Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 957.

## Wednesday, January 15, 2014

### Geometry Problem 957: Equilateral Triangle, Inscribed Circle, Incircle, Circumscribed Circle, Circumcircle, Area, Circular Segment

Labels:
area,
circular segment,
circumcircle,
circumscribed,
equilateral,
incircle,
inscribed,
triangle

Subscribe to:
Post Comments (Atom)

(∏R2/3-√3 R2/4)+ (√3 R2/4- ∏R2/12)=∏R2/4

ReplyDeleteNote that 3*yellow area=OA^2*pi-OF^2*pi=pi(OA^2-(OA/2)^2)=3/4*OA^2*pi, so yellow area=OA^2*pi/4=(OA/2)^2*pi=area of circle with radius OF

ReplyDeleteThe circumradius is double of the inradius in an equilateral triangle. Thus the blue area is 1/4 of the circumcircle. Since by symmetry, there are 3 equal yellow area, each of them must also be 1/4 of the circumcirlce.

ReplyDelete