Given the length of sides of an equilateral triangle, the task is to find the area and perimeter of Incircle of the given equilateral triangle. Examples:
Input: side = 6
Output: Area = 9.4. Perimeter = 10.88
Input: side = 9
Output: Area = 21.21, Perimeter = 16.32
Properties of an Incircle are:
- The center of the Incircle is same as the center of the triangle i.e. the point where the medians of the equilateral triangle intersect.
- Inscribed circle of an equilateral triangle is made through the midpoint of the edges of an equilateral triangle.
- The Inradius of an Incircle of an equilateral triangle can be calculated using the formula:
,
- where
is the length of the side of equilateral triangle. - Below image shows an equilateral triangle with incircle:
- Approach: Area of circle =
and perimeter of circle =
, where r is the radius of given circle. Also the radius of Incircle of an equilateral triangle = (side of the equilateral triangle)/ 3. Therefore,- The formula used to calculate the area of Incircle using Inradius is:

- The formula used to calculate the perimeter of Incircle using Inradius is:

C
#include <math.h>
#include <stdio.h>
#define PI 3.14159265
float area_inscribed( float a)
{
return (a * a * (PI / 12));
}
float perm_inscribed( float a)
{
return (PI * (a / sqrt (3)));
}
int main()
{
float a = 6;
printf ( "Area of inscribed circle is :%f\n" ,
area_inscribed(a));
printf ( "Perimeter of inscribed circle is :%f" ,
perm_inscribed(a));
return 0;
}
|
Java
import java.lang.*;
class GFG {
static double PI = 3.14159265 ;
public static double area_inscribed( double a)
{
return (a * a * (PI / 12 ));
}
public static double perm_inscribed( double a)
{
return (PI * (a / Math.sqrt( 3 )));
}
public static void main(String[] args)
{
double a = 6.0 ;
System.out.println( "Area of inscribed circle is :"
+ area_inscribed(a));
System.out.println( "\nPerimeter of inscribed circle is :"
+ perm_inscribed(a));
}
}
|
Python3
import math
PI = 3.14159265
def area_inscribed(a):
return (a * a * (PI / 12 ))
def perm_inscribed(a):
return ( PI * (a / math.sqrt( 3 ) ) )
a = 6.0
print ( "Area of inscribed circle is :% f"
% area_inscribed(a))
print ( "\nPerimeter of inscribed circle is :% f"
% perm_inscribed(a))
|
C#
using System;
class GFG {
static double PI = 3.14159265;
public static double area_inscribed( double a)
{
return (a * a * (PI / 12));
}
public static double perm_inscribed( double a)
{
return (PI * (a / Math.Sqrt(3)));
}
public static void Main()
{
double a = 6.0;
Console.Write( "Area of inscribed circle is :"
+ area_inscribed(a));
Console.Write( "\nPerimeter of inscribed circle is :"
+ perm_inscribed(a));
}
}
|
PHP
<?php
$PI = 3.14159265;
function area_inscribed( $a )
{
global $PI ;
return ( $a * $a * ( $PI / 12));
}
function perm_inscribed( $a )
{
global $PI ;
return ( $PI * ( $a / sqrt(3) ) );
}
$a = 6;
echo ( "Area of inscribed circle is :" );
echo (area_inscribed( $a ));
echo ( "Perimeter of inscribed circle is :" );
echo (perm_inscribed( $a ));
?>
|
Javascript
Javascrip
let PI = 3.14159265
function area_inscribed(a)
{
return (a * a * (PI / 12))
}
function perm_inscribed(a)
{
return ( PI * (a / Math.sqrt(3) ) )
}
let a = 6.0
console.log( "Area of inscribed circle is :" , area_inscribed(a))
console.log( "\nPerimeter of inscribed circle is :" , perm_inscribed(a))
|
Time Complexity: O(1)
Auxiliary Space: O(1)