Tuesday, November 27, 2012

Problem 825: Circle, Semicircle, Arc, Chord, Midpoint, Sector, Triangle, Area

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

Click the figure below to see the complete problem 825.

Online Geometry Problem 825: Circle, Semicircle, Arc, Chord, Midpoint, Sector, Triangle, Area.

3 comments:

  1. ∠DOA = 135/2 = 67.5 = ∠CBO
    So OD // BC.
    By sharing the same base BC and same height, ΔBCD has the same area as ΔOCB.
    S = sector(OBC) = π *(36)*(1/8) = (9/2) π

    ReplyDelete
  2. See attached sketch
    http://img706.imageshack.us/img706/5949/problem825.png
    Let Area(XYZ)= area of triangle XYZ
    Let Sector(XYZ)= area of sector XYZ
    Let R=OA=6
    We have Area of white area= Area(ADB)+2.S1
    But S1= Sector(AOD)-Area(AOD)
    And Area(ADB)=2.Area(AOD)
    So white area=2.Sector(AOD) => S= sector(AOB)-white area=(pi/2-3/8.pi)*R^2= 9/2.pi

    ReplyDelete
  3. Since ∠DOC=∠OCB=67.5°, so OD//BC.

    Hence,
    S = area of sector BOC (with angle 45°)
    = 1/2×6^2×π/4
    = 9π/2

    ReplyDelete