How to find circle area using πr ² formula in C

πr² is a common formula to us.But can you find a circle area value by C?
if not,try this.It's help you to know how to find/print the value of a circle 
area in C programming language.


Code to print the value:


#include<stdio.h>
#define PI 3.1416

int main()
{
    float circle_area, r;

    printf("Enter the value of radius: \n");
    scanf("%f", &r);

    circle_area = PI *r * r;

    printf("The circle area is: %.2f\n", circle_area);


return 0;

}


OR,


#include<stdio.h>


int main()
{
    float circle_area, r, PI=3.1416;

    printf("Enter the value of radius: \n");
    scanf("%f", &r);

    circle_area = PI * r * r;

    printf("The circle area is: %.2f\n", circle_area);


return 0;

}

The Output is:

Face problems?No problem.You can share your problems by commenting below.or you can leave a massage us.Thank you.

How to make simple calculator by C

We are going to make a simple calculator by C.By this calculator we can do summation,subtraction,multiplication and division.Lets have a look to the program.

#include <stdio.h>
#include <conio.h>

int main()
{
    float a,b,sum,sub,mul,div;
    char ch;

    printf("Enter a number: \n");
    scanf("%f", &a);

    printf("Enter another number: \n");
    scanf("%f", &b);

    sum=a+b;
    sub=a-b;
    mul=a*b;
    div=a/b;

    printf("What do you want?\n");
    printf("a: summation.\t b: subtraction.\nc: multiplication.\t d: division.\n");

    scanf(" %c",&ch);

    if(ch == 'a')
    {

        printf("summation value: %.2f\n",sum);

    }
    else if(ch == 'b')
    {

        printf("subtraction value: %.2f\n",sub);

    }
       else if(ch == 'c')
    {

        printf("multiplication value: %.2f\n",mul);

    }
       else if(ch == 'd')
    {

        printf("division value: %.2f\n",div);

    }
    else
    {

        printf("something went wrong!!\n");
    }

    return 0;
    }


Now Try it by yourself.If you face any problem and you have any question regarding the program you can comment below.
Thank you.