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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