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: 28Input: 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> |
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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