Area of AOC + Area of ABO
= √(10-1)/2 + √(10-1)/2
= √9
= 3
Answer [spoiler]{D}[/spoiler]
Geometry
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Tangents to a circle from an external point meet at right angles, ABO and ACO are both 90 degrees.
In the figure above, line segments AB and AC are tangent to circle O. If the length of OB = 1 and the length of OA = √10, what is the area of quadrilateral ABOC? (Figure not drawn to scale.)
A. 3/2
B. √3
C. 2√2
D. 3
E. 2√3
In the right angled triangle OBA,


BA = 3
Thus area of the quadrilateral = Sum of the areas of the triangles OBA and OCA = 2*(1/2)*3*1 = 3
[spoiler]Answer : D[/spoiler]
Regards,
Vivek














