Area of a circle inscribed in a rectangle which is inscribed in a semicircle

Given a semicircle with radius R, which inscribes a rectangle of length L and breadth B, which in turn inscribes a circle of radius r. The task is to find the area of the circle with radius r.

Examples:

Input : R = 2
Output : 1.57

Input : R = 5
Output : 9.8125

Approach:

We know the biggest rectangle that can be inscribed within the semicircle has, length, l=√2R/2 &
breadth, b=R/√2(Please refer)
Also, the biggest circle that can be inscribed within the rectangle has radius, r=b/2=R/2√2(Please refer)
So area of the circle, A=π*r^2=π(R/2√2)^2

C++

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// C++ Program to find the area of the circle
// inscribed within the rectangle which in turn
// is inscribed in a semicircle
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the area of the circle
float area(float r)
{
  
    // radius cannot be negative
    if (r < 0)
        return -1;
  
    // area of the circle
    float area = 3.14 * pow(r / (2 * sqrt(2)), 2);
    return area;
}
  
// Driver code
int main()
{
    float a = 5;
    cout << area(a) << endl;
    return 0;
}

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Java

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// Java Program to find the area of the circle
// inscribed within the rectangle which in turn
// is inscribed in a semicircle
  
import java.io.*;
  
class GFG {
  
  
// Function to find the area of the circle
static float area(float r)
{
  
    // radius cannot be negative
    if (r < 0)
        return -1;
  
    // area of the circle
    float area = (float)(3.14 * Math.pow(r / (2 * Math.sqrt(2)), 2));
    return area;
}
  
// Driver code
  
    public static void main (String[] args) {
            float a = 5;
    System.out.println( area(a));
    }
}
  
 // This code is contributed by ajit

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Python3

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# Python 3 Program to find the 
# area of the circle inscribed 
# within the rectangle which in
# turn is inscribed in a semicircle
from math import pow, sqrt
  
# Function to find the area 
# of the circle
def area(r):
      
    # radius cannot be negative
    if (r < 0):
        return -1
  
    # area of the circle
    area = 3.14 * pow(r / (2 * sqrt(2)), 2);
      
    return area;
  
# Driver code
if __name__ == '__main__':
    a = 5
    print("{0:.6}".format(area(a)))
  
# This code is contributed By 
# Surendra_Gangwar

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C#

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// C# Program to find the area of 
// the circle inscribed within the 
// rectangle which in turn is 
// inscribed in a semicircle 
using System; 
  
class GFG 
  
// Function to find the area
// of the circle 
static float area(float r) 
  
    // radius cannot be negative 
    if (r < 0) 
        return -1; 
  
    // area of the circle 
    float area = (float)(3.14 * Math.Pow(r / 
                        (2 * Math.Sqrt(2)), 2)); 
    return area; 
  
// Driver code 
static public void Main (String []args)
    float a = 5; 
    Console.WriteLine(area(a)); 
  
// This code is contributed
// by Arnab Kundu

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PHP

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<?php
// PHP Program to find the area 
// of the circle inscribed within 
// the rectangle which in turn
// is inscribed in a semicircle
  
// Function to find the area
// of the circle
function area($r)
{
    // radius cannot be negative
    if ($r < 0)
        return -1;
  
    // area of the circle
    $area = 3.14 * pow($r
              (2 * sqrt(2)), 2);
    return $area;
}
  
// Driver code
$a = 5;
echo area($a);
  
// This code is contributed by mits

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Output:

9.8125


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