Unable to pass the test case 3 in Module 5

Question

Hey guys.
I'm unable to pass the test case 3
It would be great if you could help me identify where i went wrong.
Below is my code

import java.util.Scanner;

abstract class Shape {
int width;
abstract void area();
}
//your code goes here
class Square extends Shape{
Square(int x)
{
width=x;
}

public void area()
{
System.out.println(width*width);
}
}
class Circle extends Shape{
Circle(int y)
{
width=y;
}
public void area()
{
System.out.println(Math.PI*(width*width));
}
}

public class Program {
public static void main(String[ ] args) {
Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
int y = sc.nextInt();

Square a = new Square(x);
Circle b = new Circle(y);
a.area();
b.area();
}
}

Answer 1

'width' is integer value.
In arithmetic operations, first the operands are converted to higher types first. so int is converted to double

double * int => results double
int * int => result int value only.
And also **rounded value for destination type.

For ex: test with values 10, 11 :

System.out.println(Math.PI* 10*10);
System.out.println(Math.PI* 11*11);

System.out.println(Math.PI* (10*10));
System.out.println(Math.PI* (11*11));

In first, case you get Math.PI*11 result is already a rounded value, then multiplied by 11 ,returns rounded result. Rounded both times.
=> it's like Math.PI*11 then rounded value multiplied by 11.
It may not equal to multiply with 121.

In second case : Math.PI multiplied by 121 then returns rounded value..

Hope it helps to understand..

Answer 2

Thank you @Jayakrishna🇮🇳
I edited the code as u have suggested and it worked. But I didn't understand how it actually worked. when I tried both of them for other sums it gave me the same results.
Can u tell me the difference between
Math.PI * ( width*width) & Math.PI * width * width

Answer 3

Use Math.PI * width * width

Instead of Math.PI * ( width*width) ; // remove braces, there you get a slite difference in praction part.