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Thread: Inner triangle problm

  1. #1
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    Inner triangle problm

    Inner triangle problm-tri.pngIf L is length and W is width, I got LW-((1/12)LW)-((1/3)LW)-((1/4)LW)=10, which means that (1/3)LW=10, and the area of rectangle PQRS is 30cm^2. Is that right?
    Last edited by Ilikebugs; Jan 28th 2017 at 06:58 PM.
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  2. #2
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    Re: Inner triangle problm

    Call $\displaystyle \begin{align*} l \end{align*}$ the length and $\displaystyle \begin{align*} w \end{align*}$ the width, then the area of the rectangle is $\displaystyle \begin{align*} l\,w\end{align*}$

    The area of triangle CPS is $\displaystyle \begin{align*} \frac{1}{2}\,\frac{w}{2}\,l = \frac{l\,w}{4} \end{align*}$

    The area of triangle CQA is $\displaystyle \begin{align*} \frac{1}{2}\,\frac{w}{2}\,\frac{l}{3} = \frac{l\,w}{12} \end{align*}$

    The area of triangle ARS is $\displaystyle \begin{align*} \frac{1}{2}\,\frac{2\,l}{3}\,w = \frac{l\,w}{3} \end{align*}$

    The area of triangle ACS is $\displaystyle \begin{align*} 10\,\textrm{cm}^2 \end{align*}$.

    All these triangles together make the full rectangle, so

    $\displaystyle \begin{align*} \frac{l\,w}{4} + \frac{l\,w}{12} + \frac{l\,w}{3} + 10 &= l\,w \\ \frac{3\,l\,w}{12} + \frac{l\,w}{12} + \frac{4\,l\,w}{12} + 10 &= l\,w \\ \frac{8\,l\,w}{12} + 10 &= l\,w \\ \frac{2\,l\,w}{3} + 10 &= l\,w \\ 10 &= \frac{l\,w}{3} \\ 30 &= l\,w \end{align*}$

    So yes I agree that the area of the rectangle is $\displaystyle \begin{align*} 30\,\textrm{cm}^2 \end{align*}$.
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  3. #3
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    Re: Inner triangle problm

    My method ignores the areas of the three surrounding triangles.

    Let m(QA) = m(AB) = m(BR) = x

    Let m(QC) = m(CP) = y

    Locate point D, the midpoint of line segment AS, which is the diagonal of a rectangle having points
    A, R, S, and some fourth unnamed point as vertices. Point D is the center of that rectangle.
    CD is parallel to both QR and PS. Then CD is perpendicular to RS. So, line segment CD is a base
    of triangles ACD and CDS.

    Area of triangle ACS = area of triangle ACD + area of triangle CDS

    = $\tfrac{1}{2}(2x)y + \tfrac{1}{2}(2x)y = $

    $xy + xy = 2xy = 10$

    Area of rectangle PQRS = (3x)(2y) = 6xy = 3(2xy) = 3(10) = 30

    Area of rectangle PQRS = $ \ 30\ cm^2$
    Last edited by greg1313; Jan 28th 2017 at 09:50 PM.
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  4. #4
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    Re: Inner triangle problm

    Well, also can be solved this way:

    ARS: 1/2 of 1/3
    CPS: 1/2 of 1/2
    ACQ: 1/2 of 1/6

    Add 'em up to get 2/3, so ACS = 10 means 1/3 of total area.
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