.pow(side, 2) +
                         Math.pow(side / 2, 2));

    double s = (a + b + c) / 2;

    // Stores the area of the triangle
    double area = Math.sqrt(s * (s - a) *
                           (s - b) * (s - c));

    // Return the resultant area
    return area;
}

// Driver code
public static void main(String[] args)
{
    int N = 10;

    System.out.print(areaOftriangle(N));
}
}

// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach
from math import sqrt

# Function to find the area of the
# triangle that inscribed in square
def areaOftriangle(side):

    # Stores the length of the first
    # side of triangle
    a = sqrt(pow(side / 2, 2) + pow(side / 2, 2))

    # Stores the length of the second
    # side of triangle
    b = sqrt(pow(side, 2) + pow(side / 2, 2))

    # Stores the length of the third
    # side of triangle
    c = sqrt(pow(side, 2) + pow(side / 2, 2))

    s = (a + b + c) / 2

    # Stores the area of the triangle
    area = sqrt(s * (s - a) * (s - b) * (s - c))

    # Return the resultant area
    return round(area, 1)

# Driver Code
if __name__ == '__main__':

    N = 10

    print (areaOftriangle(N))

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
class GFG{

// Function to find the area of the
// triangle that inscribed in square
static double areaOftriangle(int side)
{

    // Stores the length of the first
    // side of triangle
    double a = Math.Sqrt(Math.Pow(side / 2, 2) +
                         Math.Pow(side / 2, 2));

    // Stores the length of the second
    // side of triangle
    double b = Math.Sqrt(Math.Pow(side, 2) +
                         Math.Pow(side / 2, 2));

    // Stores the length of the third
    // side of triangle
    double c = Math.Sqrt(Math.Pow(side, 2) +
                         Math.Pow(side / 2, 2));

    double s = (a + b + c) / 2;

    // Stores the area of the triangle
    double area = Math.Sqrt(s * (s - a) *
                           (s - b) * (s - c));

    // Return the resultant area
    return area;
}

// Driver code
public static void Main(string[] args)
{
    int N = 10;

    Console.WriteLine(areaOftriangle(N));
}}

// This code is contributed by ukasp.

Javascript

<script>
    // Javascript program for the above approach

    // Function to find the area of the
    // triangle that inscribed in square
    function areaOftriangle(side)
    {

        // Stores the length of the first
        // side of triangle
        let a = Math.sqrt(Math.pow(side / 2, 2) +
                             Math.pow(side / 2, 2));

        // Stores the length of the second
        // side of triangle
        let b = Math.sqrt(Math.pow(side, 2) +
                             Math.pow(side / 2, 2));

        // Stores the length of the third
        // side of triangle
        let c = Math.sqrt(Math.pow(side, 2) +
                             Math.pow(side / 2, 2));

        let s = (a + b + c) / 2;

        // Stores the area of the triangle
        let area = Math.sqrt(s * (s - a) *
                               (s - b) * (s - c));

        // Return the resultant area
        return area.toFixed(1);
    }

    let N = 10;

    document.write(areaOftriangle(N));

// This code is contributed by suresh07.
</script>

Output:

37.5

Time Complexity: O(logn) because using inbuilt sqrt function
Auxiliary Space: O(1)