Algorithm
Area
Adapted by Neilor Tonin, URI 
Brazil
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 |
|
12.7 10.4 15.2 |
TRIANGULO: 96.520 |
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("QUADRADO: %.3lf\n", b * b);
printf("RETANGULO: %.3lf\n", a * b);
return 0;
}
Input
Output
#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 ("QUADRADO: %.3f\n",b*b);
printf ("RETANGULO: %.3f\n",a*b);
}
return 0;
}
Input
Output
#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("QUADRADO: %.3lf\n", b * b);
printf("RETANGULO: %.3lf\n", a * b);
return 0;
}
Input
Output
#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("QUADRADO: %.3f\n", b * b);
System.out.printf("RETANGULO: %.3f\n", a * b);
}
}
Input
Output
#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
quadrado = float(b) * float(b)
retangulo = float(a) * float(b)
print("TRIANGULO: %0.3f\nCIRCULO: %0.3f\nTRAPEZIO: %0.3f\nQUADRADO: %0.3f\nRETANGULO: %0.3f" % (triangulo, circulo, trapezio, quadrado, retangulo))
Input
Output