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

(D) Because line segments AB and AC are tangent to the circle (lines that are tangent to a circle form right angles with the radius at the point of tangency), we know that both OCA and OBA are right triangles. Furthermore, since OB = OC (since they are both radii and therefore the same length), we know that these triangles are congruent. Therefore, they have the same area.

From here, we can use the Pythagorean Theorem to solve for the length of BA. Since OB is 1 and OA is √10, the equation becomes:

1² + BA² = (√10)²

BA² = 10 – 1 = 9

BA = 3.

Remember that the base and the height of a triangle MUST be perpendicular. Therefore, we will assign OB and BA to be the base and the height of triangle OBA.

Now that we have the measurement of both the base and the height of OBA, we can find the area of the triangle:

Area = (1/2) × (base × height) =

= (1/2) × (1 × 3) = (1/2) × 3 = 3/2.

Since the two triangles are congruent, we can just double this number to get the area of quadrilateral ABOC:

3/2 × 2 = 3.

The correct answer is choice (D).

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**Since A is the common point for the two tangents AB and AC, shouldn't their lengths be equal?**

If so, we have a figure where the sides are OB = OC = 1 and AB = AC = 3. Therefore it is a Rectangle. Area of a rectangle = L × W = 3 × 1 = 3.

Would this be another approach, maybe a simpler one, to this problem?