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n the figure, BDC is a tangent to the given circle at point D such that BD = 30 to the circle and meet when produced at A making BAC a right angle triangle. Calculate (i) AF (ii) radius of the circle.

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a bit typical question, provide an easy solution to it, how should i approach???


1 Answer

  1. In the figure, BDC is a tangent to the given circle with centre O and D is a point such that
    BD = 30 cm and CD = 7 cm
    RD Sharma Class 10 Book Pdf Free Download Chapter 10 Circles
    BE and CF are other two tangents drawn from B and C respectively which meet at A on producing this and ∆BAC is a right angle so formed
    To find : (i) AF and (ii) radius of the circle
    Join OE and OF
    OE = OF radii of the circle
    OE ⊥ AB and OF ⊥ AC
    OEAF is a square
    BD and BE are the tangents from B
    BE = BD = 30 cm and similarly
    CF = CD = 7 cm
    Let r be the radius of the circle
    OF = AF = AE = r
    AB = 30 + r and AC = 7 + r and BC = 30 + 7 = 37 cm
    Now in right ∆ABC
    RD Sharma Class 10 Pdf Ebook Chapter 10 Circles

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