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Math Help - Solve the side of these shapes - Please Help -

  1. #1
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    Solve the side of these shapes - Please Help -

    Hi, I'm taking an online Math course and im really stuck on these couple of questions. My book doesn't even come close to telling me how to solve these. I'f you could please tell me how to solve these than that would be great!!!!
    Any help is appreciated.
    Thank you very much.
    Solve the side of these shapes - Please Help --q1.jpgSolve the side of these shapes - Please Help --q3.jpgSolve the side of these shapes - Please Help --q2.jpgSolve the side of these shapes - Please Help --q4.jpg






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  2. #2
    A riddle wrapped in an enigma
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    Solve the side of these shapes - Please Help --1.jpg
    Quote Originally Posted by chriss2525 View Post
    [FONT=Arial][SIZE=3]Hi, I'm taking an online Math course and im really stuck on these couple of questions. My book doesn't even come close to telling me how to solve these. I'f you could please tell me how to solve these than that would be great!!!!
    Any help is appreciated.
    Thank you very much.


    Hi chriss2525,

    I'll attack the last one.

    For this one, connect the 3 centers of the holes and you form an equilateral triangle.
    Next, draw a radius from the center of the big circle to a vertex.
    Now draw a perpendicular to the base of the triangle.

    What you have formed now is a 30-60-90 triangle.

    You can find the hypotenuse by taking half the diameter of the big circle.
    That would be 162.5.

    Using your 30-60-90 rules (or trigonometry, if you prefer), you can determine the apothem of the equilateral triangle.
    It's half the hypotenuse or 81.25

    Finally, the base of the right triangle that you formed can be determined by a number of different ways:

    Pythagorean Theorem
    Trigonometry
    30-60-90 rules

    You should find that length to be 81.25\sqrt{3}

    Multiply this answer by 2 and you have your desired result.Click image for larger version. 

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  3. #3
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    wow, Thats amazing! i would have never have thought of that. Thanks a lot for answering that last question! Now i just need to se if anyone else will answer my other ones.
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  4. #4
    Senior Member eumyang's Avatar
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    Here's the 2nd to last one: I edited the diagram and made a right triangle by adding a red line segment, which you can see here:
    Solve the side of these shapes - Please Help --trig-shape.jpg

    The length of the green line segment can be found be subtracting 0.400 from 0.930. Then, use the trig definition
    \tan\,\theta = \frac{opp}{adj}
    to find the opposite side. (The red is the opposite side, and the green is the adjacent side.) Once you find the opposite side, subtract that from 2.188 to get y.
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  5. #5
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    For the first one, try redrawing the image, and extend out the slanted edges so that they meet.

    Then draw another horizontal from the top of y and another horiztonal from the bottom of y.

    You should then have two triangles you can solve from angles you can find from "angles in parallel lines" properties.
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  6. #6
    A riddle wrapped in an enigma
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    Quote Originally Posted by chriss2525 View Post
    Hi, I'm taking an online Math course and im really stuck on these couple of questions. My book doesn't even come close to telling me how to solve these. I'f you could please tell me how to solve these than that would be great!!!!
    Any help is appreciated.
    Thank you very much.
    Click image for larger version. 

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Views:	4 
Size:	15.8 KB 
ID:	18343




    Hi chriss2525,

    If you're still working on these, here's #2:

    Solve the side of these shapes - Please Help --1.png

    I drew 2 red vertical lines to the base.

    Two right triangles were formed with 45 deg. 50 min angles in them.

    I subtracted .250 from .785 to get the height of the right triangles which is .535.

    I needed to find the bases of these triangles, so I used a little trigonometry.

    \tan (42\frac{5}{6})=\frac{.535}{b}

    Once I found the bases, I subtracted them from 2.125.

    This is the value of x.

    x=2.125-.577-.577

    x=.971

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  7. #7
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    Yes i am still working on them. Thats great Masters Thanks a lot for you're help it'll for sure help me outon my final exam in 2 weeks!
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