Q. 144.2( 15 Votes )

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

Answer :

Given : Two tangents are drawn from an external point A to the circle with center O, Tangent BC is drawn at a point R.


Radius of circle equals 5cm and OA = 13 cm



OA = 13 cm


To Find : Perimeter of ΔABC.


OPA = 90°


[tangent at any point of a circle is perpendicular to the radius through the point of contact]


(OA)2 = (OP)2 + (PA)2


(13)2 = (5)2 + (PA)2


169 - 25 = (PA)2


(PA)2= 144


PA = 12 cm


As we know that, Tangents drawn from an external point to a circle are equal.


So we have


PB = BR [1] [Tangents from point B]


CR = QC [2] [Tangents from point C]


Now Perimeter of Triangle PCD


= AB + BC + CA


= AB + BR + CR + CA


= AB + PB + QC + CA [From 1 and 2]


= PA + QA


Now,


PA = QA = 12 cm as tangents drawn from an external point to a circle are equal


So we have


Perimeter = PA + QA = 12 + 12 = 24 cm


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