URI-1012 Solve

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

Problem link: https://www.urionlinejudge.com.br/judge/en/problems/view/1012

Solution:

• [tab]
• C

• #include<stdio.h>
  int main(void){
      double a,b,c,triangle,circle,trapezium,square,rectangle,PI=3.14159;
      scanf("%lf%lf%lf",&a,&b,&c);
      triangle=0.5*a*c;
      circle=PI*c*c;
      trapezium=(a+b)/2.0*c;
      square=b*b;
      rectangle=a*b;
      printf("TRIANGULO: %.3lf\n",triangle);
      printf("CIRCULO: %.3lf\n",circle);
      printf("TRAPEZIO: %.3lf\n",trapezium);
      printf("QUADRADO: %.3lf\n",square);
      printf("RETANGULO: %.3lf\n",rectangle);
      return 0;
  }

• C++

• #include <cstdio>
  
  int main()
  {
   double a;
   double b;
   double 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;
  }