Given an integer R which denotes the radius of a circle, the task is to find the area of an equilateral triangle inscribed in this circle.
Examples:
Input: R = 4
Output: 20.784
Explanation:
Area of equilateral triangle inscribed in a circle of radius R will be 20.784, whereas side of the triangle will be 6.928Input: R = 7
Output: 63.651
Explanation:
Area of equilateral triangle inscribed in a circle of radius R will be 63.651, whereas side of the triangle will be 12.124
Approach: Let the above triangle shown be an equilateral triangle denoted as PQR.
- The area of the triangle can be calculated as:
Area of triangle = (1/2) * Base * Height
- In this case, Base can be PQ, PR or QR and The height of the triangle can be PM. Hence,
Area of Triangle = (1/2) * QR * PM
- Now Applying sine law on the triangle ORQ,
RQ OR ------ = ------- sin 60 sin 30 => RQ = OR * sin60 / sin30 => Side of Triangle = OR * sqrt(3) As it is clearly observed PM = PO + OM = r + r * sin30 = (3/2) * r
- Therefore, the Base and height of the required equilateral triangle will be:
Base = r * sqrt(3) = r * 1.732 Height = (3/2) * r
- Compute the area of the triangle with the help of the formulae given above.
Below is the implementation of the above approach:
// C++ implementation to find // the area of the equilateral triangle // inscribed in a circle of radius R #include <iostream> using namespace std;
// Function to find the area of // equilateral triangle inscribed // in a circle of radius R double area( int R) {
// Base and Height of
// equilateral triangle
double base = 1.732 * R;
double height = (1.5) * R;
// Area using Base and Height
double area = 0.5 * base * height;
return area;
} // Driver Code int main()
{ int R = 7;
cout<<(area(R));
return 0;
} // This code is contributed by 29AjayKumar |
// Java implementation to find // the area of the equilateral triangle // inscribed in a circle of radius R class GFG
{ // Function to find the area of
// equilateral triangle inscribed
// in a circle of radius R
static double area( int R) {
// Base and Height of
// equilateral triangle
double base = 1.732 * R;
double height = ( 1.5 ) * R;
// Area using Base and Height
double area = 0.5 * base * height;
return area;
}
// Driver code
public static void main(String[] args) {
int R = 7 ;
System.out.println(area(R));
}
} // This code is contributed by 29AjayKumar |
# Python 3 implementation to find # the area of the equilateral triangle # inscribed in a circle of radius R # Function to find the area of # equilateral triangle inscribed # in a circle of radius R def area(R):
# Base and Height of
# equilateral triangle
base = 1.732 * R
height = ( 3 / 2 ) * R
# Area using Base and Height
area = (( 1 / 2 ) * base * height )
return area
# Driver Code if __name__ = = '__main__' :
R = 7
print (area(R))
|
// C# implementation to find // the area of the equilateral triangle // inscribed in a circle of radius R using System;
class GFG
{ // Function to find the area of
// equilateral triangle inscribed
// in a circle of radius R
static double area( int R)
{
// Base and Height of
// equilateral triangle
double Base = 1.732 * R;
double height = (1.5) * R;
// Area using Base and Height
double area = 0.5 * Base * height;
return area;
}
// Driver code
public static void Main(String[] args)
{
int R = 7;
Console.WriteLine(area(R));
}
} // This code is contributed by 29AjayKumar |
<script> // Javascript implementation to find // the area of the equilateral triangle // inscribed in a circle of radius R // Function to find the area of // equilateral triangle inscribed // in a circle of radius R function area(R)
{ // Base and Height of
// equilateral triangle
var base = 1.732 * R;
var height = (1.5) * R;
// Area using Base and Height
var area = 0.5 * base * height;
return area;
} // Driver code var R = 7;
document.write(area(R)); // This code is contributed by todaysgaurav </script> |
63.651
Time complexity : O(1)
Auxiliary Space : O(1)