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Thread: Circle Geometry Questions

  1. #1
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    Circle Geometry Questions

    Most circle geometry questions I understand, but these are from my homework that I don't understand.
    Thanks for your help.








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  2. #2
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    Lexington, MA (USA)
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    Hello, skitty!


    Here's the first one . . .


    [quote]1. Prove that the triangles in the diagram are both right triangles.
    Code:
                  Q
                  o * *
              *     *     *
            *   *     *     *
           *            *    *
             *            *
          *                 * *
        P o * * * * * * * * * o R
          *                 * *
             *            *
           *            *    *
            *   *     *     *
              *     *     *
                  o * *
                  S

    We are given: .$\displaystyle PQ = PS,\;\;RQ = RS$

    Since $\displaystyle PR = PR,\;\Delta PQR \cong \Delta PSR\;\;\text{ (s.s.s.)}$

    Then: $\displaystyle \angle QPR = \angle SPR\quad\Rightarrow\quad \text{arc(QR)} = \text{arc(SR)}$ .[1]

    And: $\displaystyle \angle QRP = \angle SRP\quad\Rightarrow\quad \text{arc(PQ)} = \text{arc(PS)}$ .[2]

    Add [1] and [2]: .$\displaystyle \text{arc(PQ) + arc(QR)} \:=\:\text{arc(PS) + arc(SR)} $

    . . That is: .$\displaystyle \text{arc(PQR)} \:=\:\text{arc(PSR)}$

    Since $\displaystyle \text{arc(PQR)} + \text{arc(PSR)} \:=\:360^o $, then: .$\displaystyle \text{arc(PQR)} \:=\:\text{arc(PSR)} \:=\:180^o$


    Since $\displaystyle \angle Q\text{ and }\angle S$ are inscribed in semicircles, $\displaystyle \angle Q \:=\:\angle S \:=\:90^o$

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  3. #3
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    dear skitty

    HELLO SKITTY

    ANSWER TO SECOND QUESTION

    TO PROVE-
    PERIMETER (PQR)=2PA
    PROOF-
    PA=PB (Tangents from external point)
    Also;
    QX=BQ and AR=RX (Tangents from external point)
    now;
    PQ+QX=PR+XR
    PQ+QX+PR+XR=PERIMETER(PQR)
    2(PR+RX)=Perimeter(PQR)
    2PA=Perimeter (PQR)
    hence proved.
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  4. #4
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    dear skitty

    hello skitty;

    answer to fourth question-

    TO PROVE-
    angle QOS=90

    PROOF-
    anglePQS+angleRSQ=180 (CO-INTERIOR ANGLES)

    anglePQO=angleSQO and angleRSO=angleQSO

    NOW;
    1/2 (anglePQS+angleRSQ)=90
    angleSQO+angleQSO=90

    BY ANGLE SUM PROPERTY OF TRIANGLE-
    angleSQO+angleQSO+ angleQOS=180
    angleQOS=180-90=90
    So angleQOS=90

    HENCE PROVED
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