Question

A is a point at a distance 13 cm from the centre of a circle O of a radius 5cm.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 triangle.

Answer 1

BP = BR and CR = CQ

Perimeter of ΔABC = AB + BR + RC + CA

= AB + BP + QC + CA

= AP + QA (AP = QA)

= 2AP

ΔAPO

AO²=AP²+PO²

13² = AP² + 5²

AP² = 144

AP = 12

The perimeter of the ΔABC

2 x 12 = 24 cm is the answer

Answer 2

Here we have,

OA = 13cm

Radius = OP = 5cm

Since AP is a tangent to the circle with center O and OP is its radius, OP ⊥ AP

Now, In ΔOPA

∠OPA = 90°

$AP^{2} = OA^{2} - OP^{2}$  {using pythagoras theorem}

                  = $13^{2} - 5^{2}$

                  = 169 - 25

$AP^{2} = 144$

$AP = \sqrt{144}$

⇒ $AP = 12$

Now,

$AP = \frac{1}{2} \times Perimeter of ΔABC$

⇒ $12 = \frac{1}{2} \times Perimeter of ΔABC$

⇒ $Perimeter of ΔABC = 24cm$