Q. 383.7( 3 Votes )

In Fig. 10.100, a

Answer :

Given:


AB = 29 cm


AD = 23 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,


OQ = OP (radius)


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°


quadrilateral OPBQ is a square


Now,


AD = 23 cm


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


Hence, radius = 11 cm

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Area Related to Circles- Important Formula and ConceptsArea Related to Circles- Important Formula and ConceptsArea Related to Circles- Important Formula and Concepts59 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses