Answer :

Step1: Draw a line PQ = 9 cm. taking P and Q as centres draw

circles of radii 5 cm and 3 cm.

Step2: Now bisect PQ. We get midpoint to be T.

Now take T as a center , draw a circle of PT radius , this will intersect

the circle at points A, B, C, D. Join PB, PD, AQ, QC.

Justification:

It can be justified by proof that PB, PD are tangents of circle (whose centre is P and radius is 5 cm) and AQ, QC are tangents of circle (whose centre is Q and radius is 3 cm)

Join PA, PC, QB, QD

∠PBQ=90° (Angle is on semicircle)

BQ ⊥ PB

Since, BQ is radius of circle, PB has to be a tangent. Similarly, PD, QA, QC are tangents.

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