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Math Help - Area of triangle within a triangle

  1. #1
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    Area of triangle within a triangle

    Given a triangle ABC. On the sides: Let R lie one third of the way from A to B. Let S lie one third of the way from B to C. Let T lie one third of the way from C to A. Now connect points RST to make a triangle RST within triangle ABC. Prove that the area of triangle RST is equal to one third the area of triangle ABC.

    Any help please?? I've tried law of sines and such to no avail.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Area of triangle within a triangle

    The area A of triangle ABC is:

    A=\frac{1}{6}\left(bc\sin(A)+ac\sin(B)+ab\sin(C) \right)

    The area A_O within ABC but outside of RST is:

    A_O=\frac{1}{9}\left(bc\sin(A)+ac\sin(B)+ab\sin(C) \right)

    Hence the area within RST is:

    A-A_O=\frac{1}{18}\left(bc\sin(A)+ac\sin(B)+ab\sin(C  ) \right)=\frac{1}{3}A
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  3. #3
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    Re: Area of triangle within a triangle

    Where did the 1/6 and 1/9 come from?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Area of triangle within a triangle

    I used the formula for the area of a triangle:

    A=\frac{1}{2}ab\sin(C)

    Now, for triangle ABC, I included all three angles so that each term is 1/3 of the entire area, or equivalently the area of ABC can be found from:

    3A=\frac{1}{2}bc\sin(A)+\frac{1}{2}ac\sin(B)+\frac  {1}{2}ab\sin(C) and so:

    A=\frac{1}{6}\left(bc\sin(A)+ac\sin(B)+ab\sin(C) \right)

    For the 3 triangles outside of RST, I used the same formula:

    A_O=\frac{1}{2}\cdot\frac{2}{3}b\cdot\frac{1}{3}c \sin(A)+\frac{1}{2}\cdot\frac{1}{3}a\cdot\frac{2}{  3}c\sin(B)+\frac{1}{2}\cdot\frac{2}{3}a\cdot\frac{  1}{3}b\sin(C)

    A_O=\frac{1}{9}\left(bc\sin(A)+ac\sin(B)+ab\sin(C) \right)
    Last edited by MarkFL; January 9th 2013 at 03:27 PM.
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