Given a square of side **a**, the task is to find the area of the largest semi-circle that can be drawn inside the square.

**Examples:**

Input:a = 3Output:4.84865Input:a = 4Output:8.61982

**Approach**

The semicircle of maximal area inscribed in the square has its diameter parallel to a diagonal, and its radius **rmax** is given as:

- Since the figure is symmetrical in the diagonal
**BD**,**angle QPB = 45°**.OY = r cos 45 = r/ √2

- Hence
a = AB = r + r/√2 = r(1 + 1/√2)

- Thus
r = a / (1 + 1/√2) = a*√2 / (√2 + 1)

- Rationalizing the denominator, we obtain
r = a*√2*(√2-1)

- Thus
r = a*2 - a √2 = a*(2-√2)

- Therfore,
Area of the required semicircle = pi * r

^{2}/2 = 3.14*(a*(2-√2))^{2}/ 2Below is the implementation of the above approach:

## CPP

`// C++ program to find Area of`

`// semicircle in a square`

`#include <bits/stdc++.h>`

`using`

`namespace`

`std;`

`// Function to find area of semicircle`

`float`

`find_Area(`

`float`

`a)`

`{`

`float`

`R = a * (2.0 -`

`sqrt`

`(2));`

`float`

`area = 3.14 * R * R / 2.0;`

`return`

`area;`

`}`

`// Driver code`

`int`

`main()`

`{`

`// side of a square`

`float`

`a = 4;`

`// Call Function to find`

`// the area of semicircle`

`cout <<`

`" Area of semicircle = "`

`<< find_Area(a);`

`return`

`0;`

`}`

*chevron_right**filter_none*## Java

`// Java program to find Area of`

`// semicircle in a square`

`class`

`GFG {`

`// Function to find area of semicircle`

`static`

`float`

`find_Area(`

`float`

`a)`

`{`

`float`

`R = a * (`

`float`

`)(`

`2.0`

`- Math.sqrt(`

`2`

`));`

`float`

`area = (`

`float`

`)((`

`3.14`

`* R * R) /`

`2.0`

`);`

`return`

`area;`

`}`

`// Driver code`

`public`

`static`

`void`

`main (String[] args)`

`{`

`// side of a square`

`float`

`a =`

`4`

`;`

`// Call Function to find`

`// the area of semicircle`

`System.out.println(`

`" Area of semicircle = "`

`+ find_Area(a));`

`}`

`}`

`// This code is contributed by AnkitRai01`

*chevron_right**filter_none*## Python3

`# Python3 program to find Area of`

`# semicircle in a square`

`from`

`math`

`import`

`sqrt`

`# Function to find area of semicircle`

`def`

`find_Area(a) :`

`R`

`=`

`a`

`*`

`(`

`2.0`

`-`

`sqrt(`

`2`

`));`

`area`

`=`

`3.14`

`*`

`R`

`*`

`R`

`/`

`2.0`

`;`

`return`

`area;`

`# Driver code`

`if`

`__name__`

`=`

`=`

`"__main__"`

`:`

`# side of a square`

`a`

`=`

`4`

`;`

`# Call Function to find`

`# the area of semicircle`

`print`

`(`

`"Area of semicircle ="`

`,find_Area(a));`

`# This code is contributed by AnkitRai01`

*chevron_right**filter_none*## C#

`// C# program to find Area of`

`// semicircle in a square`

`using`

`System;`

`class`

`GFG {`

`// Function to find area of semicircle`

`static`

`float`

`find_Area(`

`float`

`a)`

`{`

`float`

`R = a * (`

`float`

`)(2.0 - Math.Sqrt(2));`

`float`

`area = (`

`float`

`)((3.14 * R * R) / 2.0);`

`return`

`area;`

`}`

`// Driver code`

`public`

`static`

`void`

`Main (`

`string`

`[] args)`

`{`

`// side of a square`

`float`

`a = 4;`

`// Call Function to find`

`// the area of semicircle`

`Console.WriteLine(`

`" Area of semicircle = "`

`+ find_Area(a));`

`}`

`}`

`// This code is contributed by AnkitRai01`

*chevron_right**filter_none***Output:**Area of semicircle = 8.61982

**Reference:**http://www.qbyte.org/puzzles/p153s.htmlAttention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the

**DSA Self Paced Course**at a student-friendly price and become industry ready.## Recommended Posts:

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