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Math Help - Finding the perimeter for polygons

  1. #1
    Junior Member Melancholy's Avatar
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    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...
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Melancholy View Post
    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
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  3. #3
    Junior Member Melancholy's Avatar
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    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...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Melancholy View Post
    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
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  5. #5
    Junior Member Melancholy's Avatar
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    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
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  6. #6
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    Quote Originally Posted by Melancholy View Post
    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.

    If you computed correctly each side, your answer should be right
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Melancholy View Post
    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!


    Quote Originally Posted by Krizalid View Post
    ... 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?
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  8. #8
    Junior Member Melancholy's Avatar
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    ... 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...
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Melancholy View Post
    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
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  10. #10
    Junior Member Melancholy's Avatar
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    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...
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Melancholy View Post
    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
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  12. #12
    Junior Member Melancholy's Avatar
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    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*
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  13. #13
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Melancholy View Post
    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
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