Area of a Regular Pentagram

Given a Pentagram and it’s inner side length(d). The task is find out area of Pentagram. The Pentagram is a five-pointed star that is formed by drawing a continuous line in five straight segments.

Examples:



Input: d = 5
Output: Area = 139.187
Area of regular pentagram = 139.187
Input: d = 7
Output: Area = 272.807

Idea is to use Golden Ratio between a/b, b/c, and c/d which equals approximately 1.618
Inner side length d is given so
c = 1.618 * d
b = 1.618 * c
a = 1.618 * b

AB, BC and CD are equals(both side of regular pentagram)
So AB = BC = CD = c and BD is given by d.

Area of pentgram = Area of Pentagon BDFHJ + 5 * (Area of tringle BCD)
Area of Pentagon BDFHJ = (d^2 * 5)/ (4* tan 36)
Area of tringle BCD = [s(s-d)(s-c)(s-c)]^(1/2) {Heron’s Formula}
where
s = (d + c + c)/2

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
#define PI 3.14159
using namespace std;
  
// Function to return the area of triangle BCD
double areaOfTriangle(float d)
{
    // Using Golden ratio
    float c = 1.618 * d;
    float s = (d + c + c) / 2;
  
    // Calculate area of triangle BCD
    double area = sqrt(s * (s - c) *
                          (s - c) * (s - d));
  
    // Return area of all 5 trianlge are same
    return 5 * area;
}
  
// Function to return the area of regular pentagon
double areaOfRegPentagon(float d)
{
    // Calculate the area of regular
    // pentagon using above formula
    double cal = 4 * tan(PI / 5);
    double area = (5 * d * d) / cal;
  
    // Return area of regular pentagon
    return area;
}
  
// Function to return the area of pentagram
double areaOfPentagram(float d)
{
    // Area of a pentagram is equal to the 
    // area of regular  pentagon and five times 
    // the area of Triangle
    return areaOfRegPentagon(d) + 
                             areaOfTriangle(d);
}
  
// Driver code
int main()
{
    float d = 5;
    cout << areaOfPentagram(d) << endl;
  
    return 0;
}

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Java

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// Java implemenation of above approach
public class GFG 
{
  
    static double PI = 3.14159;
  
    // Function to return the area of triangle BCD
    static double areaOfTriangle(float d) 
    {
        // Using Golden ratio
        float c = (float) (1.618 * d);
        float s = (d + c + c) / 2;
  
        // Calculate area of triangle BCD
        double area = Math.sqrt(s * (s - c)
                * (s - c) * (s - d));
  
        // Return area of all 5 trianlge are same
        return 5 * area;
    }
  
    // Function to return the area of regular pentagon
    static double areaOfRegPentagon(float d)
    {
        // Calculate the area of regular
        // pentagon using above formula
        double cal = 4 * Math.tan(PI / 5);
        double area = (5 * d * d) / cal;
  
        // Return area of regular pentagon
        return area;
    }
  
    // Function to return the area of pentagram
    static double areaOfPentagram(float d) 
    {
        // Area of a pentagram is equal to the 
        // area of regular pentagon and five times 
        // the area of Triangle
        return areaOfRegPentagon(d)
                + areaOfTriangle(d);
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        float d = 5;
        System.out.println(areaOfPentagram(d));
    }
}
  
// This code has been contributed by 29AjayKumar

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Python

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# Python3 implementation of the approach
  
import math
  
PI = 3.14159
  
# Function to return the area of triangle BCD
def areaOfTriangle(d) :
  
    # Using Golden ratio
    c = 1.618 * d
    s = (d + c + c) / 2
  
    # Calculate area of triangle BCD
    area = math.sqrt(s * (s - c) *
                        (s - c) * (s - d))
  
    # Return area of all 5 triangles are the same
    return 5 * area
  
  
# Function to return the area of regular pentagon
def areaOfRegPentagon(d) :
      
    global PI
    # Calculate the area of regular
    # pentagon using above formula
    cal = 4 * math.tan(PI / 5)
    area = (5 * d * d) / cal
      
    # Return area of regular pentagon
    return area
  
  
# Function to return the area of pentagram
def areaOfPentagram(d) :
  
    # Area of a pentagram is equal to the 
    # area of regular pentagon and five times 
    # the area of Triangle
    return areaOfRegPentagon(d) + areaOfTriangle(d)
  
  
# Driver code
  
d = 5
print(areaOfPentagram(d)) 
  
      
# This code is contributed by ihritik

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C#

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// C# implementation of the above approach
using System;
  
class GFG 
{
  
    static double PI = 3.14159;
  
    // Function to return the area of triangle BCD
    static double areaOfTriangle(float d) 
    {
        // Using Golden ratio
        float c = (float) (1.618 * d);
        float s = (d + c + c) / 2;
  
        // Calculate area of triangle BCD
        double area = Math.Sqrt(s * (s - c)
                * (s - c) * (s - d));
  
        // Return area of all 5 trianlge are same
        return 5 * area;
    }
  
    // Function to return the area of regular pentagon
    static double areaOfRegPentagon(float d)
    {
        // Calculate the area of regular
        // pentagon using above formula
        double cal = 4 * Math.Tan(PI / 5);
        double area = (5 * d * d) / cal;
  
        // Return area of regular pentagon
        return area;
    }
  
    // Function to return the area of pentagram
    static double areaOfPentagram(float d) 
    {
        // Area of a pentagram is equal to the 
        // area of regular pentagon and five times 
        // the area of Triangle
        return areaOfRegPentagon(d)
                + areaOfTriangle(d);
    }
  
    // Driver code
    public static void Main()
    {
        float d = 5;
        Console.WriteLine(areaOfPentagram(d));
    }
}
  
// This code has been contributed by ihritik

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Output:

139.187

Time Complexity : O(1)



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