Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 822.
Thursday, November 15, 2012
Problem 822: Circle, Semicircle, Arc, Chord, Midpoint, Sector, Triangle, Area
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∵ D, O are mid-points of AC, AB resp.
ReplyDelete∴ OD//BC
S = area of sector OBC (with angle 45°)
= π×6^2×1/8
= 9π/2
http://img812.imageshack.us/img812/2623/problem822.png
ReplyDeleteDraw lines per attached sketch
We have Area S=Area(DCB)+ Area S1
But Area(DCB)= Area ( OBC) …… ( Same base BC and altitude)
So S= Area of sector COB = pi x OB^2/8 = 4.5 pi
Join BC, OC
ReplyDelete(i) Area bounded by chord BC and arc BC = Area of sector OBC - Area of ΔOBC
= (1/2).6.6. (π /4)- (1/2).6.6.sin (π/4) = 9 π/2 - 9√2
(ii) Area of Δ BDC = Area of Δ ADB = (1/2) Area of Δ ABC
= (1/2)(twice the area of Δ AOC)= area of Δ AOC
= (1/2) OA.OC.sin (3 π /4)= (1/2).6.6.(1/ √ 2)= 9 √ 2
So S = Area (i) + area (ii) = 9 π /2
Jacob Ha's proof is nice.
ReplyDeleteI realized that if T denotes the area cut off by arc BC and chord BC,
S = T + area of ΔBDC = T + area of ΔBOC = area of sector BOC
http://www.mathematica.gr/forum/viewtopic.php?f=22&t=32946&p=152531
ReplyDelete