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

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∠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