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Split given isosceles triangle of height H into N equal parts

  • Difficulty Level : Easy
  • Last Updated : 18 Jun, 2021

Given an integer N and an isosceles triangle consisting of height H, the task is to find (N – 1) points on the triangle such that the line passing through these points and parallel to the base of the triangle, divide the total area into N equal parts.

Examples:

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Input: N = 3, H = 2 
Output: 1.15 1.63 
Explanation: Make cuts at point 1.15 and 1.63 as shown below: 
 



Input: N = 2, H = 1000 
Output: 70710.67

Approach: The problem can be solved by observing the following properties:

Divide the triangle such that (xi / h)2 = i / N 
=> xi = h*√(i/n) 
xi = height of ith cut from the top vertex of the triangle

Follow the steps below to solve the problem:

  1. Iterate over the range [1, N – 1].
  2. In every ith iteration, print the value of xi using the above formula.

Below is the implementation of the above approach:

C++




// C++ Code for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to divide the isosceles triangle
// in equal parts by making N-1 cuts
// parallel to the base
void findPoint(int n, int h)
{
   
    // Iterate over the range [1, n - 1]
    for (int i = 1; i < n; i++)
        printf("%.2f ", sqrt(i / (n*1.0)) * h);
}
 
// Driver code
int main()
{
  // Given N
  int n = 3;
 
  // Given H
  int h = 2;
 
  // Function call
  findPoint(n, h);
 
  return 0;
}
 
// This code is contributed by mohit kumar 29

Java




// Java Code for above approach
import java.util.*;
class GFG
{
 
    // Function to divide the isosceles triangle
    // in equal parts by making N-1 cuts
    // parallel to the base
    static void findPoint(int n, int h)
    {
 
        // Iterate over the range [1, n - 1]
        for (int i = 1; i < n; i++)
            System.out.printf("%.2f ",
                    Math.sqrt(i / (n * 1.0)) * h);
    }
 
    // Driver code
    public static void main(String[] args)
    {
       
        // Given N
        int n = 3;
 
        // Given H
        int h = 2;
 
        // Function call
        findPoint(n, h);
    }
}
 
// This code is contributed by shikhasingrajput

Python3




# Python Code for above approach
 
 
# Function to divide the isosceles triangle
# in equal parts by making N-1 cuts
# parallel to the base
def findPoint(n, h):
 
 
    # Iterate over the range [1, n - 1]
    for i in range(1, n):
        print("{0:.2f}".format(((i / n) ** 0.5) * h), end =' ')
 
 
# Driver Code
if __name__ == '__main__':
 
    # Given N
    n = 3
 
    # Given H
    h = 2
 
    # Function call
    findPoint(n, h)

C#




// C# Code for above approach
using System;
class GFG
{
 
  // Function to divide the isosceles triangle
  // in equal parts by making N-1 cuts
  // parallel to the base
  static void findPoint(int n, int h)
  {
 
    // Iterate over the range [1, n - 1]
    for (int i = 1; i < n; i++)
      Console.Write("{0:F2} ",
                    Math.Sqrt(i / (n * 1.0)) * h);
  }
 
  // Driver code
  public static void Main(String[] args)
  {
 
    // Given N
    int n = 3;
 
    // Given H
    int h = 2;
 
    // Function call
    findPoint(n, h);
  }
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
// Javascript program for the above approach
 
    // Function to divide the isosceles triangle
    // in equal parts by making N-1 cuts
    // parallel to the base
    function findPolet(n, h)
    {
  
        // Iterate over the range [1, n - 1]
        for (let i = 1; i < n; i++)
            document.write(
                    Math.sqrt(i / (n * 1.0)) * h + " ");
    }
     
// driver function
 
        // Given N
        let n = 3;
  
        // Given H
        let h = 2;
  
        // Function call
        findPolet(n, h);;
      
     // This code is contributed by souravghosh0416.
</script>   
Output: 
1.15 1.63

 

Time Complexity: O(N)
Auxiliary Space: O(1)




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