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In this mathcast we will be constructing a square using the freeware GeoGebra. The first
step is to open GeoGebra. Here we have GeoGebra open. And then we want to see the Algebra
View – that is, this part over here. Free Objects and Dependent Objects. We always have
the Drawing Pad open and for these exercises we don’t want to see the Spreadsheet view
so we go to View and see that the Algebra view is selected and we see that Spreadsheet
view is selected. We don’t want it, so we deselect it. So now we have just the Algebra
view and the Drawing pad. The second thing we want to do is to turn on the grid. So we
move our mouse pointer over onto the Drawing pad in a free place, right-click and select
grid. And we are going to be working only in the first quadrant so we go up to the Move
drawing pad button and click it and we click again in an empty place in the first quadrant
and click and drag (0,0) towards the corner. We want to have points along here and points
along here show up so don’t drag it way down in the corner. This is a pretty good
place for (0,0). And then you can use your mouse scroll button to make it bigger or smaller.
Notice that it changed the grid. You can force the grid to be 1. That is in a different mathcast.
So here we are with the grid at 1 in both directions and we have a good place to draw
our square. Now, generally speaking, the options for Point capturing Automatic gives you enough
control so that if you are close to a grid point, it will go to a grid point. If you
want it to force it, then you would click On (grid) here. We are not going to do that.
And the other thing is Labeling Automatic is a pretty good choice. Sometimes you only
want the New points. Then you do this. Now the last thing is to get this every time you
open it is to go to Options and you say Save settings. Now the next time we open GeoGebra
this will be our setup. So we have setup GeoGebra. Step 2 coming up. We are going to construct
squares that have as their bottom left point (0,0) and whose sides are horizontal or vertical
line segments, that is they are parallel to the axes. So let’s start by clicking on
the New point tool and making a point down here at (0,0). Notice that point is black.
That means it is fixed. We can’t move it because we hit it as the intersection point
of the two axes. No problem, that is where we wanted it. Look at its coordinates. The
coordinates over here in the Algebra window show (0,0) and that is exactly where it is.
Now the first square we are going to construct will have sides 3. Let’s make the bottom
right hand point. We go over 3. 1, 2, 3. So here is where we are going to put our point
and we have moved horizontally 3 and vertically 0. Horizontal before vertical. H before V
so the point is (3,0). Let’s check. Click to make our point (3,0). And now we want to
make a point 3 up. 1, 2, 3. So here is where we are going to put our point. From (0,0)
we have moved 3 horizontally to the right and 3 vertically up so the coordinates should
be (as it says there) (3,3). So let’s click. Let’s see. C is (3,3). The last point is
to move left 3. 1, 2, 3. So this is where we are going to put our point. Where is that
point with respect to (0,0)? Well this point is nothing horizontally and up 3 vertically.
Horizontal before vertical so 0 horizontal, 3 vertical. So(0,3) is the point D. There
we have our 4 points of our square. And it looks square. So let’s see if it is square.
Click on the Polygon tool. Click on the first vertex A, second corner point B, third vertex,
corner point here and then back to the first corner point again. And we have a polygon
here and it definitely looks like a perfect square. So let’s move these labels out of
here so we can check some other things. Put B down here. Put the sides inside and now let’s look at some things. The little
letters here belong to the line segments. They are linear so they have linear units.
How long is “a”? “a” is 3 long. If I click on this, it says it is the line segment
between A and B. So its length is 3, but its definition is the segment joining A and B.
The same thing for “b”. It is the line segment joining B and C. Its length is 3.
These are all of the line segments on the outside of the square. If we add them up,
what do we get? The perimeter. 3, 6, 9, 12. So the perimeter of this square is 12. On
the other hand, the area of the square is – it’s a 3 by 3 – so it’s 9. We can
count the squares: 1, 2, 3, 4, 5, 6, 7, 8, 9 square units. So 9 square units. So the
segments because they are lines show the length and the polygon because it is a shape shows
the area. Now what else can we do? Let’s see if we can move our points and make a square
that is 4 by 4. So we still have the Move tool. Where would we need to put B to make
a square that is 4 by 4? We would have to move it 1 to the right. So we are not increasing
the vertical coordinate. Just the horizontal so it should be (4,0). Now C has to go both
1 to the right and 1 up. So both the horizontal and the vertical coordinate will increase
by 1. We should get (4,4). Let’s see. (4,4). And then to make it a square we need to move
D up. So the vertical coordinate is moving to (0,4). The line segments are now all 4.
The perimeter is 16 linear units and the polygon has an area of 16 square units. Let’s check
to make sure all of these angles inside – we know they should be right angles. Let’s
check to make sure that they are. So we go to the Angle tool, click on it and then anywhere
in the square and then all of the inside angles should be marked. Click. And we have alpha
equals 90, beta equals 90, gamma equals 90 and delta equals 90. So we can see that it
really is a square. Not only are all 4 of the sides equal, but all 4 of the angles are
90 degrees. Now we certainly can have some fun while we are here and change the shape
of this quadrilateral (four sided figure) which is currently a square. We cannot move
A because it is fixed at (0,0). But we can move B along the x-axis. Let’s move it to
5 let’s say. We can move C anywhere we want to. Let’s say we move it there. And here
we have an odd shaped quadrilateral – not even something we know what it is. Just a
quadrilateral, right? We can check and add up the angles and find out that the sum is
still 360 degrees. We can find the perimeter by the lengths of the line segments and we
can find the area by looking at “poly1” and by looking at the squares inside. So we
can have some fun with that. Of course you can check this area by noticing that this
triangle has a height of 1 and a base of 3. This triangle here has a height of 3 and a
base of 2. And what is left is a square here with 3 by 3. So we have 3 by 3 is 9 and then
3 by 1 by 1/2 which would be 1.5 so we have 10.5 and then we have 2 by 3 by 1/2 which
would be 3 so altogether 13.5. Lots of fun to be had here. Smile.