Q_4.java

// Muhammad Naveed

// (Diameter, Circumference and Area of a Circle) Here’s a peek ahead.
// In this chapter, you learned about integers and the type int.
// Java can also represent floating-point numbers that contain decimal
// points, such as 3.14159. Write an application that inputs from the
// user the radius of a circle as an integer and prints the circle’s
// diameter, circumference and area using the floating-point value
// 3.14159 for π. Use the techniques shown in Fig. 2.7.
// [Note: You may also use the predefined constant Math.PI for the value of π.
// This constant is more precise than the value 3.14159. Class Math is defined in
// package java.lang. Classes in that package are imported automatically, so you
// do not need to import class Math to use it.] Use the following formulas (r is the radius):
// diameter = 2r
// circumference = 2πr
// area = πr2
//Do not store the results of each calculation in a variable. Rather,
// specify each calculation as the value that will be output in a System.out.printf
// statement. The values produced by the circumference and area calculations are floating-point
// numbers. Such values can be output with the format specifier %f in a System.out.printf
// statement.


import java.util.Scanner;

public class Q_4
{
    public static void main(String[] args)
    {
        // Creating the scanner object.
        Scanner sc = new Scanner(System.in);

        // inputs from the user the radius of a circle as an integer. (given in question)
        System.out.print("Enter the radius of the circle : ");
        int radius = sc.nextInt();

        // closing the scanner object to avoid potential memory leakage.
        sc.close();

        // printing the diameter, circumference and area of the circle.
        System.out.printf("The diameter of the circle having radius %s is %s \n", radius, (2 * radius));
        System.out.printf("The circumference of the circle having radius %s is %f \n", radius, (2 * Math.PI * radius));
        System.out.printf("The area of the circle having radius %s is %f \n", radius, (Math.PI * Math.sqrt(radius)));


    }
}