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). 
If AC=CB, AB=18, then how do we conclude that the diameter AB is 18 inches?
Attachments: 
circle.gif [2.64 KiB]
Not downloaded yet

