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Review over Segment Addition

So, to recap over what we’ve done so far, let’s go over segment addition.

Segment addition states the following:

Let A, B, and C be collinear points.

If B is between A and C, then

+ BC

= AC

So, in other words, if we have two segments that are connected, and we add their lengths together, we’ll get the bigger segment.

Makes sense right?

Well, let’s give it a try.

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Example of Segment Addition

So, let’s say you are given something like:�

Well, let’s imagine they ask you to find x.�How exactly do you set this up?

�

Well, if we look at the segment addition postulate, we know that

ET + TK + KB = EB�And also, from what they gave us, we can see that

EB = 20�So, now this just becomes an algebra 1 problem!�First, let’s compile what we know.

We know that ET = x

TK = 4

KB = 9 and

EB = 20

So, now we just substitute in the numbers instead of the segments, and we get:�x + 4 + 9 = 20

x + 13 = 20

-13 -13�x = 7

�

So, since x was the measure of ET, we know that ET = 7.

�

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So, the Perimeter is basically that

So, when it comes to the perimeter of a polygon, essentially that’s what we’re working with is just the addition of all of the segments that make up that polygon.

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How to find the Perimeter of a polygon

So, here is how we determine the perimeter of a polygon.

Let’s say we have a polygon, like such:

And, let’s say they want to know the perimeter of the shape.

Well, all we need to do is simply add up all of the sides and we’ve got it.

So, the perimeter of this polygon is:

2.67 + 4.74 + 5.14 + 3.31 + 3.19 = 19.05

So, the perimeter of this shape is 19.05

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WELL, AT LEAST THAT’S THE EASIEST VERSION

Usually what will happen is they will give you a problem asking for the perimeter of a polygon, and to find it you may need to do some algebra.

For example, let’s say we have a polygon:

And they tell us that the perimeter of this polygon is: 6x - 78

Find the length of each side of the polygon

Well, in this case, all we would need to do is still set up what we know.

We know that the perimeter is going to be:

(x) + (x + 3) + (x + 4) + (x + 6) + (x + 15) = 6x - 78

Now it’s an algebra problem! So we combine like terms and get:

5x + 28 = 6x – 78

-5x + 78 -5x + 78

And we get:

x = 106

However, now we need to make sure plug in our x because they wanted the length of the sides.

And of course, when we add those altogether (or plug in 106 for x) we get the perimeter is:

558

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WHAT IF THEY DON’T GIVE US THE NUMBERS?

So what happens when they don’t give us the lengths of each side, but instead give us just some points?

Well, this is when we need to use the distance formula!

Remember, the distance formula is:

So, by using the distance formula

We figure out the distance of each leg

And then we find the perimeter

Like:

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EXAMPLE 1:

Find the perimeter of the following points:

A (-4, 4), B (1, 4 ), C (4, 0), D (0, -3), E (-4, -3)

So, to find the perimeter of this polygon

We need to find the distance of each point from its other point.

So, we go in order:

Wow, that was a lot of work.

But now we know the distances

So we can add them together!

So:

So the perimeter of this polygon is 26

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FINDING THE AREA OF DIFFERENT POLYGONS

So now that we know how to find the perimeter of any polygon.

Let’s talk about finding the area.

So, there are a few different formulas for area, and these all depend on what kind of polygon you are dealing with

So, let’s start with the one that we know for sure, and work from there.

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The Area of a triangle.

The area of a triangle is:

Or:

But we know this, we’ve dealt with this a lot.

Now, looking at this, we were to have two triangles

And stack them on top of each other, we’d get a rectangle

Which is perfect because the next area formula is:

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The Area of a rectangle, and parallelogram

The area of a rectangle is:

Or:

But again, we know this because we’ve dealt with it.

But, another type of quadrilateral we haven’t dealt with is like this

But askew

And the area of a parallelogram is also:

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The Area of a kite or rhombus

The area of a kite is:

Or:

So this is a little weird, but it’s how we find the area.

Mainly because the polygon is a quadrilateral, but again

It’s askew

And finally the last quadrilateral we know of is:

And the area of a rhombus is also:

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The Area of a trapezoid

The area of a triangle is:

Or:

So, this is a bit to remember.

But it’s really only remembering 4 different shapes

And their formulas

So, now that we know this, the question that comes to mind is:

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HOW DO WE FIND THE AREA OF SOMETHING THAT ISN’T WHAT WE JUST TALKED ABOUT?

Definitely the biggest title I’ve ever made in a PowerPoint, but the question is true.

How do we find the area of a shape we have never seen before?

For example, what about a pentagon?

Or a hexagon?

Or an Octagon?

How do we find those?

Well, all we do is break it up into pieces, find the area of each piece

Then add them altogether!

It sounds harder to do than it is, so for example:

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EXAMPLE 1

Find the area from the following figure:

Of course, we can’t actually tell what shape this is

But, we can see that it looks like 2 triangles

And a square

And we know how to solve for those!

The square is going to be base times height.

So the area is 16.

Now let’s do the triangles.

They look like they are congruent

So, let’s see what the base would be:

So the base is 2

And the height is 4!

So, ½ 2 * 4 = 4

But since there are 2 of them

Then we have

8 + 16 = 24

So the area of this shape is 24!

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EXAMPLE 2

Find the area from the following figure:

Of course, we can’t actually tell what shape this is

But, we can see that it looks like a square and a rhombus

And we know how to solve for those!

The square is going to be base times height.

So the area is 16.

Now let’s do the rhombus.

The area of a rhombus is d1 * d2

And since the diagonals are 3 and 2, then

½ * 2 * 3 = 3

So the area of the rhombus is 3

3 + 16 = 19

So the area of this shape is 19!