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.
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