Write a programme that calculates length and width of rectangle in c plus plus?

Answer 1

Commented code for a win32 console application:

#include //Include stdio.h for console input/output funcs

#include //Include conio.h for getch function (lol)

void main( )

{

int nWidth, nHeight; //Variables to hold width and height of rectangle

while( 1 ) //Infinite loop - can only be exited using "break;" (among others)

{

printf( "Enter rectangle width: " ); //Print text...

scanf( "%d", nWidth ); //Scan for user input

fflush( stdin ); //Flush user input

printf( "Enter rectangle height: " ); //Print text...

scanf( "%d", nHeight ); //Scan for user input

fflush( stdin ); //Flush user input

printf( "Area of rectangle with width of %d and height of %d is %d\n", nWidth, nHeight, nWidth*nHeight ); //Print text...

fflush( stdout ); //Flush output

//Now program will jump back to start of loop

}

}

Answer 2

Assuming the width and height are natural values (whole numbers greater than 0), the vast majority of rectangles will have areas that are greater than their perimeters. The exceptions are finite:

1. where the width is 1 because 1*height is always less than (1+height)*2;

2. where the width is 2 because 2*height is always less than (2+height)*2;

3. where the width is 3 units and the height is no more than 6 units because 3*6 is the same as (3+6)*2, and;

4. when both the width and height are 4.

The following program will show this to be true, testing all rectangles in the range 1x1 though 10x10. Note that duplicates are not tested because 7x8 is the same as 8x7.

#include<iostream>

int main()

{

size_t max = 10;

for (size_t w=1; w<=max; w+=1)

for (size_t h=w; h<=max; h+=1)

if (h*w > (h+w)*2)

std::cout << "A rectangle of " << w << " by " << h << " has an area greater than its perimeter" << std::endl;

}

When width and height are real numbers, however, the range of exceptions is infinite. The following program tests widths in the range 0.01 to 5.00 in increments of 0.01, locating the smallest rectangles for a given width. When width reaches 4.0, the first rectangles that meet the criteria will be squares (such as 4.01x4.01) which are all excluded.

#include<iostream>

int main()

{

double max = 5.00, step = 0.01;

for (double w=step; w<=max; w+=step)

{

bool found = false;

for (double h=w; h<=max && !found; h+=step)

if (found = (h!=w) && (h*w > (h+w)*2))

std::cout << "A rectangle of " << w << " by " << h << " has an area greater than its perimeter" << std::endl;

}

}

Answer 3

#include <iostream>

using namespace std;

int main()

{

double L, W, area;

cout<<"Enter length: ":

cin>>L;

cout<<"Enter width: ":

cin>>W;

area=L * W;

cout<<endl<< "Area of the square is: "<<area;

system("pause");

return 0;

}

Answer 4

#include <iostream>

using namespace std;

int main()

{

double x, y;

cout << "Enter the length: ";

cin >> x;

cout << "Enter the height: ";

cin >> y;

cout << "The perimeter is: " << (x + y) * 2 << endl;

char wait;

cin >> wait;

return 0;

}