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Thread: I'm stuck with this question about circle

  1. #1
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    I'm stuck with this question about circle

    Only (a) I need help with.

    Thank you very much .
    Attached Thumbnails Attached Thumbnails I'm stuck with this question about circle-img_0124.jpg  
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  2. #2
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    Re: I'm stuck with this question about circle

    Hint: draw 2 radius lines from center of circle to ends of water line;
    then you'll get a 3-7-7 triangle...
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  3. #3
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    Re: I'm stuck with this question about circle

    You can represent the pipe as $\displaystyle x^2+ y^2= 49$ by setting up a coordinate system with origin at the center of the pipe. Since the maximum depth of the water is 2 cm, its surface is given by y= -7+ 2= -5. The lower half of the pipe is given by $\displaystyle y= -\sqrt{49- x^2}$. The surface of the water crosses the circle when $\displaystyle y= -5= -\sqrt{49- x^2}$ so $\displaystyle 25= 49- x^2$, $\displaystyle x^2= 49- 25= 24$. $\displaystyle x= -2\sqrt{6}$ and $\displaystyle x= 2\sqrt{6}$. Integrate $\displaystyle -5- (-\sqrt{49-x^2})= \sqrt{49- x^2}- 5$ from $\displaystyle x= -2\sqrt{6}$ to $\displaystyle x= 2\sqrt{6}$ to find the area.
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  4. #4
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    Re: I'm stuck with this question about circle

    http://mathworld.wolfram.com/CircularSegment.html

    Area = $7^2\arccos\left(\dfrac{5}{7}\right)-5\sqrt{24}$
    Last edited by SlipEternal; May 10th 2018 at 05:31 AM.
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  5. #5
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    Re: I'm stuck with this question about circle

    Thanks from Syndy
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