# Small Basic • ### Question

• Hey i'm having trouble with a certain part of the Small Basic Curriculum, i cant figure out how to write a program that calculates the area and circumference of a circle based on its diameter. Can anyone help me understand what to do?

Thanks, ScikoticRage

• Moved by Thursday, May 16, 2013 3:34 AM
Wednesday, May 15, 2013 11:15 PM

• Ok, i will explain this as best i can.

Ask the user for the diameter of the circle. then, to find the circumference, use this:

You could go:

circum = 3.14 * diameter (or what ever you name the variables)

Then,Find the area:

area = 3.14 * (diameter / 2)

(note, i used parentheses to that it wouldn't go like 3.14 * diameter then that answer divided by 2.

code:

```TextWindow.WriteLine("Enter in the diameter of a circle: ")
circum = 3.14 * diameter
area = 3.14 * (diameter / 2)
TextWindow.WriteLine("The area is: " + area)
TextWindow.WriteLine("The circumference is: " + circum)```

Help? If it doesn't, i'll gladly re-make this and explain it. :3

I am a 12 year old learning how to code using small basic.

Thursday, May 16, 2013 4:09 AM
• For the area you should square the diameter/2 as:

diameter*diameter/4

Jan [ WhTurner ] The Netherlands

Thursday, May 16, 2013 8:06 AM
• For the area you do: A = ((d / 2) * (d / 2)) * 3.14

"D" would be the diameter

Its not "area = 3.14 * (diameter / 2)" Its radius squared times pi.

See here: http://www.mathgoodies.com/lessons/vol2/circle_area.html

It is written: "'As surely as I live,' says the Lord, 'every knee will bow before me; every tongue will acknowledge God.'" Romans 14:11

Thursday, May 16, 2013 6:18 PM
• So like this:

```TextWindow.Write("Enter the diameter: ")
Area = ((Dia / 2) * (Dia / 2)) * 3.14
TextWindow.WriteLine("The area is " + Area)```

It is written: "'As surely as I live,' says the Lord, 'every knee will bow before me; every tongue will acknowledge God.'" Romans 14:11

Thursday, May 16, 2013 9:44 PM

### All replies

• Ok, i will explain this as best i can.

Ask the user for the diameter of the circle. then, to find the circumference, use this:

You could go:

circum = 3.14 * diameter (or what ever you name the variables)

Then,Find the area:

area = 3.14 * (diameter / 2)

(note, i used parentheses to that it wouldn't go like 3.14 * diameter then that answer divided by 2.

code:

```TextWindow.WriteLine("Enter in the diameter of a circle: ")
circum = 3.14 * diameter
area = 3.14 * (diameter / 2)
TextWindow.WriteLine("The area is: " + area)
TextWindow.WriteLine("The circumference is: " + circum)```

Help? If it doesn't, i'll gladly re-make this and explain it. :3

I am a 12 year old learning how to code using small basic.

Thursday, May 16, 2013 4:09 AM
• For the area you should square the diameter/2 as:

diameter*diameter/4

Jan [ WhTurner ] The Netherlands

Thursday, May 16, 2013 8:06 AM
• For the area you do: A = ((d / 2) * (d / 2)) * 3.14

"D" would be the diameter

Its not "area = 3.14 * (diameter / 2)" Its radius squared times pi.

See here: http://www.mathgoodies.com/lessons/vol2/circle_area.html

It is written: "'As surely as I live,' says the Lord, 'every knee will bow before me; every tongue will acknowledge God.'" Romans 14:11

Thursday, May 16, 2013 6:18 PM
• oh thanks. i don't really like finding the area's and circumferences...i kinda forgot how to, so yes, thanks!

I am a 12 year old learning how to code using small basic.

Thursday, May 16, 2013 6:21 PM
• So like this:

```TextWindow.Write("Enter the diameter: ")
Area = ((Dia / 2) * (Dia / 2)) * 3.14
TextWindow.WriteLine("The area is " + Area)```

It is written: "'As surely as I live,' says the Lord, 'every knee will bow before me; every tongue will acknowledge God.'" Romans 14:11

Thursday, May 16, 2013 9:44 PM
• This is my conclusion.
Sample of Small Basic Curriculum "The Pi Property" is

```TextWindow.Write("Enter the radius of the circle:")
Area = Math.Pi * Math.Power(Radius, 2)
TextWindow.WriteLine("Area of the Circle is " + Area)```

So base on diameter:

```TextWindow.Write("Enter the diameter of the circle:")
Area = Math.Pi * Math.Power(Radius, 2)
TextWindow.WriteLine("Area of the Circle is " + Area)
Circum = 2 * Math.Pi * Radius
TextWindow.WriteLine("Circumference of the Circle is " + Circum)```

Nonki Takahashi

Friday, May 17, 2013 10:34 AM
• This is my conclusion.
Sample of Small Basic Curriculum "The Pi Property" is

```TextWindow.Write("Enter the radius of the circle:")
Area = Math.Pi * Math.Power(Radius, 2)
TextWindow.WriteLine("Area of the Circle is " + Area)```

So base on diameter:

```TextWindow.Write("Enter the diameter of the circle:")
Area = Math.Pi * Math.Power(Radius, 2)
TextWindow.WriteLine("Area of the Circle is " + Area)
Circum = 2 * Math.Pi * Radius
TextWindow.WriteLine("Circumference of the Circle is " + Circum)```

Nonki Takahashi

Should we make a change to the curriculum to make it easier to understand?

Thanks!

Ed Price (a.k.a User Ed), SQL Server Customer Program Manager (Blog, Small Basic, Wiki Ninjas, Wiki)