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
// C program to find the area of Inscribed circle
// of equilateral triangle
#include <math.h>
#include <stdio.h>
#define PI 3.14159265
// function to find area of inscribed circle
float
area_inscribed(
float
a)
{
return
(a * a * (PI / 12));
}
// function to find Perimeter of inscribed circle
float
perm_inscribed(
float
a)
{
return
(PI * (a /
sqrt
(3)));
}
// Driver code
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
// Java code to find the area of inscribed
// circle of equilateral triangle
import
java.lang.*;
class
GFG {
static
double
PI =
3.14159265
;
// function to find the area of
// inscribed circle
public
static
double
area_inscribed(
double
a)
{
return
(a * a * (PI /
12
));
}
// function to find the perimeter of
// inscribed circle
public
static
double
perm_inscribed(
double
a)
{
return
(PI * (a / Math.sqrt(
3
)));
}
// Driver code
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
# Python3 code to find the area of inscribed
# circle of equilateral triangle
import
math
PI
=
3.14159265
# Function to find the area of
# inscribed circle
def
area_inscribed(a):
return
(a
*
a
*
(PI
/
12
))
# Function to find the perimeter of
# inscribed circle
def
perm_inscribed(a):
return
( PI
*
(a
/
math.sqrt(
3
) ) )
# Driver code
a
=
6.0
print
(
"Area of inscribed circle is :% f"
%
area_inscribed(a))
print
(
"\nPerimeter of inscribed circle is :% f"
%
perm_inscribed(a))
C#
// C# code to find the area of
// inscribed circle
// of equilateral triangle
using
System;
class
GFG {
static
double
PI = 3.14159265;
// function to find the area of
// inscribed circle
public
static
double
area_inscribed(
double
a)
{
return
(a * a * (PI / 12));
}
// function to find the perimeter of
// inscribed circle
public
static
double
perm_inscribed(
double
a)
{
return
(PI * (a / Math.Sqrt(3)));
}
// Driver code
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
// PHP program to find the
// area of inscribed
// circle of equilateral triangle
$PI
= 3.14159265;
// function to find area of
// inscribed circle
function
area_inscribed(
$a
)
{
global
$PI
;
return
(
$a
*
$a
* (
$PI
/ 12));
}
// function to find perimeter of
// inscribed circle
function
perm_inscribed(
$a
)
{
global
$PI
;
return
(
$PI
* (
$a
/ sqrt(3) ) );
}
// Driver code
$a
= 6;
echo
(
"Area of inscribed circle is :"
);
echo
(area_inscribed(
$a
));
echo
(
"Perimeter of inscribed circle is :"
);
echo
(perm_inscribed(
$a
));
?>
Output:Area of inscribed circle is :9.424778 Perimeter of inscribed circle is :10.882796
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- The formula used to calculate the area of Incircle using Inradius is: