Answer of the question is 5 cm
Answers
Given AB, BC and AC are tangents to the circle at E, D and F. BD = 30 cm and DC = 7 cm and ∠BAC = 90° Recall that tangents drawn from an exterior point to a circle are equal in length Hence BE = BD = 30 cm Also FC = DC = 7 cm Let AE = AF =�x� → (1) Then AB = BE + AE = (30 +�x) AC = AF + FC = (7 +�x) BC = BD + DC = 30 + 7 = 37 cm Consider right Δ ABC, by Pythagoras theorem we have BC2�= AB2�+ AC2� ⇒ (37)2�= (30 +�x)2�+ (7 +�x)2� ⇒ 1369 = 900 + 60x�+ x2�+ 49 + 14x�+�x2� ⇒ 2x2�+ 74x�+ 949 – 1369 = 0 ⇒ 2x2+ 74x�– 420 = 0 ⇒�x2�+ 37x�– 210 = 0 ⇒�x2�+ 42x�– 5x�– 210 = 0 ⇒�x�(x�+ 42) – 5 (x�+ 42) = 0 ⇒ (x�– 5) (x�+ 42) = 0 ⇒ (x�– 5) = 0 or (x�+ 42) = 0 ⇒�x�= 5 or�x�= – 42 ⇒�x�= 5 [Since�x�cannot be negative] ∴ AF = 5 cm [From (1)] Therefore AB =30 +x = 30 + 5 = 35 cm AC = 7 + x = 7 + 5 = 12 cm Let ‘O’ be the centre of the circle and ‘r’ the radius of the circle. Join point O, F; points O, D and points O, E. From the figure, Area of (ΔABC) = Area (ΔAOB) + Area (ΔBOC) + Area (ΔAOC) ∴ r = 5 Thus the radius of the circle is 5 cm
