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Thread: Radian Measure

  1. #1
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    Radian Measure

    Hi

    Two circles, each of radius length 10 cm, have their centres 16 cm apart. Calculate the area common to each circle.

    I've drawn a diagram:



    I keep on getting 126.45 cm2 (2 d.p.) for my answer but the answers says 32.69 cm2. I simply divided the shaded area into two (they become congruent minor segments and then I simply found the area of one and multiplied it by two).

    Could someone please show me how to do this question?
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  2. #2
    Moo
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    Hello,

    Let H be the intersection point of AB and O1O2.


    - First compute the measure (in radians) of $\displaystyle \alpha=\widehat{AO_1H}$

    Consider the triangle $\displaystyle AO_1H$, which is a right angle triangle.
    We know that $\displaystyle O_1H=8 \text{ cm}$
    Hence $\displaystyle \cos \alpha=\frac{8}{10}=\frac 45$

    So $\displaystyle \alpha=\arccos \left(\frac 45 \right) \approx 0.6435$ (radian)

    - Now compute the area $\displaystyle \mathcal{A}_1$ of the arc of circle $\displaystyle AO_1B$
    Note that $\displaystyle \widehat{AO_1B}=2 \alpha$

    I'm too rusty to remember if there is a formula for it, but here is the reasoning.
    The whole circle has a center angle of $\displaystyle 2 \pi$ radians.
    The arc of circle we consider has a center angle of $\displaystyle 2 \alpha$ radians.
    So basically, we're considering $\displaystyle \frac{2 \alpha}{2 \pi}=\frac{\alpha}{\pi}$ th of the circle.

    It can be used for the area : $\displaystyle \mathcal{A}_1=\frac \alpha \pi \cdot \underbrace{(\pi \cdot 10^2)}_{\text{total area of the circle}}=\alpha \cdot 10^2 \approx 64.35 \text{ cm}^2$


    - You now have to substract the area of the triangle $\displaystyle AO_1B$ to get half of the shaded area.

    Note that $\displaystyle AO_1B$ has twice the area of $\displaystyle AO_1H$

    So let's consider the triangle $\displaystyle AO_1H$.
    It's a right angle triangle.
    By the Pythagorean theorem, $\displaystyle AO_1^2=AH^2+O_1H^2$
    That is $\displaystyle 10^2=8^2+AH^2$
    Hence $\displaystyle AH=6 \text{ cm}$

    So the area of this triangle is $\displaystyle \frac 12 \cdot 6 \cdot 8=24 \text{ cm}^2$
    Twice this area = area of triangle $\displaystyle AO_1B$ = 48 cm².



    So the area of half the shaded area is $\displaystyle \approx 64.35-48=16.35 \text{ cm}^2$

    Hence the shaded area is $\displaystyle \approx 32.70 \text{ cm}^2$



    I hope it's clear enough
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