A pair of tangents AB and AC are drawn from a point which is at a distance of 10 cm from the centre O of a circle of radius 6 cm, then the area in cm² of quadrilateral ABOC, is:

48

We know that,

Area of right angled triangle = \(\frac{1}{2}\) × Base × Height

Area of quadrilateral ABOC is double of the Area of Δ ABO.

In Δ ABO, OA = 10 cm

OB = 6 cm

We know that,

=  AB² = AO² – OB²

= AB = \(\sqrt {AO^2 – OB^2}\)

AB = \(\sqrt {100 - 36}\) = \(\sqrt {64}\) = 8 cm

Area of Δ ABO = \(\frac{1}{2}\) × 8 × 6 = 24 cm²

Area of quadrilateral ABOC = 2 × Area of Δ ABO = 2 × 24 = 48 cm²