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    #1
    I have trouble in determining the ratio of the area of $\triangle PST$ in terms of $\triangle PQR$

    In the triangle PQR $QT=TR$, $PS=1 cm$ , $SQ=2 cm$ , How should I be writing the area of $\triangle PST$ in terms of $\triangle PQR $



    What is known by me :

    Since $|\overline{QT}|=|\overline{TR}|$ it follows that $\triangle PQT$ and $\triangle PTR$ have equal areas, so $$[PQT]=\frac12[PQR]$$

    Has It got to do something with,

    Quote Quote:
    If two polygons are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
    So what must be done from here?
    Last edited by mathlearn; September 28th, 2016 at 21:46.

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    #2
    Hi mathlearn,

    A key idea is that $\triangle PST$ and $\triangle PQT$ share the same altitude from vertex $T$. Using this fact, deduce that $\operatorname{Area}(\triangle PST) = \frac{1}{3}\operatorname{Area}(\triangle PQT)$.

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    #3 Thread Author
    Quote Originally Posted by Euge View Post
    Hi mathlearn,

    A key idea is that $\triangle PST$ and $\triangle PQT$ share the same altitude from vertex $T$. Using this fact, deduce that $\operatorname{Area}(\triangle PST) = \frac{1}{3}\operatorname{Area}(\triangle PQT)$.
    Hi Euge ,

    What I still don't understand is that ,

    How did $\operatorname{Area}(\triangle PST) = \frac{1}{3}\operatorname{Area}(\triangle PQT)$ get derived?

    From $|\overline{PS}|=\frac12|\overline{SQ}|$ it follows that the area of $\triangle PST$ is a half of the area of $\triangle PQT$ , From there I don't get it that much

  5. Pessimist Singularitarian
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    #4
    I would use this formula for the area $A$ of a triangle:

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

    To derive the relationship Euge mentioned.

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    #5 Thread Author
    Quote Originally Posted by MarkFL View Post
    I would use this formula for the area $A$ of a triangle:

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

    To derive the relationship Euge mentioned.
    Thanks but the issue is I don't know the use of this formula

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    #6
    Draw a line from $S$ to $\overline{PT}$ parallel to $\overline{QT}$ and use similarity...

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    #7
    Quote Originally Posted by mathlearn View Post
    Hi Euge ,

    What I still don't understand is that ,

    How did $\operatorname{Area}(\triangle PST) = \frac{1}{3}\operatorname{Area}(\triangle PQT)$ get derived?

    From $|\overline{PS}|=\frac12|\overline{SQ}|$ it follows that the area of $\triangle PST$ is a half of the area of $\triangle PQT$ , From there I don't get it that much
    Hey mathlearn,

    I suggested that you deduce this from the fact that $\triangle PST$ and $\triangle PQT$ share the same altitude from vertex $T$. Suppose their common altitude is $h$. Since the area of a triangle is $\dfrac{1}{2}\times (\text{base})\times (\text{height})$, then $\operatorname{Area}(\triangle PQT) = \dfrac{1}{2}(3)(h)$ and $\operatorname{Area}(\triangle PST) = \dfrac{1}{2}(1)(h)$. From here you can see how the result is derived.

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