Smile 發問於 科學及數學數學 · 1 月前

Maths problem: how to do, thanks?

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  • 匿名
    1 月前
    最愛解答

    Question:

    In the figure, DE = DB and ∠CED = ∠CBD = 90°. If AC:BC = 13:5, find the ratio of the area of ΔACD to that ΔBCD.

    https://s.yimg.com/tr/i/0f4a8672f5e141a391ca57846d...

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    Solution:

    ∵ ΔCDE ≅ ΔCDB ...... ( RHS )

    ∴ ∠ACD = ∠BCD ...... ( corr. ∠s, ≅ Δs )

    The area of ΔACD = (1/2)(AC)(CD) sin∠ACD

    The area of ΔBCD = (1/2)(BC)(CD) sin∠BCD

    Consider the result of the following question,

    https://hk.answers.yahoo.com/question/index?qid=20...

    The ratio of the area of ΔACD to that ΔBCD = AD:BD

    Therefore,

    AD:BD = (1/2)(AC)(CD) sin∠ACD:(1/2)(BC)(CD) sin∠BCD

    AD:BD = AC:BC

    AD:BD = 13:5

    ∴ The ratio of the area of ΔACD to that ΔBCD

    = (AD)(BC)/2:(BD)(BC)/2

    = AD:BD

    = 13:5

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    Remark:

    Using the above method, we can derive a theorem.

    If ∠BAD = ∠CAD, then BD/CD = AB/AC.

    https://s.yimg.com/tr/i/4fafe5e6f9994cff8105068667...

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