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Math Help - few geometry problems

  1. #1
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    few geometry problems

    1. given parallelogram ABCD and AB is 6, the bisector of angle A bisects BC, find BC

    2. Given triangle ABC with a perimeter of 25, what is the perimeter of the smaller triangle made by the lines of the midpoints

    3. how many combinations can be made if 8 people sat together on one table and three of them always stayed together

    4. given a bag of 5 red marbles and 3 yellow marbles, you pick one marble out and throw it back inthe bag and pick out another marble. what is the possiblity that both marbles will be the same color. what happens to the possiblity if the marble you took out the first time was not put back into the bag.

    Thanks
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  2. #2
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    Quote Originally Posted by anime_mania View Post
    1. given parallelogram ABCD and AB is 6, the bisector of angle A bisects BC, find BC
    Try to visualize me.

    1)Draw parallelogram ABCD.

    2)Draw bisector of A and let it intersect at E.

    3)Now <BAE = <EAD (because it is a bisector).

    4)But <EAD = <BEA (parallel postulate, in simpler terms, alternate interiror angles).

    5)Thus, <BAE = < BEA

    6)Thus, triangle ABE is isoseles with AB = BE.

    7)But AB=6

    8)Thus BE=6

    9)Thus, BC=2(BE)=2(6)=12

    2. Given triangle ABC with a perimeter of 25, what is the perimeter of the smaller triangle made by the lines of the midpoints
    Half of that.

    Because each length of side is half the corresponding one to the larger triangle.
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  3. #3
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    Hello, anime_mania!

    4. Given a bag of 5 red marbles and 3 yellow marbles.
    You pick one marble out and throw it back in the bag and pick out another marble.
    (a) What is the probability that both marbles will be the same color?
    (b) What happens to the probability if the marble you took out the first time
    was not put back into the bag?

    (a) With replacement

    P(RR) \:=\:\frac{5}{8}\cdot\frac{5}{8} \:=\:\frac{25}{64}\qquad\qquad P(YY) \:=\:\frac{3}{8}\cdot\frac{3}{8} \:=\:\frac{9}{64}

    Therefore: . P(\text{same color}) \:=\:\frac{25}{64} + \frac{9}{64} \:=\:\frac{34}{64} \:=\:\frac{17}{32}


    (b) Without replacement

    P(RR) \:=\:\frac{5}{8}\cdot\frac{4}{7}\:=\:\frac{20}{56}  \qquad\qquad P(YY) \:=\:\frac{3}{8}\cdot\frac{2}{7}\:=\:\frac{6}{56}

    Therefore: . P(\text{same color}) \:=\:\frac{20}{56} + \frac{6}{56} \:=\:\frac{26}{56} \:=\:\frac{13}{28}

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  4. #4
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    Quote Originally Posted by anime_mania View Post
    3. how many combinations can be made if 8 people sat together on one table and three of them always stayed together
    Treat the 3 people sitting together as a single entity. That is, there are 6 people total (3 grouped into 1).

    But the 3 can sit in any order, so there are 6 ways of seating them.

    Thus the number of ways is (6P_6)/6 x 6 = 6!*6/(6*(6-6)!) = 5!*6 = 720 ways.
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  5. #5
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    Hello, anime_mania!

    2. Given triangle ABC with a perimeter of 25, what is the perimeter
    of the smaller triangle made by the lines of the midpoints?
    Code:
                C
                *
               *  *
              *     *
            D*- - - - *E
            *           *
           *              *
        A * - - - -*- - - - * B
                   F

    We are expected to know that:
    . . the line joining the midpoints of two sides of a triangle
    . . is parallel to and one-half the length of the third side.

    Hence: . DE \,=\,\frac{1}{2}AB
    Similarly: . DF \,=\,\frac{1}{2}BC,\;\;EF\,=\,\frac{1}{2}AC


    Therefore: . DE +DF + EF \:=\:\frac{1}{2}AB + \frac{1}{2}BC + \frac{1}{2}AC

    . . . . . . =\:\frac{1}{2}(AB + BC + AC) \:=\:\frac{1}{2}(25)\:=\:12.5

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  6. #6
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    Hello, anime_mania!

    1. Given parallelogram ABCD and AB = 6..
    The bisector of angle A bisects BC. .Find BC.

    Here's the diagram for ThePerfectHacker's excellent solution.
    Code:
                B                 E               C
                *-----------------*---------------*
               /         θ    *                  /
            6 /           *                     /
             /        *                        /
            / θ   *                           /
           /  *  θ                           /
          *---------------------------------*
          A                                 D
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