Q. 445.0( 3 Votes )

Two concentric ci

Answer :

Given:


OA = 5 cm


OQ = 3 cm


Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.


Property 2: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.


By property 1, ∆OAQ is right-angled at OQA (i.e., OQA = 90°).


By Pythagoras theorem in ∆OAQ,


OA2 = QA2 + OQ2


QA2 = OA2 – OQ2


QA2 = 52 – 32


QA2 = 252 – 92


QA2 = 16


QA = 16


QA = 4 cm


By property 2,


BQ = BP (tangent from B)


And,


AQ = BQ = 4 cm [ Q is midpoint of AB]


PB = PC = 4 cm [ P is midpoint of BC]


Now,


BC = BP + PC


BC = 4 cm + 4 cm


BC = 8 cm


Hence, BC = 8 cm

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