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Area of a leaf inside a square

Given an integer a as the side of the square ABCD. The task is to find the area of the leaf AECFA inside the square as shown below:

Examples: 

Input: a = 7 
Output: 28

Input: a = 21 
Output: 252 

Approach: To calculate the area of the leaf, first find the area of the half leaf AECA, which can be given as:  

Area of half leaf = Area of quadrant AECDA – Area of right triangle ACD
Thus, Area of half leaf = ( PI * a * a / 4 ) – a * a / 2 where PI = 22 / 7 and a is the side of the square. 
Hence, the area of full leaf will be ( PI * a * a / 2 ) – a * a 
On taking a * a common we get, Area of leaf = a * a ( PI / 2 – 1 )

Below is the implementation of the above approach:  

C




// C program to find the area of
// leaf inside a square
#include <stdio.h>
#define PI 3.14159265
 
// Function to find area of
// leaf
float area_leaf(float a)
{
    return (a * a * (PI / 2 - 1));
}
 
// Driver code
int main()
{
    float a = 7;
    printf("%f",
           area_leaf(a));
    return 0;
}


Java




// Java code to find the area of
// leaf inside a square
import java.lang.*;
 
class GFG {
 
    static double PI = 3.14159265;
 
    // Function to find the area of
    // leaf
    public static double area_leaf(double a)
    {
        return (a * a * (PI / 2 - 1));
    }
 
    // Driver code
    public static void main(String[] args)
    {
        double a = 7;
        System.out.println(area_leaf(a));
    }
}


Python3




# Python3 code to find the area of leaf
# inside a square
PI = 3.14159265
     
# Function to find the area of
# leaf
def area_leaf( a ):
    return ( a * a * ( PI / 2 - 1 ) )
     
# Driver code
a = 7
print(area_leaf( a ))


C#




// C# code to find the area of
// leaf
// inside square
using System;
 
class GFG {
    static double PI = 3.14159265;
 
    // Function to find the area of
    // leaf
    public static double area_leaf(double a)
    {
        return (a * a * (PI / 2 - 1));
    }
 
    // Driver code
    public static void Main()
    {
        double a = 7;
        Console.Write(area_leaf(a));
    }
}


PHP




<?php
// PHP program to find the
// area of leaf
// inside a square
$PI = 3.14159265;
 
// Function to find area of
// leaf
function area_leaf( $a )
{
    global $PI;
    return ( $a * $a * ( $PI / 2 - 1 ) );
}
 
// Driver code
$a = 7;
 
echo(area_leaf( $a ));
?>


Javascript




<script>
 
// Javascript program to find the area of
// leaf inside a square
const PI = 3.14159265;
 
// Function to find area of
// leaf
function area_leaf(a)
{
    return(a * a * (PI / 2 - 1));
}
 
// Driver code
let a = 7;
 
document.write(Math.round(area_leaf(a)));
 
// This code is contributed by souravmahato348
 
</script>


Output: 

28

 

Time complexity: O(1), since there is no loop or recursion.

Auxiliary Space: O(1), since no extra space has been taken.

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