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.

**Explanation:**

∠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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