smc

Question

A quadrilateral is inscribed in a circle of radius 2002. Three of the sides of this quadrilateral have length 200. What is the length of the fourth side?

Accepted Answer

FIGURE Let AD intersect OB at E and OC at F. AB= BC= CD= BAD=12 BCD==AOB From there, OAB ABE, thus: OAAB = ABBE = OBAE OA = OB because they are both radii of O. Since OAAB = OBAE, we have that AB = AE. Similarly, CD = DF. OE = 1002 = OB2 and EF=BC2=100 , so AD=AE + EF + FD = 200 + 100 + 200 = 500