# Thread: Finding the perimeter for polygons

1. ## Finding the perimeter for polygons

Here I am again. And heeeeeeere's the problem:

Find the perimeter of this polygon:
triangle PQR with vertices P(-6,-3), Q(1, -1), and R(1,-5).

So first I drew a triangle (maybe I should have graphed it, but let's just take it from there), then I marked its points (P,Q,R), and used the distance formula to find the perimeter of each side. First I did $P$ and $Q$, then $Q$ and $R$, then $P$ and $R$ and came up with these answers:

Distance of P&Q=7.28, Distance of Q&R=4, and distance of P&R=7.28

Then I added them together and got 18.56 or 18.6.

Is the answer right? I know I'm not as schooled as some of you, but you have to admit this is a little complicated...

2. Originally Posted by Melancholy
Here I am again. And heeeeeeere's the problem:

Find the perimeter of this polygon:
triangle PQR with vertices P(-6,-3), Q(1, -1), and R(1,-5).

So first I drew a triangle (maybe I should have graphed it, but let's just take it from there), then I marked its points (P,Q,R), and used the distance formula to find the perimeter of each side. First I did $P$ and $Q$, then $Q$ and $R$, then $P$ and $R$ and came up with these answers:

Distance of P&Q=7.28, Distance of Q&R=4, and distance of P&R=7.28

Then I added them together and got 18.56 or 18.6.

Is the answer right? I know I'm not as schooled as some of you, but you have to admit this is a little complicated...
you are correct. however, it is good practice to leave answers in their exact form, unless other wise instructed, so the answer here is $2 \sqrt {53} + 4 = 2 (\sqrt {53} + 2) \approx 18.6$

3. Oh okay! Gracias, Senor...

Another:

Find the perimeter of pentagon ABCDE with vertices A(-6,2), B(-4,7), C(0,4), D(0,0), and E(-4,-3).

I've graphed it, and I now have to use the distance formula for finding the perimeter of each side, correct? That takes such a long time though... is there an easier way for the distance to be found? (I already know what the distance of side CD are [4])... I need a quicker way of finding the perimeter though.... if there is one...

4. Originally Posted by Melancholy
Oh okay! Gracias, Senor...

Another:

Find the perimeter of pentagon ABCDE with vertices A(-6,2), B(-4,7), C(0,4), D(0,0), and E(-4,-3).

I've graphed it, and I now have to use the distance formula for finding the perimeter of each side, correct? That takes such a long time though... is there an easier way for the distance to be found? (I already know what the distance of side CD are [4])... I need a quicker way of finding the perimeter though.... if there is one...
nope, no shortcut really, at least not that i'm aware of. well, there is one thing, we can tell immediately what the length of a vertical or horizontal line is

for example, CD is a vertical line, the x's stay fixed, so the length of the line is determined by the change in y. if the y goes from 0 to 4, obviously the length is 4, no distance formula needed.

for the rest, use the distance formula

5. Hmm... okay, so the perimeter = 24.77?

For each side I got $AB$ = 5.385, $BC$=5, $CD$=4, $DE$= 5, and $AE$= 5.385

6. Originally Posted by Melancholy
Oh okay! Gracias, Senor...
Sorry to change a little the thread... but ¿hablas español?

Comunícame vía MP ^^

Well... as you know just apply the distance formula.

7. Originally Posted by Melancholy
Hmm... okay, so the perimeter = 24.77?

For each side I got $AB$ = 5.385, $BC$=5, $CD$=4, $DE$= 5, and $AE$= 5.385
correct, good job!

Originally Posted by Krizalid
... but ¿hablas español?

Comunícame vía MP ^^
I know what you said here! ok, yeah, so it's kind of obvious, but i still know. how do you get that upside down question mark?

8. ... but ¿hablas español?
Hablo cierto español, pero no soy muy bueno. (I think I said it right... lol)

Thank you all so much for you help... now I have to remember all of these things!

Here's a word problem (having to do with the same thing) that is stumping me:

The city of Springfield is 5 miles west and 3 miles south of Capital City, while Brighton is 1 mile east and 4 miles north of Capital City. Highway 1 runs straight between Brighton and Springfield; highway 4 connects Springfield and Capital City.

Find the length of Highway 1.
How long is Highway 4?

Again, I do not know how to set up the problem... or where to begin.

They gave me a diagram that looks somewhat like this:

... help...

9. Originally Posted by Melancholy
Hablo cierto español, pero no soy muy bueno. (I think I said it right... lol)

Thank you all so much for you help... now I have to remember all of these things!

Here's a word problem (having to do with the same thing) that is stumping me:

The city of Springfield is 5 miles west and 3 miles south of Capital City, while Brighton is 1 mile east and 4 miles north of Capital City. Highway 1 runs straight between Brighton and Springfield; highway 4 connects Springfield and Capital City.

Find the length of Highway 1.
How long is Highway 4?

Again, I do not know how to set up the problem... or where to begin.

They gave me a diagram that looks somewhat like this:

... help...
well, since they want you to apply the distance formulas, you can setup coordinates for the cities, and use the distance formula to find the length of highways connecting them. here's how:

Let the Capital City be located at the origin, that is, at the point C(0,0)

We move 5 units to the left and 3 units down, and lo and behold, there is Springfield at the point S(-5,-3)

We go back to Capital City and then move 1 unit to the right and 4 units up, and we're smack dab in the middle of Brighton at B(1,4)

Thus, the length of Highway 1 is the length of BS (haha, BS)
and the length of Highway 4 is the length of SC

I think you can take it from here

10. Oh! Okay! So then the length of highway 1 = 9.22, and the length of highway 4 = 5.38? You make it seem so easy...

11. Originally Posted by Melancholy
Oh! Okay! So then the length of highway 1 = 9.22, and the length of highway 4 = 5.38?
Highway 4 is 5.83 miles long, not 5.38
You make it seem so easy...
If i can do it, then it is easy. you're just not used to it yet, you will be after you do a couple hundred problems or so... that's how long it took me

12. Highway 4 is 5.83 miles long, not 5.38
Typo, you are correct.
If i can do it, then it is easy. you're just not used to it yet, you will be after you do a couple hundred problems or so... that's how long it took me
But see, they only give me about one of these problems to work and study, not 100. It's nice that we don't have much homework, but I think we need to learn it..... not to mention I probably should know this since I'm in Geometry already. *sigh* It's so stressful! I hate tests... one is coming up shortly... and I don't think I'll do too well.

*cries*

13. Originally Posted by Melancholy
Typo, you are correct.
of course, i am

But see, they only give me about one of these problems to work and study, not 100. It's nice that we don't have much homework, but I think we need to learn it..... not to mention I probably should know this since I'm in Geometry already. *sigh* It's so stressful! I hate tests... one is coming up shortly... and I don't think I'll do too well.

*cries*
well, you don't have to be assigned 100 problems to do 100 problems, you can do them on your own, especially if you're not comfortable with the material. i'm sure your professor said that to you at some point