# 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++

`// 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;` `}` |

## Java

`// 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` |

## Python3

`# 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` |

## C#

`// 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` |

## PHP

`<?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` |

## Javascript

`<script>` `// javascript 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` ` ` `var` `area = (3.14 * Math.pow(r / (2 * Math.sqrt(2)), 2));` ` ` `return` `area;` `}` `// Driver code` `var` `a = 5;` `document.write( area(a).toFixed(6));` `// This code contributed by shikhasingrajput` `</script>` |

**Output:**

9.8125

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