# In Fig. 10.100, a

Given:

AB = 29 cm

B = 90°

DS = 5 cm

Property 1: 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.

Property 2: 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 3: Sum of all angles of a quadrilateral = 360°.

By property 1,

BP = BQ (tangent from B)

DS = DR = 5 cm (tangent from D)

AR = AQ (tangent from A)

Also,

By property 2, ∆OQB is right-angled at OQB (i.e., OQB = 90°) and OPB is right-angled at OPB (i.e., OPB = 90°).

Now by property 3,

PBC + BQO + QOP + OPB = 360°

90° + 90° + QOP + 90°= 360°

270° + QOP = 360°

QOP = 360° - 270°

QOP = 90°

adjacent sides (i.e., BP = BQ and OQ = OP) are equal and all angles are 90°

Now,

AR + RD = 23 cm [ AD = AR + RD]

AR + 5 cm = 23 cm

AR = 23 cm 5 cm

AR = 18 cm

AQ = AR = 18 cm

Now,

AB = 29 cm

AQ + QB = 29 cm [ AD = AR + RD]

18 cm + QB = 29 cm

QB = 29 cm 18 cm

QB = 11 cm

OPBQ is a square

OP = BQ = 11 cm