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Math Help - Equation help

  1. #1
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    Equation help

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  2. #2
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    Re: Equation help

    Dear all, this question has stumped me. If anyone has the formula, I would be very grateful.
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  3. #3
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    Re: Equation help

    Hello, fleurdel!

    The figure below is made up of a square of side 28 cm.
    One semicircle and one quadrant are drawn inside the square.
    Given that the total area of portions A and C is two times the area of B,
    find the area of B. . \left(\text{Take }\pi = \tfrac{22}{7}\right)

    Code:
            *-*-*-------------*-*
            |        *  A *     |
            |           *    B  |
            |          *   *    | 14
            |                   |
            |         *      *  |
            |         *       * +
            |         *   C     |
            |                  *|
            |     D    *        | 14
            |           *       *
            |             *     *
            *-----------------*-*
                     28

    \text{(Area of }A+B) \;=\;\text{(Area of the square)} - \text{(Area of the quadrant)}

    . . . . . . . A + B \;=\;28^2 - \tfrac{1}{4}\pi(28^2) \;=\;784 - 196\pi .[1]


    \text{(Area of }B+C) \;=\;\text{(Area of semicircle)}

    . . . . . . . B + C \;=\;\tfrac{1}{2}\pi(14^2) \;=\;98\pi .[2]


    Add [1] and [2]:

    . .  \underbrace{A + C}_{\text{This is }2B} + 2B \;=\;(784-196\pi) + 98\pi

    . . . . . . 4B \;=\;784 - 98\pi


    \text{Therefore: }\:B \:=\:196 - \tfrac{49}{2}\pi \:=\:196 - \tfrac{49}{2}\left(\tfrac{22}{7}\right)
    . . . . . . . . B \;=\;119\text{ cm}^2

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  4. #4
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    Re: Equation help

    Thank You, Soroban, that was awesome!!!
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