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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