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The biggest possible circle that can be inscribed in a rectangle

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  • Last Updated : 18 Sep, 2022
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Given a rectangle of length l & breadth b, we have to find the largest circle that can be inscribed in the rectangle. 
Examples: 
 

Input  : l = 4, b = 8
Output : 12.56

Input  : l = 16 b = 6
Output : 28.26

 

 

From the figure, we can see, the biggest circle that could be inscribed in the rectangle will have radius always equal to the half of the shorter side of the rectangle. So from the figure,
 

radius, r = b/2
Area, A = π * (r^2)

 

C++




// C++ Program to find the biggest circle
// which can be inscribed  within the rectangle
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the area
// of the biggest circle
float circlearea(float l, float b)
{
 
    // the length and breadth cannot be negative
    if (l < 0 || b < 0)
        return -1;
 
    // area of the circle
    if (l < b)
        return 3.14 * pow(l / 2, 2);
    else
        return 3.14 * pow(b / 2, 2);
}
 
// Driver code
int main()
{
    float l = 4, b = 8;
    cout << circlearea(l, b) << endl;
    return 0;
}

Java




// Java Program to find the
// biggest circle which can be
// inscribed within the rectangle
 
class GFG
{
 
// Function to find the area
// of the biggest circle
static float circlearea(float l,
                        float b)
{
 
// the length and breadth
// cannot be negative
if (l < 0 || b < 0)
    return -1;
 
// area of the circle
if (l < b)
    return (float)(3.14 * Math.pow(l / 2, 2));
else
    return (float)(3.14 * Math.pow(b / 2, 2));
}
 
// Driver code
public static void main(String[] args)
{
    float l = 4, b = 8;
    System.out.println(circlearea(l, b));
}
}
 
// This code is contributed
// by ChitraNayal

Python 3




# Python 3 Program to find the
# biggest circle which can be
# inscribed within the rectangle
 
# Function to find the area
# of the biggest circle
def circlearea(l, b):
 
    # the length and breadth
    # cannot be negative
    if (l < 0 or b < 0):
        return -1
 
    # area of the circle
    if (l < b):
        return 3.14 * pow(l // 2, 2)
    else:
        return 3.14 * pow(b // 2, 2)
 
# Driver code
if __name__ == "__main__":
    l = 4
    b = 8
    print(circlearea(l, b))
 
# This code is contributed
# by ChitraNayal

C#




// C# Program to find the
// biggest circle which can be
// inscribed within the rectangle
using System;
 
class GFG
{
 
// Function to find the area
// of the biggest circle
static float circlearea(float l,
                        float b)
{
 
// the length and breadth
// cannot be negative
if (l < 0 || b < 0)
    return -1;
 
// area of the circle
if (l < b)
    return (float)(3.14 * Math.Pow(l / 2, 2));
else
    return (float)(3.14 * Math.Pow(b / 2, 2));
}
 
// Driver code
public static void Main()
{
    float l = 4, b = 8;
    Console.Write(circlearea(l, b));
}
}
 
// This code is contributed
// by ChitraNayal

PHP




<?php
// PHP Program to find the
// biggest circle which can be
// inscribed within the rectangle
 
// Function to find the area
// of the biggest circle
function circlearea($l, $b)
{
 
    // the length and breadth
    // cannot be negative
    if ($l < 0 || $b < 0)
        return -1;
 
    // area of the circle
    if ($l < $b)
        return 3.14 * pow($l / 2, 2);
    else
        return 3.14 * pow($b / 2, 2);
}
 
// Driver code
$l = 4;
$b = 8;
echo circlearea($l, $b)."\n";
 
// This code is contributed
// by ChitraNayal
?>

Javascript




<script>
 
// javascript Program to find the
// biggest circle which can be
// inscribed within the rectangle
 
// Function to find the area
// of the biggest circle
function circlearea(l, b)
{
 
    // the length and breadth
    // cannot be negative
    if (l < 0 || b < 0)
        return -1;
     
    // area of the circle
    if (l < b)
        return (3.14 * Math.pow(l / 2, 2));
    else
        return (3.14 * Math.pow(b / 2, 2));
}
 
// Driver code
 
var l = 4, b = 8;
document.write(circlearea(l, b));
 
// This code is contributed by Amit Katiyar
 
</script>

Output: 

12.56

 

Time complexity: O(1) as it is doing constant operations

Auxiliary Space: O(1)


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