Perimeter and Area of Varignon’s Parallelogram
Given a and b are the lengths of the diagonals AC and BD of a quadrilateral ABCD with the area of quadrilateral as s. The task is to find the perimeter and area of the Varignon’s parallelogram PQRS.
Note: When we join the mid-points of the sides of any quadrilateral, the new quadrilateral formed inside will always be a parallelogram and this parallelogram is known as the Varignon’s parallelogram named upon the French Mathematician Pierre Varignon. Thus, PQRS will be a parallelogram since it is formed by joining the mid-points of quadrilateral ABCD as shown below:
Example:
Input: a = 7, b = 8, s = 10
Output: Perimeter = 15, Area = 5
Approach: The perimeter of Varignon’s parallelogram PQRS is equal to the sum of the length of the diagonals of quadrilateral ABCD.
Hence, Perimeter = a + b, where a and b are lengths of diagonals AC and BD.
Also, the area of the Varignon’s parallelogram is always half the area of quadrilateral ABCD.
Hence, Area = s / 2, where s is the area of quadrilateral ABCD.
Below is the implementation of the above approach:
C
// C program to find the perimeter and area #include <stdio.h> // Function to find the perimeter float per( float a, float b ) { return ( a + b ); } // Function to find the area float area( float s ) { return ( s/2 ); } // Driver code int main() { float a = 7, b = 8, s = 10; printf ( "%f\n" , per( a, b )); printf ( "%f" , area( s )); return 0; } |
Java
// Java code to find the perimeter and area import java.lang.*; class GFG { // Function to find the perimeter public static double per( double a, double b) { return (a + b); } // Function to find the area public static double area( double s) { return (s / 2 ); } // Driver code public static void main(String[] args) { double a = 7 , b = 8 , s = 10 ; System.out.println(per(a, b)); System.out.println(area(s)); } } |
Python3
# Python3 code to find the perimeter and area # Function to find the perimeter def per( a, b ): return ( a + b ) # Function to find the area def area( s ): return ( s / 2 ) # Driver code a = 7 b = 8 s = 10 print ( per( a, b )) print ( area( s )) |
C#
// C# code to find the perimeter and area using System; class GFG { // Function to find the perimeter public static double per( double a, double b) { return (a + b); } // Function to find the area public static double area( double s) { return (s / 2); } // Driver code public static void Main() { double a = 7.0, b = 8.0, s = 10.0; Console.WriteLine(per(a, b)); Console.Write(area(s)); } } |
PHP
<?php // PHP program to find perimeter and area // Function to find the perimeter function per( $a , $b ) { return ( $a + $b ); } // Function to find the area function area( $s ) { return ( $s / 2 ); } // Driver code $a =7; $b =8; $s =10; echo (per( $a , $b ) "" ); echo "\n" ; echo (area( $s )); ?> |
Javascript
<script> // javascript code to find the perimeter and area // Function to find the perimeter function per(a , b) { return (a + b); } // Function to find the area function area(s) { return (s / 2); } // Driver code var a = 7, b = 8, s = 10; document.write(per(a, b)); document.write(area(s)); // This code is contributed by shikhasingrajput </script> |
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