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Thread: How should the radius MP be drawn?

  1. #1
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    How should the radius MP be drawn?

    I would really appreciate help with this one




    The coloured parts space is limited by three semicircle archs of which the smaller has the radius r.
    How should the radius MP be drawn, if you want it to divide the coloured space into two parts with
    equally big areas?
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  2. #2
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    Hello, Kevin!



    The colored space is limited by three semicircle arcs of which the smaller has radius $\displaystyle r.$
    How should the radius $\displaystyle MP$ be drawn, if you want it to divide the colored space
    into two parts with equal areas?

    Draw radius $\displaystyle MQ$ downward, perpendicular to $\displaystyle AB.$
    Let $\displaystyle \theta = \angle PMQ.$

    Ignore the two smaller semicircles for the moment.
    If the shaded region is the semicircle $\displaystyle APB$, then $\displaystyle MQ$ bisects the area.

    But a semicircle with area $\displaystyle \tfrac{1}{2}\pi r^2$ has been subtracted from quadrant $\displaystyle MQB$
    . . and added to quadrant $\displaystyle MQA.$

    For the new area is to be bisected by radius $\displaystyle MP$,
    . . the area of sector $\displaystyle PMQ$ must equal $\displaystyle \tfrac{1}{2}\pi r^2$

    The area of sector $\displaystyle PMQ \:=\:\tfrac{1}{2}(2r)^2\theta \:=\:2r^2\theta$

    So we have: .$\displaystyle 2r^2\theta \:=\:\tfrac{1}{2}\pi r^2 \quad\Rightarrow\quad \theta \:=\:\tfrac{\pi}{4}$

    Hence: .$\displaystyle \angle PMQ \:=\:\angle AMP \:=\:\tfrac{\pi}{4}$

    Therefore, we must draw radius $\displaystyle MP$ so that angle $\displaystyle AMP = 45^o.$

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  3. #3
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    Thank U so much!
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