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Exercise 6.3 question no. 4

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Exercise 6.3 question no. 4

    • Posted by Yuvraj Dangi 6 years, 10 months ago

    CBSE > Class 10 > Mathematics
    • 3 answers

    Paras ? ? Shah ? ? 6 years, 10 months ago

    QR / QS = QT / PR ( given) Qt / Qr = Pr / Ps (1) <1 = <2 ( given) Pr = Pq (2). ( side opp. To equal < are equal) From 1 and 2 Qt / Qr= Pq / Qs = Pq / Qt = Qs/ Qr (3) In ∆ PQS and TQR we get Pq /Qt = Qs / Qr < PQS = <TQR = <Q By Side angle side criterion of similarity ∆ PQS ~ ∆ TQR Hence proved ? ?

    1Thank You

    Paras ? ? Shah ? ? 6 years, 10 months ago

    In ∆ RPQ and RTS < RPQ = <RTS. (given ) <PRQ = < TRS ( common) Therefore By AA- criterion of similarity, ∆ RPQ ~ ∆ RTS

    4Thank You

    Ishita Jain 6 years, 10 months ago

    QR/QS=Qt/PR(angle1=angle2 so in triangle PQR PQ=PR becoz sides opposite to equal angles are equal ) So QR/QS=QT/QP So by the converse of bpt theoram PS is parallel to TR Then in triangle PQS and tria.TQR Angle 1 =angle 1( common) Angle QPS=angle QTR ( corresponding angle ) SO BY AA~ CRITERIA Triangle PQS~tri.TQR

    2Thank You

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