Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 825.
Tuesday, November 27, 2012
Problem 825: Circle, Semicircle, Arc, Chord, Midpoint, Sector, Triangle, Area
Labels:
arc,
area,
chord,
circle,
circular sector,
midpoint,
semicircle,
triangle
Subscribe to:
Post Comments (Atom)
∠DOA = 135/2 = 67.5 = ∠CBO
ReplyDeleteSo 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
See attached sketch
ReplyDeletehttp://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
Since ∠DOC=∠OCB=67.5°, so OD//BC.
ReplyDeleteHence,
S = area of sector BOC (with angle 45°)
= 1/2×6^2×π/4
= 9π/2