Perimeter and Area of Varignon's Parallelogram

Perimeter and Area of Varignon’s Parallelogram

Last Updated : 07 Apr, 2021

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

// 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>

Output:

15
5

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