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:
Answer & explanation
Correct answer: option 4
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