## Algorithm

Problem Name: beecrowd | 1012

# Area

Adapted by Neilor Tonin, URI Brazil

Timelimit: 1

Make a program that reads three floating point values: A, B and C. Then, calculate and show:
a) the area of the rectangled triangle that has base A and height C.
b) the area of the radius's circle C. (pi = 3.14159)
c) the area of the trapezium which has A and B by base, and C by height.
d) the area of ​​the square that has side B.
e) the area of the rectangle that has sides A and B.

## Input

The input file contains three double values with one digit after the decimal point.

## Output

The output file must contain 5 lines of data. Each line corresponds to one of the areas described above, always with a corresponding message (in Portuguese) and one space between the two points and the value. The value calculated must be presented with 3 digits after the decimal point.

 Input Samples Output Samples 3.0 4.0 5.2 TRIANGULO: 7.800 CIRCULO: 84.949 TRAPEZIO: 18.200 QUADRADO: 16.000 RETANGULO: 12.000

 12.7 10.4 15.2 TRIANGULO: 96.520 CIRCULO: 725.833 TRAPEZIO: 175.560 QUADRADO: 108.160 RETANGULO: 132.080

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````
#include <stdio.h>

int main()
{
double a, b, c;

scanf("%lf %lf %lf", &a, &b, &c);

printf("TRIANGULO: %.3lf\n", (a * c) / 2);
printf("CIRCULO: %.3lf\n", c * c * 3.14159);
printf("TRAPEZIO: %.3lf\n", ((a + b) / 2) * c);
printf("RETANGULO: %.3lf\n", a * b);

return 0;
}
``````
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Input

cmd
3.0 4.0 5.2

Output

cmd

### #2 Code Example with C Programming

```Code - C Programming```

``````
#include <stdio.h>
#define pi 3.14159
int main ()
{
float a,b,c;
while (scanf("%f %f %f",&a,&b,&c) != EOF)
{

printf ("TRIANGULO: %.3f\n",.5*(a*c));
printf ("CIRCULO: %.3f\n",pi*(c*c));
printf ("TRAPEZIO: %.3f\n",.5*(a+b)*c);
printf ("RETANGULO: %.3f\n",a*b);

}
return 0;

}
``````
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Input

cmd
3.0 4.0 5.2

Output

cmd

### #3 Code Example with C++ Programming

```Code - C++ Programming```

``````

#include <cstdio>

int main()
{
double a, b, c;

scanf("%lf", &a);
scanf("%lf", &b);
scanf("%lf", &c);

printf("TRIANGULO: %.3lf\n", (a * c) / 2);
printf("CIRCULO: %.3lf\n", c * c * 3.14159);
printf("TRAPEZIO: %.3lf\n", ((a + b) / 2) * c);
printf("RETANGULO: %.3lf\n", a * b);

return 0;
}

``````
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Input

cmd
12.7 10.4 15.2

Output

cmd

### #4 Code Example with Java Programming

```Code - Java Programming```

``````
import java.util.Scanner;

public class Main {

public static void main(String[] args) {
double a, b, c;

Scanner sc = new Scanner(System.in);
a = sc.nextDouble();
b = sc.nextDouble();
c = sc.nextDouble();

System.out.printf("TRIANGULO: %.3f\n", (a * c) / 2);
System.out.printf("CIRCULO: %.3f\n", c * c * 3.14159);
System.out.printf("TRAPEZIO: %.3f\n", ((a + b) / 2) * c);
System.out.printf("RETANGULO: %.3f\n", a * b);

}

}
``````
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Input

cmd
12.7 10.4 15.2

Output

cmd

### #5 Code Example with Python Programming

```Code - Python Programming```

``````
valor = input().split(" ")

a, b, c = valor
pi = 3.14159

triangulo = (float(a) * float(c))/2
circulo = pi * (float(c)* float(c))
trapezio = float(c) *(float(a) + float(b)) / 2
retangulo = float(a) * float(b)

``````
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Input

cmd
3.0 4.0 5.2

Output

cmd