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). -------------

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?

The lengths of AB and AC are equal. It is true. (In paragraph 1 we've proven that triangles ABO and ACO are equal).

But the figure ABOC is NOT a rectangle. The sides that are equal are contiguous not the opposite ones. Beside that it has only two angles that equal 90 degrees. The other two angles (BAC and BOC) are not.

Let us go over the logic of explanation in brief: 1. We've proven that triangles ABO and ACO are both right and equal. 2. Then we've found the length of AB. 3. Then we've calculated the area of triangle ABO. 4. At last the area of ABOC is twice the area triangle ABO.

If the Figure is not Drawn to Scale, how can we infer that point O is the center of the circle, and thus, that OB and OC are radii?

If point O is not the center of the circle, OB and OC can still have lengths of 1. The difference is that they won't be perpendicular to the tangent, so they can't used as base or height of the triangle.

Let me ask you. If you put the 2 right triangles together, back to back (they share AB and AC), then it becomes an equilateral triangle with base 2 times OB(1) and height OA (1/2) × 2 × √10 and the answer is √10, not 3.

Let me ask you. If you put the 2 right triangles together, back to back (they share AB and AC), then it becomes an equilateral triangle with base 2 times OB(1) and height OA (1/2) × 2 × √10 and the answer is √10, not 3.

It will be the isosceles triangle with the base twice OB indeed, but the height will be AB, not OA. Take a look at where the right angles are.

AB equals 3, so we get the same answer: (1/2) × 2 × 3 = 3

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