Finding area of circle, rectangle and triangle is a common basic question in all most programming languages. Here let’s solve the question together.
Manually we can find the area of a rectangle by using equation:
Area = Breadth * Length.
Also we can find the area of a circle by using equation:
Area = 3.14 * (radius)2
Also we can find the area of a triangle by using equation:
Area = sqrt(s*(s-a)*(s-b)*(s-c))
a = length of first side
b = length of second side
c = length of third side
s = (a + b+ c)/2
-where sqrt stands for square root
Class Diagram
Class Name: areacalc |
Data members: arr[MAX]:int count:int |
Methods: area(float) area(int,int) area(float,float,float) |
Program Code
#include<math.h>
#include<iostream.h>
#include<conio.h>
class areacalc
{
private:
public:void area(float r);
void area(int a,int b);
void area(float a,float b,float d);
};
void areacalc::area(float r)
{
cout<<"Area of Circle="<<(3.14*r*r);
}
void areacalc::area(int a,int b)
{
cout<<"\n Area of Rectangle="<<(a*b);
}
void areacalc::area(float a,float b,float c)
{
float s,ar;
s=(a+b+c)/2;
ar= sqrt(s*(s-a)*(s-b)*(s-c));
cout<<"\nArea of Triangle="<<ar;
}
void main()
{
clrscr();
areacalc ac;
int r,a,b,c,d;
cout<<"\n Choose your option";
cout<<"\n...............................\n 1-rectangle\n 2-circle\n 3-triangle";
cout<<"\n Enter your choice :";
cin>>c;
switch(c)
{
case 1:cout<<"\n enter breadth and length:";
cin>>a>>b;
ac.area(a,b);
break;
case 2:cout<<"\n enter the radious of circle:";
cin>>r;
ac.area(r);
break;
case 3:cout<<"\n enter the three side of triangle:";
cin>>a>>b>>d;
ac.area(a,b,d);
break;
default:cout<<"invalid choice";
}
getch();
}
Sample Output
Choose your option
..............................
1-rectangle
2-circle
3-triangle
Enter your choice : 1
Enter Breadth and Length : 10
20
Area of rectangle = 200

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