Geometry Problem

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

Click the figure below to see the complete problem 808.

## Saturday, September 29, 2012

### Problem 808: Parallelogram, Perpendicular, Right Triangle, Semicircle Area, Diameter

Labels:
area,
diameter,
parallelogram,
perpendicular,
right triangle,
semicircle

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Let the area of segment FG be k.

ReplyDeleteFrom Pythagoras theorem,

AE^2 + DE^2 = AD^2 = BC^2

Multiply both side by π/4,

(π/4)AE^2 + (π/4)DE^2 = (π/4)BC^2

S1 + (S2 + k) = S + k

S1 + S2 = S

http://imageshack.us/a/img141/4056/problem808.png

ReplyDeleteLet S3 is the red area ( see sketch)

In right triangle AED we have pi/4(AD^2)=pi/4(AE^2+ED^2)

So S+S3=S1+S2+S3 => S=S1+S2