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A is a point at a distance 13 cm from the centre ‘O’ of a circle of radius 5 cm. AP and AQ are the tangents to circle at P and Q. If a tangent BC is drawn at point R lying on minor arc PQ to intersect AP at B and AQ at C. Find the perimeter of ΔABC.

Answer:

OA = 13 cm

OP = OQ = 5 cm

OP and PA are radius and tangent respectively at contact point P.

∴ ∠OPA = 90°

In right angled ΔOPA by Pythagoras theorem

PA^{2}=OA^{2}–OP^{2}=13^{2}–5^{2}=169–25=144

⇒ PA = 12 cm

Points A, B and C are

exterior to the circle and tangents drawn from an external point to a circle

are equal so

PA=QA

BP=BR

CR=CQ

Perimeter of ΔABC

= AB + BC + AC

= AB + BR + RC + AC ......... [From figure]

= AB+BP+CQ+AC=AP+AQ

= AP + AP

= 2AP

= 2 × 12

= 24 cm

So, the perimeter of ΔABC = 24 cm.

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