# URI Online Judge Solution 1012 Area – URI 1012 Solution in C, C++, Java, Python and C#

### URI Online Judge Solution 1012 Area – URI 1012 Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1012 Area | Beginner

Problem Name: 1012 Area code
Problem Number : URI – 1012 Area solution
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)

### URI Solution 1012 Area Code in C/ URI 1012 solution in C:

URI 1012 in simple way:
`#include <stdio.h>int main(){ double a, b, c; scanf("%lf %lf %lf", &a, &b, &c);`
` printf("TRIANGULO: %.3lfn", (a * c) / 2); printf("CIRCULO: %.3lfn", c * c * 3.14159); printf("TRAPEZIO: %.3lfn", ((a + b) / 2) * c); printf("QUADRADO: %.3lfn", b * b); printf("RETANGULO: %.3lfn", a * b); return 0;}`

URI 1012 in advanced way (Using while loop):

`#include <stdio.h>#define pi 3.14159int main (){    float a,b,c;    while (scanf("%f %f %f",&a,&b,&c) != EOF)    {         printf ("TRIANGULO: %.3fn",.5*(a*c));        printf ("CIRCULO: %.3fn",pi*(c*c));        printf ("TRAPEZIO: %.3fn",.5*(a+b)*c);        printf ("QUADRADO: %.3fn",b*b);        printf ("RETANGULO: %.3fn",a*b);     }    return 0; }`

### URI Solution 1012 Area Code / URI 1012 Area solution in CPP:

`#include <cstdio>int main(){ double a, b, c; scanf("%lf", &a); scanf("%lf", &b); scanf("%lf", &c); printf("TRIANGULO: %.3lfn", (a * c) / 2); printf("CIRCULO: %.3lfn", c * c * 3.14159); printf("TRAPEZIO: %.3lfn", ((a + b) / 2) * c); printf("QUADRADO: %.3lfn", b * b); printf("RETANGULO: %.3lfn", a * b); return 0;}`

### URI Solution 1012 Area Code / URI 1012 Area solution in Java:

`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: %.3fn", (a * c) / 2);  System.out.printf("CIRCULO: %.3fn", c * c * 3.14159);  System.out.printf("TRAPEZIO: %.3fn", ((a + b) / 2) * c);  System.out.printf("QUADRADO: %.3fn", b * b);  System.out.printf("RETANGULO: %.3fn", a * b); }}`

### URI Solution 1012 Area Code / URI 1012 Area solution in  Python:

`valor = input().split(" ")a, b, c = valorpi = 3.14159triangulo = (float(a) * float(c))/2circulo = pi * (float(c)* float(c))trapezio = float(c) *(float(a) + float(b)) / 2quadrado = float(b) * float(b)retangulo = float(a) * float(b)print("TRIANGULO: %0.3fnCIRCULO: %0.3fnTRAPEZIO: %0.3fnQUADRADO: %0.3fnRETANGULO: %0.3f" % (triangulo, circulo, trapezio, quadrado, retangulo))`

### URI Solution 1012 Area Code / URI 1012 Area solution in  C# (C Sharp):

Demonstration:

By the rules of all of the rectangles, just do this..
Just implement this in coding. Since having any problem just put a comment below. Thanks

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#### By Maniruzzaman Akash

Maniruzzaman Akash is a freelance web developer with most popular Laravel PHP frameork and Vue JS