Input x 
3
Input y 
4
Input z 
5
result = 1.77800712886037

{0.2} Task 1:

To do: Calculate an average of two variables a and b (formula for calculation: a + b)/2). Values of variables are provided (a=5, b=6). You should do this task twice with different ways of assigning and output.

Expected output:

(5 + 6) / 2 = 5.5

[Program name: L1task00.pas and L1task01.pas]

please enter the side length of a square:
5.6
Perimeter P = 22.4

        begin
        // Variable declaration to store the value of the side length
        var a := ReadReal('please enter the side length of a square:');
        var p := 4 * a; // Perimeter calculation
        Print($'Perimeter P = {p}');
        end.
    

        begin
        PrintLn('please enter a side length of a square:');
        // Variable declaration to store the value of the side length
        var a: real;
        readln(a);
        var p := 4 * a; // Perimeter calculation
        Print('Perimeter P = ', p);
        end.

{0.3} Task 3:

To do: The side of the square (variable name is side) is entered. Calculate an area of the square: S = a². You should use different methods of assigning, input and output.

Note: To calculate square of a number you can use sqr() standard function, for example:

sqrX := sqr(x);

Expected output:

enter a side length of a square:
2.90
Area S = 8.41

[Program name: L1task03.pas]

{0.3} Task 4:

To do: The sides of the rectangle are entered (a and b). Calculate an area of the rectangle (S = a*b) and its perimeter (P = 2*(a + b)).

Note: To specify a particular number of digits after the floating point you can use format expression of writeln function:

writeln('S = ', S:0:2);
// :2 means the number of digits to output after floating point

Expected output:

Enter the values of two sides:
12
13
result:
S = 156.00
P = 50.00

[Program name: L1task04.pas]

{0.4} Task 5:

To do: A diameter of a circle (variable name is d) is entered. Calculate its length (formula L = π·d). The value of π is 3.14. Use different methods of assigning, input and output.

Note 1: π has a constant value. In PascalABC we can declare constant before the begin section of the program:

const
  pi = 3.14;
begin
 // ...
end.

Note 2: Make the program using the same style of coding as in sample 2.

Expected output:

please enter a diameter of a circle:
6.7
the length of a circle is: 21.038

[Program name: L1task05.pas]

sqrtX := sqrt(x);

Input the values of triangle legs:
3.0
6.0
hypotenuse = 6.70820393249937
perimeter = 15.7082039324994

Here is an example of right program which is clear for user:

{0.4} Task 6:

To do: A length of a cube side is entered (a). Calculate a volume of the cube (V = a³) and its surface area (S = 6·a²). Give the program log in the form of a comment.

Note: To specify a particular number of digits after the floating point you can use format expression of writeln function:

writeln('V = ', v:5:3) 5 means total number of signs to output the number,
3 means the number of digits to output after floating point.

Expected output:

enter a cube side length:
9.000
V = 729.000
S = 486.000

[Program name: L1task06.pas]

{0.4} Task 7:

To do: Assign a value to integer variable x (x = 5). Calculate the value of the function:

y = 4(x-3)^6 - 7(x-3)^3 + 2

Note 1: To calculate the power of a number you can use the power(x:real, y:real) function. For example:

//2 in the power of 5 =
powNumb = power (2,5);

Note 2: It is better to use an auxiliary variable for (x-3)^3.

Expected output:

for x = 5 we have y = 202

[Program name: L1task07.pas]

{0.4} Task 8:

To do: Calculate a distance between two points with the given coordinates x₁ and x₂ on the number axis; the coordinates are entered. The formula is |x₂ − x₁|.

Note: To calculate the absolute value of a number you can use abs(x:real) standard function:

abs(x2 - x1);

Expected output:

x1 = 3.2
x2 = 2.5
the distance between two points: 0.7

[Program name: L1task08.pas]

{0.4} Task 9:

To do: Calculate a distance between two points on the plane; coordinates (x₁,y₁) and (x₂,y₂) are entered. The distance is calculated by the formula:

Note 1: Verify the correction of your program by using “simple” values that are easy to calculate. For example:

d((0,  0); (6, 0)) = 6;   
d((0,  -4); (0, 1)) = 5;   
d((-1,  1); (2, 5)) = 5:

Note 2: Display the results of your program (log) in the form of a comment after the program code. It's easy to do by copying and pasting. You should use curly brackets for comments:

Expected output:

enter x1 of the first point:
0
enter y1 of the first point:
0
enter x2 of the second point:
6
enter y2 of the second point:
0
The distance equals 6

[Program name: L1task09.pas]

{0.3} Task 10:

To do: The temperature in Celsius is entered, convert temperature to Fahrenheit. Celsius and Fahrenheit scales are related by the ratio:

(°F = °C × 9/5 + 32)

Expected output:

enter the temperature in Celsius 
56
The temperature in Fahrenheit  132.8

[Program name: L1task10.pas]

{0.2} Task 11:

To do: Swap the values of variables A and B and print out the new values to the console.

Expected output:

Enter A: 5.7
Enter B: 3
Result:
 A = 3, B = 5.7

[Program name: L1task11.pas]

{0.2} Task 12:

To do: The values of variables A, B, C are entered. Swap their values to make A = B, B = C, C = A, and display the results.

Expected output:

A = 3.4
B = 2
C = 1.5
Result:
 A = 1.5, B = 3.4, C = 2

[Program name: L1task12.pas]

c = √(a² + b²),
P = a + b + c.

<< a=4.90
<< b=9.90
results:
c=11.05
P=25.85

<< A=-3.80
<< B=3.40
<< C=0.50
results:
AC=4.30
BC=2.90
AC+BC=7.20

<< A=2.47
<< B=1.41
<< C=9.50
results:
A=9.50
B=2.47
C=1.41

<< A=3.20
results:
A²=10.24  A⁴=104.86  A⁸=1095.12

<< A=1.57
results:
A²=2.46  A³=3.87  A⁵=9.54  A¹⁰=90.99  A¹⁵=867.95

A₁·x + B₁·y = C₁,
A₂·x + B₂·y = C₂

x = (C₁·B₂ − C₂·B₁)/D,       
y = (A₁·C₂ − A₂·C₁)/D,
where D = A₁·B₂ − A₂·B₁.

<< A₁=-3.00  << B₁=-2.00  << C₁=4.00
<< A₂=-1.00  << B₂=-4.00  << C₂=-2.00
results:
x = -2.00  y = 1.00