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In the given figure, AD and AE are the tangents to a circle with center O and BC touches the circle at F. If AE = 5 cm then perimeter of ΔABC is

Given : From an external point A, two tangents, AD and AE are drawn to a circle with center O. At a point F on the circle tangent is drawn which intersects AE and AD at B and C, respectively. And AE = 5 cm

To Find : Perimeter of △ABC

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

So we have

BE = BF …[1]

[Tangents from point B]

CF = CD …[2]

[Tangents from point C]

Now Perimeter of Triangle abc

= AB + BC + AC

= AB + BF + CF + AC

= AB + BE + CD + AC …[From 1 and 2]

= AE + AD

Now,

AE = AD = 5 cm as tangents drawn from an external point to a circle are equal

So we have

Perimeter = AE + AD = 5 + 5 = 10 cm

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