Points A, B and C lie on a circle. Point O is the center of the circle and lies on the straight line AB. AC = CB. AB = 18 inches. What is the area of triangle ABC? (Figure not drawn to scale.)

A. 9 square inches

B. 18 square inches

C. 49 square inches

D. 72 square inches

E. 81 square inches

Point O is the center of the circle, so OA, OB, and OC are radii. AC = CB is given. Since their sides are equal, triangle AOC is congruent to triangle BOC. (AC = CB, OA = OB, OC = OC) Triangle AOC is congruent to triangle BOC, so angle AOC = angle BOC. Since

point O is on segment AB, angle AOC + angle BOC = 180 degrees. So each angle is equal to 90 degrees, and triangles AOC and BOC are right triangles.

Diameter AB = 18 inches, so a radius is 9 inches.

The area of triangle AOC is (1/2) × AO × OC = (1/2) × 9 × 9 = 81/2 square inches. The area of triangle BOC is also 81/2 square inches. The area of triangle ABC is the sum othe areas of triangles AOC and BOC, so it equals 81/2 + 81/2 = 81 square inches.

The correct answer is choice (E).

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**If AC=CB, AB=18, then how do we conclude that the diameter AB is 18 inches?**