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,

OA^{2} = QA^{2} + OQ^{2}

⇒ QA^{2} = OA^{2} – OQ^{2}

⇒ QA^{2} = 5^{2} – 3^{2}

⇒ QA^{2} = 25^{2} – 9^{2}

⇒ QA^{2} = 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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