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

(a) 15 cm
(b) 10 cm
(c) 22.5 cm
(d) 20 cm

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Solution

(b) 10 cm
Since the tangents from an external point are equal, we have:
$A D = A E, C D = C F, B E = B F Perimeter of ∆ A B C = A C + A B + C B = (A D - C D) + (C F + B F) + (A E - B E) = (A D - C F) + (C F + B F) + (A E - B F) = A D + A E = 2 A E = 2 \times 5 = 10 cm$

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