Python TypeError: 'int' object is not callable

To start the entry must be an integer (int) not a string (str), by default python handles only the 'input' as 'str' or string. Now having the equation s = (a + b + c) / 2 (you must bear in mind that this is a float)

math.sqrt(s(s-a)*(s-b)*(s-c))

Here you must remember that all the variables must be a number either float or int.

s(s-a)

Here is the error, because as I mention everything is an integer, and here it is as if you handled it as a function or class, it marks you that it is not 'suscriptable' since the integers are not handled in this way, if what you wanted was to multiply you should write it like that.

s*(s-a)

Here I send you the modified code.

# -*- coding: cp1252 -*-
import math
a = int(input("Ingrese a "))
b = int(input("Ingrese a "))
c = int(input("Ingrese a "))
s = (a+b+c)/2
print(s)
if a+b >= c:
    raiz = math.sqrt(s*(s-a)*(s-b)*(s-c))
    print (raiz)
else:
    print ("El triángulo no es posible")

replace:

s(s-a)*(s-b)*(s-c)

with:

s*(s-a)*(s-b)*(s-c)

you are having to identify the input value as an integer, try as follows:

import math
a = int(input("Ingrese a "))
b = int(input("Ingrese a "))
c = int(input("Ingrese a "))
s = (a+b+c)/2
if a+b >= c:
    raiz = math.sqrt(s(s-a)*(s-b)*(s-c))
    print raiz
else:
    print "El triángulo no es posible"

So that in this way you indicate the type of Python data that you will enter

The cause of the error is already more than clear, the operator * lack in the Heron formula in s(s-a)... when it should be s*(s-a)... . When doing s() you are making a call attempt, when s is an integer you get the commented error. However the code shown is for Python 2 and this implies having two more things in mind:

• You should never use input in Python 2 for user entries lightly. input evaluates all valid Python code that is entered, which means a wide-open door for code injection attacks. Maybe in cases like this it seems not to have much importance, but it is still a very bad practice. Instead, in Python 2.x you should use raw_input (which returns a string similar to what input does in Python 3) and make an explicit cast.

• The division operator ( / ) works very differently in Python 2 and in Python 3. In Python 3, the result of the actual division (5/2 => 2.5) always returns. In Python 2 the return depends on the type of the operands, for example 5/2 results in 2 (whole division) because both operands are integers, whereas 5 / 2.0 returns us 2.5 . If you cast a int of your entries and enter a = 3, b= 3, c = 5 you would have a problem since the semiperimeter will be 5 and the area will be 0.0, which is far from the 4.1457 ...

On the other hand, according to the triangular inequality theorem, a triangle is valid if the sum of any two sides is always greater than the remaining side. Just check one of the three possibilities in the if , the triangle a = 3, b = 12, c = 3 meets a + b >= c but it is an invalid triangle.

With all this, if you want to be able to enter decimal values for the measures of the sides you can do:

# -*- coding: cp1252 -*-

import math

a = float(raw_input("Ingrese a: "))
b = float(raw_input("Ingrese b: "))
c = float(raw_input("Ingrese c: "))


if ((a + b) > c) and ((a + c) > b) and ((b + c) > a):
    s = (a + b + c) / 2
    raiz = math.sqrt(s*(s-a)*(s-b)*(s-c))
    print raiz
else:
    print "El triángulo no es posible"

If you only want to accept integers then do:

# -*- coding: cp1252 -*-

import math

a = int(raw_input("Ingrese a: "))
b = int(raw_input("Ingrese b: "))
c = int(raw_input("Ingrese c: "))


if ((a + b) > c) and ((a + c) > b) and ((b + c) > a):
    s = (a + b + c) / 2.0  # Observa el 2.0 en vez de 2
    raiz = math.sqrt(s*(s-a)*(s-b)*(s-c))
    print raiz
else:
    print "El triángulo no es posible"