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Math Help - Quadrilaterals

  1. #1
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    Quadrilaterals

    I'm pretty stuck on some of these quadrilaterals... Imageshack - 38928686.jpg

    The first one is a parallelogram, the only angle I can find is angle b which is 80, how would find angle a and c?


    The second one is also a parallelogram, but I don't know what to do here. I don't know how to find y or x in this problem.


    The last one is a kite. FH = 16 in, FH & EG intersects at L, and EG = 20 in, angle FEH = 60degrees.

    The main thing I don't get here is find the measure of the lines what does line EF =? And what does FG =? Also how did you find this? One last thing would angle EFG = 60?

    Thanks for all your help.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by Jubbly View Post
    I'm pretty stuck on some of these quadrilaterals... Imageshack - 38928686.jpg

    The first one is a parallelogram, the only angle I can find is angle b which is 80, how would find angle a and c?


    The second one is also a parallelogram, but I don't know what to do here. I don't know how to find y or x in this problem.


    The last one is a kite. FH = 16 in, FH & EG intersects at L, and EG = 20 in, angle FEH = 60degrees.

    The main thing I don't get here is find the measure of the lines what does line EF =? And what does FG =? Also how did you find this? One last thing would angle EFG = 60?

    Thanks for all your help.
    Hi jubbly,

    In the first one, it seems to me that angle c is part of a linear pair with 80 degrees, so it must be supplementary to 80 degrees.

    And angle a is an alternate interior angle to the 47 degree angle and we know that the alternate interior angles are congruent.

    On the second parallelogram, remember that the diagonals bisect each other.

    3y + 13 = 2y + 37

    Solve the above one for y, and substitute it into this one.

    3y = 2x - 4

    In the kite, triangle FEH is isosceles and since the vertex angle FEH = 60, it is also equilateral. This means FH = EF = EH = 16.

    Also, FL = HL = 8 because EG bisects FH.

    Using the 30-60-90 rule, you can easily find EL = 8\sqrt{3}. Subtract that from 20 to get LG. Then use the pythagorean theorem to find FG.

    I think I've answered your last question, but angle EFG does not equal 60 degrees. Angel EFH = 60 degrees. Using trig, it's somewhere around 97.5 degrees, but that's not important here.
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