## Thursday, November 15, 2012

### Problem 822: 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 822.

1. ∵ D, O are mid-points of AC, AB resp.
∴ OD//BC

S = area of sector OBC (with angle 45°)
= π×6^2×1/8
= 9π/2

2. http://img812.imageshack.us/img812/2623/problem822.png

Draw 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

3. Join BC, OC
(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

4. Jacob Ha's proof is nice.
I 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

5. http://www.mathematica.gr/forum/viewtopic.php?f=22&t=32946&p=152531