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 paralleogram is known as the Varignon’s paralleogram 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

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// 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;
}

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Java














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// 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)); 


    } 


} 



















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Python3














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# 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 )) 



















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C#














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// 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)); 


    } 


} 



















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PHP














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<?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 )); 


?> 



















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Output:


15
5







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