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Question 130
C and C are two concentric circles with centres at 0. Their radii are 12 cm. and 3 cm. respectively. B and C are the points of contact of two tangents drawn to C2 from a point A lying on the circle C1. Then the area of the quadrilateral ABOC is
Solution
AB = AC = tangents from the same point
OB = OC = 3 and OA = 12
$$\angle$$ABO = 90
=> AB = $$\sqrt{12^2 - 3^2} = 3\sqrt{15}$$
Now, area of $$\triangle$$OAB = $$\frac{1}{2}$$ OB * AB
= $$\frac{1}{2} * 3 * 3\sqrt{15} = \frac{9\sqrt{15}}{2}$$
$$\therefore$$ area of OABC = $$9\sqrt{15}$$ sq. cm
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